Kirchhoff's Laws & Ohm's Law
In this video we're going to talk about how to use Kirchhoff's junction rule and loop rule to calculate the current in a complex circuit.
Two rules carry the entire deck:
- Junction rule — current in equals current out at every node.
- Loop rule — voltages around any closed loop sum to zero.
Tying both together is Ohm's law: every resistor's voltage drop is V = IR.
Gustav Kirchhoff published these two rules in 1845 when he was a 21-year-old student — and they turned out to be so powerful that every circuit in every device on Earth, from a TV remote to a satellite, is still analyzed with them today. They're not approximations or special cases. They work for any DC circuit with any number of components.
What makes them universal: both rules are just conservation laws in disguise. The junction rule is charge conservation at a point (charge can't be created or destroyed). The loop rule is energy conservation for a single charge making a round trip (you can't gain or lose energy by returning to where you started). Since these conservation laws are absolute, the rules apply everywhere.
The big picture of this deck:
- Slides 2–6 establish the tools: Ohm's Law and sign conventions.
- Slides 7–9 walk through a simple 2-loop example, showing how to set up and solve systems of equations.
- Slides 10–11 add a second battery to show the substitution trick.
- Slides 12–13 handle the case where you don't know current directions ahead of time.
- Slide 14 scales up to a 3-loop circuit — the exam-grade difficulty.
Why you learn this before Faraday: circuits are the language in which induced EMFs express themselves. Once Faraday's Law hands you a voltage source, you still need Kirchhoff to figure out the currents it drives through the rest of the circuit. Master this deck and everything in the Faraday deck becomes "just another battery problem."
Ohm's Law — V = IR
When you're moving across a resistor, the voltage drop will be the current times the resistor.
Current always flows from high potential to low potential. So whichever side of a resistor the current enters from is the + terminal, and the side it leaves from is the − terminal.
V = I · RThis single equation is the link between every resistor in the deck and the loop-rule equations on the next slides.
Diagram
V = IR looks like a fundamental law of nature, but it's actually an empirical relationship — an observation that many materials (called "ohmic" conductors) happen to follow. It's not a law of physics in the same sense as Newton's laws or Maxwell's equations.
What it says: for ohmic materials, the current that flows through a resistor is directly proportional to the voltage across it. Double the voltage → double the current. The constant of proportionality is 1/R, where R is the resistance in ohms (Ω). One ohm = one volt per amp.
What determines R? Four things: the material's resistivity ρ, the wire's length L, its cross-sectional area A, and temperature. The formula is R = ρL/A. Longer wire = more resistance (longer obstacle course for electrons). Thicker wire = less resistance (more lanes for electrons). Hotter wire = more resistance (atoms vibrate more and scatter electrons).
Non-ohmic devices exist: diodes, transistors, light bulbs, and gas discharge tubes all have V-I curves that are not straight lines. For a diode, current is nearly zero until voltage exceeds a threshold, then it shoots up exponentially. Ohm's Law doesn't apply to those. But for simple metal resistors at normal temperatures, V = IR is extremely accurate and is the foundation of all circuit analysis.
The three derived forms:
V = IR— find voltage drop if you know I and RI = V/R— find current if you know V and RR = V/I— find resistance if you know V and I
Power dissipated: P = VI = I²R = V²/R. All three forms are equivalent by substituting V = IR. Memorize one, derive the others.
Junction Rule — ΣIin = ΣIout
The current that enters the junction is equal to the total current that leaves the junction. So the only current that's entering the junction is I₁, the other two currents are leaving the junction, so I₁ has to equal the sum of I₂ plus I₃.
This is just charge conservation at every wire intersection. No charge piles up at a node, so whatever flows in must flow back out.
I₁ = I₂ + I₃Diagram
Charge in = charge out. No accumulation at the node.
The junction rule — ΣI_in = ΣI_out — is just charge conservation applied to a point. Charge cannot be created, destroyed, or stored at a wire junction. Whatever flows in must flow out, instantaneously.
Why no accumulation? If charge did pile up at a node, the node's potential would keep rising until it pushed the incoming current back out. In steady-state DC circuits (which is all you'll see in exam 2), the system has long since reached equilibrium where inflow exactly equals outflow at every junction.
Water analogy: imagine a pipe T-junction where 6 gallons/sec flow in and split into two branches. If 2 gal/s goes down one branch, then exactly 4 gal/s must go down the other. Anything else would mean water is either disappearing (impossible) or piling up at the junction (which is what blocked pipes look like — not a steady state).
How to use it in problems: at every node, write one equation of the form (sum of currents entering) = (sum of currents leaving). This immediately gives you a relationship between the branch currents, letting you eliminate one unknown per node. For N nodes in a circuit, you get N−1 independent junction equations (the Nth one is redundant — it's implied by the others).
Sign convention — arbitrary but consistent: you can pick any direction for each current arrow before writing the equation. If your guess is wrong, the solution will have a negative value — no re-derivation needed. Just remember: a negative current means the real flow is opposite your arrow.
Loop Rule — ΣV around any closed loop = 0
The sum of all the voltages around the loop or around the closed circle must add to zero.
Pick a starting point, walk all the way around any closed loop in your circuit, and add up every voltage change you encounter. The total must be zero — energy is conserved.
Each step is either a lift (potential increases) or a drop (potential decreases). Slides 5 and 6 cover the sign rules.
ΣVloop = 0Diagram
Walk the loop, sum every ΔV, set total to zero.
The loop rule — ΣV around any closed loop = 0 — looks like a rule about electronics, but it's really just energy conservation applied to a single charge. If an electron starts at some point in a circuit and returns to the same point, it must have the same potential energy it started with. Anything else would violate the conservation of energy.
What the loop rule says in plain English: add up every voltage rise (from batteries or any EMF source) and subtract every voltage drop (across resistors). The sum is always exactly zero. The electron gets "lifted" at batteries and "falls" through resistors, but after a full trip, the net elevation change is 0.
Hydraulic analogy: imagine a water loop with a pump and some paddlewheels. The pump lifts water up (battery adds energy); the paddlewheels convert water's kinetic energy into something else — heat, light, sound (resistors dissipate energy). Water must return to the pump at the same height it left. If the pump lifts it by 10 m, the paddlewheels must collectively drop it by 10 m. That's KVL.
Why it works for any closed loop, even imaginary ones: the rule applies to any path that closes back on itself — you don't have to pick "real" physical loops. You can draw an imaginary loop diagonally through a circuit, and the sum of voltages along it still equals 0. That's because potential is path-independent — electric fields are conservative (no curl).
Practical takeaway: once you've assigned currents and directions, write one loop equation per independent loop, combine with junction equations, and solve. The method is almost mechanical; the hard part is just tracking the signs carefully.
Sign Conventions — Resistors
If you're going in a direction of the current, that's associated with a voltage drop. Now if you're going opposite to the direction of the current, then you're traveling from a low potential to a high potential.
Pin this to the wall:
- Walk with the current across a resistor → −IR (drop).
- Walk against the current across a resistor → +IR (lift).
Why? Current flows from high to low potential, so going downstream means going downhill in voltage.
Anytime you travel from a high potential to a low potential, there's going to be a voltage drop, and it makes sense, because you're going from high to low. Now if you're going from a low potential to, let's say, a high potential, if you're traveling in that direction, then the voltage is increasing, so there should be a voltage lift.
Now we know that current flows from high potential to low potential, so notice that we're going against the current and we're traveling towards the positive side. So this is going to be associated with a voltage lift.
Now let's say if we have a resistor and the current is flowing in this direction and using the loop rule we're traveling in this direction. Will this be associated with a voltage drop or voltage lift? Now we know that current flows from positive to negative, so we're traveling towards a low potential, so this will be associated with a voltage drop.
Diagram
Resistor + side = current entry · − side = current exit.
A resistor is the circuit equivalent of a slope you slide down. Current naturally flows from high potential to low potential through a resistor, "losing" voltage equal to IR as it goes. This lost electrical energy becomes heat — which is why resistors get warm.
The two voltages on a resistor:
- The side current enters = + side (higher potential)
- The side current exits = − side (lower potential)
Unlike a battery, a resistor doesn't have a fixed polarity. Its polarity depends entirely on which direction current happens to be flowing. Flip the current and the + and − flip with it.
Loop-walking sign rule:
- Walking in the direction of the assumed current through R: you move from high to low potential → voltage drops → write −IR in the loop sum.
- Walking against the assumed current through R: you move from low to high potential → voltage rises → write +IR.
Why IR and not something else? Ohm's Law: V = IR. For any resistor, the voltage difference between its two ends equals the current times the resistance. So the "potential change" as you cross a resistor is exactly IR, with a sign that depends on which way you're walking relative to I.
Physical intuition for dissipated power: P = VI = I²R = V²/R. The energy comes from the battery (where chemistry pushes charges uphill) and gets dumped as heat in the resistor. Every joule of electrical energy lost in a resistor ends up as thermal vibration in the atoms of the resistor material. A 60-watt incandescent bulb, for instance, is literally a high-resistance wire getting so hot it glows.
Sign Conventions — Batteries
We're going towards the positive sign, so this is going to be associated with a voltage lift. Now what if we're going in this direction using the loop rule? Notice that we're going towards the negative sign, and so that should be associated with a voltage drop.
- Entering the + terminal as you walk → +ε (lift).
- Entering the − terminal as you walk → −ε (drop).
Now let's focus on a battery. So let's say this is positive and this is negative, and I'm going to draw another battery in this direction. So let's say if we're using the loop rule and we're going in this direction, should we apply a positive voltage or negative voltage in the voltage law equation? Is this a voltage lift or a voltage drop? Now we're going towards the positive sign, so this is going to be associated with a voltage lift.
Get one sign wrong and the whole problem unravels. The transcript warns: "All it takes is one single mistake to make a long problem one big headache."
So hopefully that makes sense, and understanding this will be very useful when solving these types of problems, because if you get, let's say, a negative sign wrong, the whole question is gone — it's going to be very hard to find the right answer.
Diagram
Long line = + terminal · short line = − terminal.
A battery is a charge pump. Chemical reactions inside the cell take positive charges from the low-potential (−) terminal and physically carry them through the cell to the high-potential (+) terminal, against the electric field. Once at the + terminal, those charges are eager to flow back through the external circuit to the − terminal, dissipating their potential energy in resistors.
Why crossing − → + adds ε to the loop sum: as you walk the loop in a specific direction, crossing a battery from − to + means you're moving from low potential to high potential. Voltage increases; you add ε.
Why crossing + → − subtracts ε: the other direction, from + to −, is a potential drop; subtract ε. Batteries don't "know" which way you're walking; they pump charge one specific direction, and your loop walk either moves with it (gain) or against it (loss).
The EMF vs. terminal voltage subtlety (not in this problem, but worth knowing): real batteries have a small internal resistance r. Under load, the voltage at the terminals is V_terminal = ε − Ir, which is less than the open-circuit EMF. In exam 2 problems you'll usually treat batteries as ideal (r = 0), so V_terminal = ε exactly.
Mnemonic: think "climbing up" to + (add voltage) and "climbing down" to − (subtract voltage). A battery is just a hill in your voltage landscape — which direction you're walking determines whether you're going up or down the hill.
Example 1 — 24 V, 3 Ω in series with 4 Ω ‖ 12 Ω
Let's start with this problem. We have a resistor in series with two other resistors that are parallel to each other, and let's say this resistor has a value of 3 ohms and let's call it R₁. This is going to be R₂ and it has a value of 4 ohms, and R₃ has a value of 12 ohms. Calculate the current flowing through each resistor using Kirchhoff's rules.
So let's say that the current that's flowing through the first resistor — we're going to call it I₁. The current flowing through the second resistor, let's call it I₂, and the current flowing through the third resistor, we're going to call it I₃.
Now, using Kirchhoff's junction rule, or his current law, let's focus on this junction here. So notice that we have a current I₁ which flows into the junction, and I₂ leaves the junction, I₃ also leaves it to go this way. Now, according to his junction rule, the current that enters the junction is equal to the total current that leaves the junction, so the only current that's entering the junction is I₁ — the other two currents are leaving the junction, so I₁ has to equal the sum of I₂ plus I₃. So that's one equation that we have so far; we're going to use it later.
I₁ = I₂ + I₃ (equation ①)Diagram
Single junction at node a · I₁ in (top wire) · I₂ and I₃ split into the two parallel branches.
If a circuit has only one battery and resistors that reduce cleanly into series/parallel combinations, you don't need full Kirchhoff's laws — just use R_series = R₁ + R₂ + ... and 1/R_parallel = 1/R₁ + 1/R₂ + ... to collapse the circuit down to a single equivalent resistance, then V = IR.
When Kirchhoff is mandatory:
- Multiple batteries in different branches (like Example 2 coming up)
- Bridge circuits where resistors aren't purely series or parallel
- You need to know currents in every branch, not just the total
This example is a hybrid: it has only one battery, so you could simplify the 4 Ω ‖ 12 Ω first to 3 Ω (1/R_p = 1/4 + 1/12 = 1/3), then add the 3 Ω series for a total of 6 Ω, then I_total = 24/6 = 4 A. But the transcript walks through it with full Kirchhoff as a teaching exercise — and the answer (I₁ = 4 A, I₂ = 3 A, I₃ = 1 A) matches exactly.
Both methods must give the same answer. If they don't, you have an arithmetic mistake. This is one of the best self-checks in circuit analysis: solve the same problem two different ways and compare.
Parallel combination math check: 4 Ω ‖ 12 Ω = (4·12)/(4+12) = 48/16 = 3 Ω. Shortcut: when you have only two parallel resistors, use R_p = (R₁·R₂)/(R₁+R₂) — the "product over sum" trick. For three or more, you must use the reciprocal formula.
Example 1 — Loop 1 and Loop 2 equations
Now, current flows from high potential to low potential, so this is going to be the positive side of the resistor and that's the negative side; this is going to be the positive side of R₂ and that's the negative side of R₂, because current is flowing in that direction. And the same is true for R₃.
So now, using the loop rule, we're going in this direction through R₁, so we're going in the direction of the current towards a lower potential. So because we're going towards the lower potential, that's going to be a voltage drop. So the voltage drop across R₁ is V₁, and V₁ is equal to the current that flows through it times R₁; and R₁ is 3, so then this voltage drop is simply the current times 3 ohms, which we can write as 3·I₁.
As we follow the direction of the loop, we're going through R₂ towards a low potential, so that's going to be another voltage drop, and so that's going to be minus I₂ times the resistance of 4. So we could say 4·I₂. And so now we've completed loop one, so we're going to set this equal to zero.
+24 − 3I₁ − 4I₂ = 0 → 24 = 3I₁ + 4I₂ (equation ②)Now let's move on to the second loop. So we're going to go in this direction — that's going to be loop two. So initially we're going up through R₂ towards the positive sign, we're going against the current. And so since we're going from negative to positive, that's going to be a voltage lift, and so the voltage lift is going to be I₂ times the resistance of 4, so it's 4·I₂.
Then, following the loop, we're going in this direction through R₃, so we're going towards the negative sign, we're going from positive to negative, so that's going to be a voltage drop. So the current is I₃, the resistance is 12, so that's going to be 12·I₃, and so we have a negative sign for a voltage drop but a positive sign for a voltage lift. And this is equal to zero.
+4I₂ − 12I₃ = 0 → 4I₂ = 12I₃ (equation ③)Diagram
For any multi-loop circuit, you have to write as many independent loop equations as there are unknowns (after using KCL to eliminate one variable). "Independent" means each loop should include at least one branch the others don't touch.
The mesh-loop picking rule: imagine the circuit drawn flat on paper. Each "window" (open enclosed region surrounded by wires) is one mesh loop. A circuit with 2 windows needs 2 loop equations; 3 windows needs 3. Using the windows automatically guarantees independent loops.
Sign conventions — the part that trips everyone up:
- Resistor, crossed in the direction of the assumed current: voltage drops → write −IR
- Resistor, crossed against the assumed current: voltage rises → write +IR
- Battery, crossed from − to +: voltage rises → write +ε
- Battery, crossed from + to −: voltage drops → write −ε
The loop direction is arbitrary. You can walk a loop clockwise or counter-clockwise — either is valid, and both give the same physics after the signs work out. Pick whichever makes the signs easiest to track. The total must equal 0 because potential is single-valued at any point (you can't "lose" voltage by walking around and returning).
The deep reason loops sum to zero: electric potential is like gravitational potential energy. If you climb a hill and come back to where you started, your net change in height is 0. Same for voltage: whatever rises you take, you must descend by the same amount to return. Loop rule = energy conservation for circuits.
Example 1 — Solving the system & walking the potentials
Now, notice the two equations that we have. We have two equations and we have three variables, so in order to solve a system with three variables we need three equations, and that's why this equation is important. So now let's combine those three equations.
So what I'm going to do in this case is I'm going to replace I₁ with I₂ + I₃, because it's equal to that. So 24 is equal to 3 times (I₂ + I₃) plus 4·I₂. And so 24 is equal to 3·I₂ + 3·I₃ + 4·I₂. And now we can combine like terms; so far we have 24 is equal to 7·I₂ + 3·I₃.
Now notice that these two equations have the same variables I₂ and I₃, so I'm going to solve by elimination, which means that I want to change this to negative 3·I₃ so that it can cancel with positive 3·I₃. So I'm going to multiply this equation by ¼, which is equivalent to dividing everything by 4. So 0/4 is 0, and then 4·I₂/4 is I₂, and then −12·I₃/4 is −3·I₃.
Now let's add these two equations. So these will cancel: 24 + 0 is 24, and 7 + 1 is 8. So now let's divide both sides by 8, and 24 ÷ 8 is 3. So I₂ is equal to 3 amps — that's the first answer.
Now let's find a second answer. So let's use this equation here to calculate I₃ using I₂. We have 0 = I₂ − 3·I₃, so I'm going to move this to that side, so 3·I₃ is equal to I₂, so 3·I₃ is equal to 3 amps. So let's replace it with 3, and then let's divide both sides by 3, so 3/3 is 1. Therefore I₃ is equal to 1 amp.
Now we can calculate I₁. I₁ is simply the sum of I₂ and I₃, so I₁ is equal to 3 + 1, so that's 4 amps.
It makes sense: if 4 amps is flowing into this junction, 4 amps should leave it. We have 3 going here and 1 going there, so the numbers add up.
Let's assume that this is 0 volts, so that's the potential anywhere along this portion of the wire. Now here we have a difference of 24 volts across that battery, so the potential at, let's say, this point is 24 volts — it's 24 volts higher than this point. Now what is the potential at point a? Let's calculate the voltage drop across this resistor: V = IR, so it's the current that flows through it times the resistance. 4 × 3 gives us a voltage drop of 12, and 24 − 12 tells us that the potential at this point is 12 volts.
Now if we take 12 volts — because 12 − 0 gives us a voltage of 12 volts across this resistor — 12 volts divided by 4 ohms will give us the current of 3 amps, and 12 volts divided by 12 ohms will give us a current that flows through that resistor, which is 1 amp. So all the numbers make sense; it's good to check your work to see if you have the right answer.
Diagram — currents & node potentials
One of the most important circuit facts — and the source of endless confusion — is: parallel branches share the same voltage, but split the current. Series elements do the opposite: they share the same current but split the voltage.
Why parallel resistors see the same voltage: both ends of every parallel branch touch the same two nodes. Since each node has a well-defined potential, the voltage across each branch = V_node₁ − V_node₂ = same for all. The voltage is determined by the nodes, not the resistor.
Why they split the current: at the node, total current arriving must equal total current leaving (KCL). The current chooses paths based on resistance: lower R branches get more current. For two parallel resistors, I₁/I₂ = R₂/R₁ (inverse ratio — less resistance gets more current).
Sanity check from this problem: 4 Ω and 12 Ω in parallel both sit across 12 V. So I through 4 Ω = 12/4 = 3 A and I through 12 Ω = 12/12 = 1 A. Total = 4 A → matches the current through the series 3 Ω resistor feeding the junction. KCL holds.
Common mistake: students think "the 12 Ω gets more voltage because it's bigger." Wrong — in parallel, they share voltage. The 12 Ω gets less current, not more voltage. Keep repeating: parallel shares V, series shares I.
Example 2 — Two batteries & the substitution trick
This time we're going to have a second battery in this problem. So here's the positive terminal and here is the negative terminal. On this side we have a 30 volt battery and on this side a 10 volt battery. Because the voltage of this battery is higher, chances are that the current will flow in this direction as opposed to that direction.
Now even if you use the wrong arrow and guess it incorrectly, the current will simply be negative instead of positive. So if you get a negative current, it just means you have to reverse the arrow and then it's going to be positive. So really it doesn't matter what direction you choose the current to be — you can still get the right answer.
Now we're going to say that this resistor is a 2 ohm resistor — we'll call it R₁ — and this is going to be a 5 ohm resistor which we'll call R₂, and this will be a 3 ohm resistor and that's R₃. Now once again this current will be I₁, and the current flowing through this resistor we're going to call I₂. Now, like before we used to call this I₃, but this time we're going to do something different.
In this junction we have I₁ flowing in, I₃ leaving the junction, and I₂ also leaving it. So based on Kirchhoff's current law or junction rule, the current that enters the junction must equal the current that leaves the junction, so I₁ — which enters the junction — must add up to I₂ and I₃. So if we wish to solve for I₃, I₃ is simply the difference between I₁ and I₂.
Now why is this important? Instead of writing I₃ as the current flowing through this branch, we could express it as I₁ − I₂, and so instead of dealing with three variables I₁, I₂ and I₃, we can deal with two variables I₁ and I₂.
For more complicated examples you can make the process a lot easier if, instead of writing this as I₃, you can write it as I₁ − I₂, and so from this point on that's what I'm going to do for any future problems that I solve in this video.
I₁ = I₂ + I₃ ⇒ I₃ = I₁ − I₂ (only 2 unknowns now)Diagram
When a circuit has three branches meeting at a node, there are three currents — but only two of them are independent. The junction rule (KCL) ties the third to the first two. The substitution trick is: use KCL to eliminate that third variable before writing any loop equations.
Before substitution: 3 unknowns (I₁, I₂, I₃) → need 3 equations (1 junction + 2 loops). Annoying.
After substitution: KCL gives I₃ = I₁ − I₂ (or similar). Now you have 2 unknowns → only 2 loop equations needed. Cleaner, faster, fewer chances for arithmetic mistakes.
When to apply it: always. In every multi-loop problem, identify the junction with the most branches and use KCL to eliminate one current. The only exception is when the circuit has more than one independent node (rare in textbook problems).
Intuition for why it works: current is a conserved quantity, like water flow. At a T-junction in a pipe, if you know how much water enters and how much leaves one branch, you automatically know how much leaves the other branch. You never need to measure all three — the physics enforces the relationship. Kirchhoff's Current Law is nothing more than "charge is conserved at every node."
Example 2 — Loops, algebra, and verification
Let's start with loop one. Initially we're going in this direction towards the positive part of the battery, so that's going to be a voltage lift of 30. Based on the direction of the current, this is going to be positive and this is going to be negative. So now as we travel this way towards the resistor, we're going in a direction of I₁ towards the lower potential, so that's going to be a voltage drop of 2 times I₁ — so I₁ times the resistance of 2.
Now, following the loop, we're going to go in this direction — that's in the direction of the current towards the low potential — so that's another voltage drop of I₂ times the resistance of 5, so that's going to be −5·I₂. And so that's equal to zero. Now I'm going to take these two terms and move them to the other side, so now I have this equation: 30 is equal to 2·I₁ + 5·I₂.
30 = 2I₁ + 5I₂ (equation A)Now let's focus on loop two. Here we're going against I₂ towards a higher potential, so that's going to be a voltage lift of 5 times I₂. Next, as we follow the loop, we need to go through this battery towards the higher potential, so that's another voltage lift of positive 10. And then we need to travel in this direction — that is, in the direction of this current as it goes this way towards the lower potential — so that's going to be a voltage drop, and so that's going to be −3 times this current, which is I₁ − I₂.
All of this is equal to zero. Let's distribute the −3, so 0 = 5·I₂ + 10 − 3·I₁ + 3·I₂. Let's combine like terms, and so we have 0 = 10 − 3·I₁, and 5 + 3 is 8, so that's plus 8·I₂. Take these two and move them to the other side, so now I have positive 3·I₁ − 8·I₂ that's equal to 10.
3I₁ − 8I₂ = 10 (equation B)Now let's do some algebra. Let's try to get rid of I₁. The least common multiple of 2 and 3 is 6, so I'm going to multiply this by positive 3 and this equation by negative 2. So 3 × 30 is 90, 2·I₁ × 3 is 6·I₁, and 5·I₂ × 3 is 15·I₂. Now −2 × 10 is −20, and 3·I₁ × −2 is −6·I₁, and −8·I₂ × −2 is +16·I₂.
Now let's add up these two equations. These will cancel: 90 + (−20) is positive 70, and 15 + 16 is 31. So I₂ is equal to 70/31, so I₂ is 2.258 amps.
Now let's calculate I₁. Let's use this formula, so 3·I₁ − 8·I₂ — or 8 × 2.258 — is equal to 10. Negative 8 × 2.258 is −18.064, so let's add 18.064 to both sides, so this is going to be 3·I₁ that's equal to 28.064, and so I₁ is 28.064 ÷ 3, and so it's about 9.3 amps.
Let's say that point a is the ground — let's say it's at zero volts. So given the potential at point a, calculate the potential at b, c and d. Going from a to b we have a voltage lift of 30 volts, so the potential at b is going to be 30 volts. Now going from b to c, we know that the current flowing through the 2 ohm resistor is I₁ and that's a current of 9.355 amps. The voltage drop is the current times the resistance — 9.355 × 2, so that gives us a voltage drop of 18.71 volts. So 30 − 18.71 means the potential at point c is 11.29 volts.
As we go from c to a, the potential drops to zero, so we can calculate the current in the 5 ohm resistor by taking the voltage across the 5 ohm resistor — 11.29 — and dividing it by the resistance of 5 ohms; that gives us a current of 2.258 amps, which is in agreement with this answer I₂. Now let's move from c to d: as we travel in this direction we're going towards the positive terminal, so that's a voltage lift of 10 volts, so 11.29 + 10 — the potential at d is 21.29 volts. The voltage across the 3 ohm resistor is 21.29 ÷ 3 = 7.097 amps, the same as I₁ − I₂. All the numbers are in agreement.
Diagram — final currents & node potentials
Once you've written the junction and loop equations, you're left with linear algebra. For two unknowns, the cleanest method is substitution: solve one equation for one variable, then plug it into the other. For three unknowns, elimination is usually faster — add/subtract equations to cancel variables.
The substitution trick in action: from Loop 1 you get something like 30 = 2·I₁ + 5·(I₁ − I₂). Expand: 30 = 7I₁ − 5I₂. From Loop 2: 10 = 3·I₁ − 8·I₂. Now you have two equations in two unknowns — solve normally (multiply rows to align coefficients, subtract, solve for one variable, back-substitute).
Why the middle branch carries I₁ − I₂: this is the junction rule quietly doing its job. At the top node, I₁ comes in from the left and I₂ goes out to the right; by KCL, whatever's left (I₁ − I₂) flows down the middle branch. By using this expression in your loop equations from the start, you only need two variables instead of three.
Verification checklist (do this on every problem):
- Plug currents back into every loop equation — each should sum to 0 (or match the EMFs)
- Check the junction: ΣI_in = ΣI_out at every node
- Compute power in vs. power dissipated — they must match
- Walk node potentials — any loop you traverse must return to its starting V
These four checks are independent, so if you pass all of them, your answer is definitely right.
Example 3 — Predicting current directions
Now let's work on a more complicated example. This is going to be a 20 volt battery, this is going to be a 3 ohm resistor and this is going to be a 2 ohm resistor; and then we have a 4 ohm resistor and a 12 volt battery, and then this is going to be an 18 volt battery, a 10 ohm resistor and a 6 volt battery and a 6 ohm resistor. Calculate the current flowing through the circuit, and we're going to say that the potential at this point is zero volts.
The first thing I like to do is try to predict the direction of the currents in this circuit. So let's focus on this loop. The 20 volt battery wants to shoot out a current in that direction, and so that current wants to travel in this direction in the clockwise direction; and the 12 volt battery wants to shoot out a current in this direction which is going counterclockwise. However, this battery is stronger, so therefore comparing those two circuits I believe the net current in this part will be going in that direction.
Looking at this loop, the 12 volt battery wants to send a current going in this direction, and these two batteries they add up to make a 24 volt battery. The combined effect of those two batteries will want to send the current going in this direction as opposed to that, and this voltage wants to send a current going through this direction. So therefore it makes sense that the current in this branch is going in that direction towards the left, because the 24 volt battery wants to send current in this direction which is greater than the 12 volt battery.
Now let's assume that we have a current going in this direction which we'll call I₁, and there's a current going in this direction which we'll call I₂. Now we can say that there's a current going in this direction which is I₁ − I₂. If we're wrong, we're going to get a negative answer for this result, which means that the current is going this way and it's going to be I₂ − I₁ instead.
Perhaps you've noticed that there's two junctions, and we really don't need to take into consideration this junction because the current that flows through the 2 ohm resistor is the same current that flows in that part of the circuit. So this is also I₁ and this is I₂, and the current that flows here is the difference between I₁ and I₂, so we get the same currents in that junction so we don't have to worry about it.
Diagram
Kirchhoff's method has one beautiful feature that trips up beginners: you don't need to know the actual current directions before you start. You just assign arbitrary arrows and let the math sort it out.
How it works: label currents I₁, I₂, I₃ with arrows however you want. Write junction and loop equations using those assumed directions. Solve. If a current comes out negative, nature is telling you "you guessed the direction wrong — the real current flows the other way, but the magnitude is correct."
Concrete example: suppose you assume I₁ flows right-to-left through a 6 Ω resistor. You solve and get I₁ = −0.8 A. Reality: 0.8 A flowing left-to-right. The voltage drop across the resistor is still IR = 4.8 V, and everything downstream works out. You don't rewrite the equations. You just reinterpret the sign.
Why this is unavoidable in multi-battery circuits: when you have two batteries pushing in opposite directions, it's often impossible to tell by inspection which wins. The stronger EMF determines net current flow, and that depends on all the resistors too. Rather than trying to figure it out ahead of time, just guess, solve, and trust the sign.
The deeper lesson: in physics, negative answers usually contain real information. Negative work means energy flowed the other way. Negative acceleration means decelerating. Negative current means the flow is reversed from your assumption. Don't discard minus signs — read them.
Example 3 — Loops, solution & full potential walk
Let's focus on the top loop. So let's define the loop going in that direction — we'll call it loop one. In loop one we're going towards a higher potential, and so that's going to be a voltage lift of 20. I₁ is going in this direction so therefore this is going to be positive, negative, positive, negative.
As we flow through the 3 ohm resistor as we follow the loop, we're going in the direction of the current towards a negative potential, so that's going to be a voltage drop across the 3 ohm resistor, and that's going to be −3·I₁. As we follow the loop through the 2 ohm resistor we're still going in a direction of I₁, so that's another voltage drop which is −2·I₁. Then we're going in this direction through the 12 volt battery but towards the negative terminal, so we're going from positive to negative, so that's going to be a voltage drop of 12. Then we're going to travel through the 4 ohm resistor from positive to negative, and so the current in that branch is I₂, so that's going to be −4·I₂.
So we've covered all the elements in loop one. Let's combine like terms: 20 − 12 is 8, and then we have −3·I₁ − 2·I₁ which is −5·I₁, and then this is −4·I₂. Take these terms and move them to this side, so now we have this equation: 8 = 5·I₁ + 4·I₂.
8 = 5I₁ + 4I₂ (loop 1)Now let's focus on the next loop — loop two. Let's start with this direction: we're going towards the negative terminal of the battery, so from positive to negative, that's a voltage drop of 18 volts. Then we're following this resistor from negative to positive, so we're going from low to high potential against I₂, so that's a voltage lift of 4·I₂. Then, as we follow loop 2, we're traveling through the 12 volt battery from negative to positive — towards the higher potential — so that's a voltage lift of 12 volts.
Then we're going to follow through this resistor which is in the direction of this current which we defined that way, so we're going from positive to negative in the direction of the current, so that's a voltage drop, and that's going to be −6·(I₁ − I₂). Next we're going to go through this 6 volt battery from positive to negative, so we're going towards the lower potential, and that's going to be a voltage drop of 6. And then since the current is going in this direction, this is going to be the positive terminal of the resistor and the negative terminal of the resistor, so we're going from high potential to low potential in the direction of the current, so that's a voltage drop of 10·(I₁ − I₂), and all of this is equal to zero.
Now let's simplify: −18 + 12 is −6, plus another −6 so that's −12, and then plus 4·I₂. Now distribute, so that's −6·I₁ + 6·I₂, and then distribute the −10 so that's −10·I₁ + 10·I₂, and that's equal to zero. Combine like terms: 4·I₂ + 6·I₂ + 10·I₂ is 20·I₂, and we have −6·I₁ − 10·I₁ which is −16·I₁. So we have −16·I₁ + 20·I₂ = 12.
−16I₁ + 20I₂ = 12 (loop 2)Let's cancel I₂. Multiply the first equation by −5, so it becomes −40 = −25·I₁ − 20·I₂, and rewrite the other equation right beneath it. Add them: those two terms cancel, −40 + 12 = −28, and that's equal to −25 − 16 = −41 times I₁. So I₁ is −28/−41, so I₁ = 0.6829 amps.
Now let's calculate I₂. So 8 is equal to 5·I₁ + 4·I₂; 5 × 0.6829 is 3.4145, so 8 − 3.4145 is 4.5855 = 4·I₂, so I₂ = 4.5855 ÷ 4 = 1.1464 amps.
I₁ is positive, which means there is a current flowing in this direction. I₂ is also positive. Now what about I₁ − I₂? If we take 0.6829 − 1.1464 that gives us a negative answer of −0.4635, so therefore this direction is not correct — the correct direction is I₂ − I₁, so we have a current of 0.4635 amps going the other way.
As we travel from a to b we're going from low potential to high potential, so that's going to be a voltage lift of 20 volts, so the potential at b is 20 volts. Now let's go from b to c — I₁ flows from positive to negative, so as we go from b to c this is going to be a voltage drop of 3·I₁ — 3 × 0.6829 = 2.0487 — so 20 − 2.0487 gives us 17.95 V at c. From c to d we're going towards a lower potential — 2 × 0.6829 = 1.3658 — so 17.95 − 1.37 gives us 16.58 V at d. From d to e we're going from positive to negative, so that's a voltage drop of 12 volts: 16.58 − 12 = 4.58 V at e.
From d to f: we're going against the current and towards a higher potential — from low to high — so that's a voltage lift. The voltage across the 6 ohm resistor is 6 × 0.4635 = 2.781 volts, so 16.58 + 2.781 = 19.36 V at f. From f to g we're going from positive to negative — a voltage drop of 6 volts — so 19.36 − 6 = 13.36 V at g. From a to h we're going from negative to positive, that's a voltage lift of 18 volts, so the potential at h is 18 V. Current flows from high to low — 18 is greater than 13.36 — so the current is flowing through the 10 ohm resistor in the direction we defined. The voltage difference 18 − 13.36 = 4.64 V; divided by 10 ohms gives 0.464 amps, in agreement with I₂ − I₁.
Diagram — currents & node potentials a–h
Once you have all the branch currents, the node potential walk is the single most reliable way to catch mistakes. Here's why it's so powerful:
The idea: pick one node, call it ground (V = 0). Now walk from that node along any path to any other node, keeping track of the voltage. Every time you cross a resistor with the current, subtract IR. Every time you cross against the current, add IR. Every time you cross a battery from − to +, add ε. From + to −, subtract ε. The potential at each node is a single well-defined number.
The magic: no matter which path you take to a node, you must get the same potential. This is just the loop rule restated — any closed path sums to zero. So if two different paths from ground to node X give different numbers, one of your currents is wrong.
Why negative currents are OK: when your math gives I = −2 A, don't panic. It means your assumed arrow pointed the wrong way. Just interpret the current as flowing in the opposite direction with magnitude 2 A. The physics is the same; only your mental arrow was wrong.
Power dissipation check: after solving, verify energy conservation: total power delivered by batteries (ΣεI) should equal total power dissipated in resistors (ΣI²R). If they don't match, you have a sign error somewhere.
Example 4 — Three-loop system & negative-current handling
This is going to be the final problem that we're going to work on. Let's say this is a 2 ohm resistor and here we have an 8 ohm resistor and a 4 ohm resistor; this one's going to be a 3 ohm resistor, 5, and 6. On the left we have a 20 volt battery and this is going to be a 6 volt battery, and there's an 8 volt, a 12 volt, and the last one's a 16 volt battery. Calculate the current flowing through every branch of the circuit, and let's define this as point a, b, c, d, e, f, g, h. At point a the potential will be zero volts.
Let's define the current that flows through I₂ as I₁, and the current that flows through the 8 ohm resistor — that's going to be I₂. The current flowing through the 3 ohm resistor — that's going to be I₁ − I₂. Through the 5 ohm resistor we're going to call that I₃. The current that flows in this branch which is basically the same as through the 6 ohm resistor — that's going to be I₁ − I₂ − I₃.
Let's start with the first loop — loop one. As we go in this direction we have a voltage lift of 20 volts. As we flow in this direction — that is, in the direction of I₁ — that's going to be a voltage drop, so that's −2·I₁. Then as we follow the loop through the 8 ohm resistor in the direction of I₂, that's going to be −8·I₂, that's another voltage drop. Then as we go through the 6 volt battery we're going from a high potential to a low potential, so that's a voltage drop of 6 volts. Then, following I₁ which goes through the 4 ohm resistor, that's another voltage drop of −4·I₁.
Combine like terms: 20 − 6 = 14, −2·I₁ − 4·I₁ = −6·I₁, then minus 8·I₂. Move them, so 14 = 6·I₁ + 8·I₂. Divide everything by 2: 7 = 3·I₁ + 4·I₂.
7 = 3I₁ + 4I₂ (loop 1)Loop two. We have a voltage lift of 6 volts; then through the 8 ohm resistor going from negative to positive against I₂, so that's +8·I₂. Then we travel in the direction of I₁ − I₂ — that's a voltage drop of −3·(I₁ − I₂). Through the 5 ohm resistor in the direction of I₃, that's a voltage drop of −5·I₃. Then through the 8 volt battery towards the lower potential, that's a voltage drop of 8 volts, and that should equal 0. Simplify: 6 − 8 = −2, plus 8·I₂, distribute the −3 → −3·I₁ + 3·I₂, minus 5·I₃; combine: 8·I₂ + 3·I₂ = 11·I₂. So −2 − 3·I₁ + 11·I₂ − 5·I₃ = 0, i.e. −3·I₁ + 11·I₂ − 5·I₃ = 2.
−3I₁ + 11I₂ − 5I₃ = 2 (loop 2)Loop three. As we travel in this direction we have a voltage lift of 5·I₃ — going against I₃ — so that's +5·I₃. Then we're traveling towards the positive terminal of the battery — a voltage lift of 16 volts. Then we're following the direction of this current, so that's a voltage drop of −6·(I₁ − I₂ − I₃). Then we're traveling towards the positive terminal — a voltage lift of 12 volts. Combine: 16 + 12 = 28, then distribute the −6 to get −6·I₁ + 6·I₂ + 6·I₃, plus the 5·I₃ from before makes 11·I₃. So 28 = 6·I₁ − 6·I₂ − 11·I₃.
28 = 6I₁ − 6I₂ − 11I₃ (loop 3)We have three equations with three variables, so we need to use the system-of-equations process to get the answer. The best way to approach this particular situation is to combine equations two and three; our goal is to cancel I₃, because that gives us an equation in terms of I₁ and I₂ which we can combine with equation one.
Multiply the second equation by −11 and the third equation by +5. So +55·I₃ from one cancels with −55·I₃ from the other. −3·I₁ × −11 = +33·I₁; 11·I₂ × −11 = −121·I₂; −5·I₃ × −11 = +55·I₃; 2 × −11 = −22. Then 6·I₁ × 5 = 30·I₁; −6·I₂ × 5 = −30·I₂; −11·I₃ × 5 = −55·I₃; 28 × 5 = 140. Add: 33 + 30 = 63·I₁; −121 + (−30) = −151·I₂; 140 − 22 = 118.
Now combine this with equation one: cancel I₁. Multiply the first equation by −21: 3·I₁ × −21 = −63·I₁, 4·I₂ × −21 = −84·I₂, 7 × −21 = −147. Add: −151 − 84 = −235·I₂; 118 − 147 = −29. So I₂ = −29 / −235 = 0.1234 amps.
Now use the first equation before multiplying by 7 to calculate I₁: 3·I₁ + 4·I₂ = 7, so 4 × 0.1234 = 0.4936; 7 − 0.4936 = 6.5064 = 3·I₁; I₁ = 6.5064 / 3 = 2.1688 amps.
Now calculate I₃ using the third equation before multiplying by 5: 6·I₁ − 6·I₂ − 11·I₃ = 28. 6 × 2.1688 = 13.0128; 6 × 0.1234 = 0.7404; 13.0128 − 0.7404 = 12.2724; 28 − 12.2724 = 15.7276 = −11·I₃; so I₃ = −1.4298 amps. Because it's negative, we need to reverse the direction of I₃.
I₁ flows through this branch — this is 2.1688 amps. I₂ is 0.1234 amps. I₁ − I₂ = 2.0454 amps through the 3 ohm resistor. The current that flows in this direction is I₁ − I₂ − I₃ = 2.1688 − 0.1234 − (−1.4298) = 3.4752 amps. So now let's calculate the potential at every point — let's assume point a is 0.
Going in this direction we have a voltage lift of 20 volts, so the potential at b is 20. Going from b to c we have a voltage drop because we're going towards a lower potential, so 20 − 2.1688 × 2; the potential at c is 15.66 volts. Going this way we're going towards a higher potential — the current through the 4 ohm resistor is I₁ — so 0 + 4 × 2.1688, the potential at h is 8.675 volts. As we go from h to i we have a voltage lift toward the positive terminal of 6 volts, so 8.675 + 6 = 14.675 volts.
The voltage across the 8 ohm resistor is the difference between the potentials at c and i: 15.66 − 14.675 = 0.985 volts; divided by 8 ohms gives a current of 0.1231 amps, which is very close to I₂ — so the information that we have in this loop is correct.
From c to d we have a voltage drop, so 15.66 − 2.0454 × 3, and the potential at d is 9.524 volts. From h to g is a voltage increase of 8 volts, so 8.675 + 8 = 16.675 volts at g. The potential difference 16.675 − 9.524 = 7.151 volts; dividing by 5 gives a current of 1.4302 amps — keep in mind I₃ is negative 1.43 going this way, but once we reverse it it becomes positive 1.43.
Now we have the potential at d, so let's calculate the potential at e. Going from d to e we have a voltage lift of 16 volts, so 9.524 + 16 = 25.52 volts. Now from g to f, going from a high potential to a low potential, that's a voltage drop of 12 volts, so 16.675 − 12 = 4.675 volts at f. The potential difference between e and f is 25.524 − 4.675 = 20.849, that's the voltage across the 6 ohm resistor; divide by 6 to get a current of 3.475 amps — very close to the I₁ − I₂ − I₃ branch current.
Now you know how to use Kirchhoff's junction rule and loop rule to calculate all the currents in the circuit, and how to calculate the electric potential at any point in a circuit — but you need to be given a reference value first.
Diagram — three-loop ladder, reversed I₃
Once a circuit has more than two loops, blindly solving three simultaneous equations by hand becomes painful. The trick is to use the junction rule first to eliminate one unknown before you even write loop equations.
The procedure in 5 steps:
- Step 1 — Label every branch current. Pick arbitrary directions. If you guess wrong, the math will give you a negative number and you just know the real current flows the other way. It never matters which direction you pick.
- Step 2 — Apply the junction rule at each node. At any node, ΣI_in = ΣI_out. Use this to express one current in terms of the others (e.g., I₃ = I₁ − I₂). This reduces your unknowns from 3 to 2.
- Step 3 — Pick independent loops. For an N-loop circuit, you need N−1 junction equations and enough loop equations to close the system. Always pick loops that each include at least one element the other loops don't.
- Step 4 — Walk each loop in one direction (CW or CCW). Resistors crossed with the assumed current give −IR; against gives +IR. Batteries from − to + give +ε; + to − give −ε.
- Step 5 — Solve and check. Plug solutions back to verify every loop sums to zero and every node balances.
Negative currents don't mean you failed. A negative answer just means the real current flows opposite to your assumed arrow. You don't have to redo the problem — just reinterpret the sign.
Sanity check via node potentials: once you know all the currents, pick one node as ground (V = 0) and walk around the circuit computing each node's potential by adding/subtracting IRs and EMFs. If you come back to the starting node and don't get 0, your current values are wrong. This is how you catch algebra mistakes.