Kirchhoff’s rules

Two rules for any circuit

Narrated video. Transcript:

Here is a circuit with two cells and three resistors, joined in two loops. It is neither a simple series nor a simple parallel combination. How do we find the current in each wire?

Gustav Kirchhoff gave two simple rules that work for any circuit, however complicated. First, some words. A junction is a point where three or more wires meet. A loop is any closed path in the circuit.

Step one: in each resistor, mark a current with a symbol and an arrow. Pick any direction you like. If you guess wrong, the answer simply comes out negative.

Rule one, the junction rule. At any junction, the sum of the currents entering is equal to the sum of the currents leaving. Here, 3 amperes and 2 amperes flow in, so 5 amperes must flow out.

Why? Because charge is conserved. In a steady current, charge cannot pile up at a junction, so whatever charge arrives each second must leave each second.

Rule two, the loop rule. Go once round any closed loop and add up all the changes in potential, across the cells and across the resistors. The algebraic sum is zero.

Why? Potential is like height. Walk round a hill path and come back to where you started, and your total rise and fall is zero. Coming back to the same point, you come back to the same potential.

Now the signs. Crossing a cell from its negative plate to its positive plate, the potential rises by ε: write plus ε. Crossing from plus to minus, write minus ε.

Crossing a resistor in the same direction as its current, the potential falls by IR: write minus IR. Crossing it against the current, write plus IR. A cell’s internal resistance r is treated just like any other resistor.

Check with one loop: a 6 volt cell and a 3 ohm resistor. Going round with the current, plus 6 minus 3 I equals zero, so I is 2 amperes. Exactly what Ohm’s law told us.

Solving a two-loop circuit

Narrated video. Transcript:

Let us solve a real network. Two junctions, A at the top and B at the bottom, are joined by three branches. On the left, a 10 volt cell with a 2 ohm resistor. In the middle, a 4 ohm resistor. On the right, a 4 volt cell with a 4 ohm resistor.

Both cells have their positive plate at the top. Mark I₁ going up the left branch, I₂ going up the right branch, and I₃ coming down the middle.

Junction rule at A: I₁ and I₂ come in, I₃ goes out. So I₃ equals I₁ plus I₂. That is our first equation.

Loop rule round the left loop. Up through the cell, minus to plus: plus 10. Through the 2 ohm resistor with I₁: minus 2 I₁. Down the middle with I₃: minus 4 I₃. So 10 minus 2 I₁ minus 4 I₃ equals zero.

Loop rule round the right loop, the same way: 4 minus 4 I₂ minus 4 I₃ equals zero. Three equations, three unknown currents.

From the right loop, I₂ equals 1 minus I₃. Then I₁ equals I₃ minus I₂, which is 2 I₃ minus 1. Put this in the left loop: 10 minus 4 I₃ plus 2 minus 4 I₃ equals zero.

So 12 equals 8 I₃, and I₃ is 1.5 amperes. Then I₁ is 2 amperes, and I₂ is 1 minus 1.5: minus 0.5 ampere.

A negative current! It is not a mistake. It means I₂ really flows the other way: 0.5 ampere goes down through the right branch, into the positive plate of the 4 volt cell. The stronger cell is charging the weaker one.

Always check. At junction A: 2 amperes in from the left, 0.5 ampere out down the right, 1.5 amperes out down the middle. 2 equals 0.5 plus 1.5. The junction rule holds.

And the outer loop, through both cells: 10 minus 2 × 2, minus 4, minus 4 × 0.5 gives 10 minus 4 minus 4 minus 2, which is zero. Every rule is satisfied.

Try it: two loops, three currents

Interactive simulation. Things to try:

• The grey arrows are the directions we assumed. The moving dots show where the current really goes.
• With the starting values I₂ comes out −0.5 A: the dots on the right run against the arrow.
• Press “Loop 1” or “Loop 2” to walk round a loop and see each +ε and −IR term add up to zero.
• Raise ε₂ until I₂ becomes positive. At what value is I₂ exactly zero?
• Reverse cell 2 and predict which currents grow before you look.

Four wires at a junction

Four wires meet at a junction. Currents of 3 A and 2 A flow into it along two of the wires, and 4 A flows out along the third. What is the current in the fourth wire?

1. 1 A, flowing out of the junction
2. 1 A, flowing into the junction
3. 5 A, flowing out of the junction
4. 9 A, flowing out of the junction

A cell with no current

In the two-loop circuit, take ε₁ = 12 V, R₁ = 2 Ω, ε₂ = 6 V, R₂ = 2 Ω and R₃ = 2 Ω (both cells with + at the top). What current flows through cell 2?

1. 3 A
2. 1.5 A
3. 0 A
4. 2 A

Summary

• Junction rule: Σ I (in) = Σ I (out)
• Loop rule: Σ ΔV = 0 round any closed loop
• Signs: cell − to +: +ε; resistor along I: −IR

• The junction rule comes from conservation of charge: in a steady current, charge does not pile up at a junction.
• The loop rule comes from the fact that potential has one value at each point: round a closed loop the rises and falls cancel.
• Assume a direction for each current. A negative answer means the current flows the other way; the size is still right.
• Use the junction rule at junctions and the loop rule on enough loops to get as many independent equations as unknown currents. Internal resistances count as resistors.