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Kirchhoff's voltage law: round any loop, the volts add to zero

Going once around a closed loop, the rises through sources and the drops across resistors sum to zero: ΣV = 0.

Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.

What it is for

A control loop has a 24 V supply, a fuse, a relay coil and a long cable, and the relay will not pull in. Adding the drops around the loop shows where the volts went, and that is the whole of fault-finding by meter.

How to think about it

Pick a direction and go round once. A source you pass from minus to plus is a rise; a resistor passed in the direction of the current is a drop of I × R. Set the sum to zero and solve for the one current.

Worked example

Around any closed loop the voltages add to zero: ΣV = 0
The law.
A 10 V source, then 2 Ω and 3 Ω in series
One loop, one current.
10 − I × 2 − I × 3 = 0 → I = 10 / 5 = 2 A
Rise through the source, drops across each resistor, in the direction of travel.
V across the 3 Ω = 2 × 3 = 6 V; across the 2 Ω = 4 V; 6 + 4 = 10 OK
The drops add up to the source, as they must.

Your turn

A 9 V source with 1 Ω and 7 Ω in series. Write the loop equation.

9 − I × 1 − I ×  = 0

The trap

Losing a sign. A second source facing the other way is a drop, not a rise, and adding it instead of subtracting can double the current. Go round in one direction and never change your mind halfway.

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