Practice problems

Kirchhoff's laws practice problems

Kirchhoff's two laws let you work out the currents and voltages a single Ohm's law sum can't reach. These ten problems balance currents at a junction and voltages around a loop, then find a missing resistor and finish with a circuit that has two batteries and two loops.

The method

Kirchhoff's current law (KCL): at any junction, current in equals current out.

ΣI_in = ΣI_out

Kirchhoff's voltage law (KVL): around any closed loop, the rises equal the drops.

ΣV_rises = ΣV_drops

A missing current: add up the known currents in and out at the junction, and the difference is the one you want
A missing voltage: take the known drops off the supply
Two loops: call the shared node's voltage V, write each branch current in terms of V, and use KCL at that node

The problems

Type your answer and tap Check. Every one has the working, step by step.

Problem 1 of 10
50 mA flows into a junction. One wire carries 20 mA out. How much current leaves on the only other wire?
Show the working
1. KCL: what flows in flows out. 50 mA in = 20 mA + I out.
2. I = 50 − 20 = 30 mA
Answer: 30 mA
Problem 2 of 10
Three branches leave a junction, carrying 12 mA, 8 mA and 25 mA. How much current flows into the junction on the supply wire?
Show the working
1. The current in equals the total current out.
2. 12 + 8 + 25 = 45 mA
Answer: 45 mA
Problem 3 of 10
A 12 V battery drives three resistors in series. Two of them drop 3.5 V and 5.2 V. What does the third drop?
Show the working
1. KVL: around the loop, the drops add up to the battery's 12 V rise.
2. 12 − 3.5 − 5.2 = 3.3 V
Answer: 3.3 V
Problem 4 of 10
Four wires meet at a node. Two carry current in: 15 mA and 40 mA. A third carries 22 mA out. How much flows out on the fourth?
Show the working
1. Total in: 15 + 40 = 55 mA. Total out must match: 22 mA + I.
2. I = 55 − 22 = 33 mA out.
Answer: 33 mA
Problem 5 of 10
A green LED (2.2 V) and a 150 Ω resistor are in series across 5 V. What current flows?
Show the working
1. KVL first: the resistor drops what the LED leaves, 5 − 2.2 = 2.8 V.
2. Then Ohm's law on the resistor: I = 2.8 V over 150 Ω ≈ 0.0187 A, which is 18.7 mA.
Answer: 18.7 mA
Problem 6 of 10
A 1.2 kΩ resistor R1 and an unknown resistor R2 are in series across 6 V. A meter shows 2.4 V across R1. What is R2?
Show the working
1. The current through R1: I = 2.4 V over 1.2 kΩ = 2 mA. In series, the same 2 mA flows through R2.
2. KVL gives R2's drop: 6 − 2.4 = 3.6 V.
3. R2 = 3.6 V over 2 mA = 1.8 kΩ
Answer: 1.8 kΩ
Problem 7 of 10
A 12 V battery, a 1 kΩ resistor and an open switch are wired in a loop. What voltage does a meter read across the switch?
Show the working
1. No current flows, so the resistor drops I × R = 0 V.
2. KVL still holds: 12 V = 0 V + V_switch, so the switch has the full 12 V across it.
Answer: 12 V
Problem 8 of 10
A 5 V USB supply charges a 3.6 V rechargeable pack through a 68 Ω resistor, the two facing each other in the loop. What charging current flows?
Show the working
1. Facing each other, the two sources subtract. KVL leaves the resistor 5 − 3.6 = 1.4 V.
2. I = 1.4 V over 68 Ω ≈ 0.0206 A, about 20.6 mA, flowing into the pack's + terminal.
Answer: 20.6 mA
Problem 9 of 10
A 9 V battery feeds node A through a 1 kΩ resistor. A 6 V battery also feeds node A through its own 1 kΩ resistor. A third 1 kΩ resistor runs from A to ground, where both batteries' − terminals connect. What current flows through the third resistor?
Show the working
1. Call node A's voltage V. Each resistor's current is the voltage across it divided by 1 kΩ, so in milliamps it's just the volts.
2. KCL at A, current in = current out: (9 − V) + (6 − V) = V, so 15 = 3 × V and V = 5 V.
3. The third resistor carries 5 V over 1 kΩ = 5 mA.
Answer: 5 mA
Problem 10 of 10
In the same two-battery circuit, how much current does the 6 V battery supply?
Show the working
1. Node A sits at 5 V, so the 6 V battery's resistor has 6 − 5 = 1 V across it.
2. I = 1 V over 1 kΩ = 1 mA. The 9 V battery supplies 9 − 5 over 1 = 4 mA, and KCL checks out: 4 + 1 = 5 mA.
Answer: 1 mA

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“Kirchhoff's current law” is lesson 4 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 10 before it first.

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Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Kirchhoff's current law” is lesson 4 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 10 before it first.

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The ideas behind it

Questions people ask

What are Kirchhoff's two laws?

Kirchhoff's current law (KCL): the current flowing into any junction equals the current flowing out. Kirchhoff's voltage law (KVL): around any closed loop, the voltage rises add up to the voltage drops. KCL is conservation of charge, and KVL is conservation of energy.

When do I use KCL and when KVL?

Use KCL at a point where wires meet, to find a missing current. Use KVL around a loop, to find a missing voltage. Many problems need both: KVL to find a resistor's voltage, Ohm's law for its current, then KCL where the currents meet.

What if my answer comes out negative?

Nothing is wrong. A negative current means it flows the other way from the direction you guessed. Keep the minus sign while you work, then say which way it really flows.