Kirchhoff's laws

Physics · Senior school and first-year physics · about 10 min

Many circuits cannot be solved by combining resistors in one simple series or parallel chain. Kirchhoff's laws let you write reliable equations for currents at junctions and voltage changes around closed loops. The challenge is usually keeping track of signs rather than choosing a complicated formula.

The key idea

At every junction, current is conserved; around a closed circuit loop, potential changes sum to zero:

\sum I_{\mathrm{in}}=\sum I_{\mathrm{out}},\qquad\sum\Delta V=0.

Currents are measured in $\mathrm{A}$ and voltage changes in $\mathrm{V}$ .

Two small separated circuit sketches in one horizontal composition. Left: a junction drawn as a solid dot with two incoming arrows from left and above labelled I_1 and I_2, and one outgoing arrow to the right labelled I_3. Right: a single rectangular closed loop containing an ideal battery on its left vertical edge and one resistor R on its top edge, with a clockwise traversal arrow I. On the battery clearly mark plus at the upper terminal and minus at the lower terminal; show an upward traversal arrow on the left edge and a right-pointing arrow along the top edge. No numerical values and no additional branches. — Current balances at a junction, and signed potential changes balance around a loop.

Junction rule for current

At a junction, charge cannot build up indefinitely in a steady circuit. Therefore, the total current flowing into the junction equals the total current leaving it. Start by drawing a direction for each unknown current; the direction can be chosen arbitrarily. If an answer is negative, the actual current flows the other way. Current is the same through series components, but it can split between parallel branches. In equations, either write the incoming currents equal to the outgoing currents or give each current an algebraic sign and set their sum to zero. Do not mix the two conventions.

Loop rule for voltage

Travel once around any closed loop. Across a resistor in the direction of conventional current, the potential change is $-IR$ ; moving against the current gives $+IR$ . Across an ideal battery from its negative to its positive terminal, the change is $+\mathcal E$ , and the reverse traversal gives $-\mathcal E$ . Add these changes and set the total to zero. Label battery polarity and choose a loop direction before writing terms. An ideal wire has no voltage drop; a real circuit may need extra resistance terms if the problem includes them.

Shared resistors and mesh currents

For a planar circuit with two adjacent loops, choose one mesh current for each loop, perhaps both clockwise. In an outer resistor, the branch current is just its mesh current. In a resistor shared by both loops, the mesh contributions run in opposite directions. If the first mesh crosses it downwards, use $I_1-I_2$ for downward branch current and a drop $R(I_1-I_2)$ when traversing downwards. Apply the loop rule to each mesh and solve the simultaneous equations. Finally, use the junction rule to check the directions of the branch currents.

Worked example: Two loops with one shared resistor

A circuit has two rectangular meshes side by side. The left outside branch contains an ideal $12\,\mathrm{V}$ battery with its positive terminal at the top and a $2.0\,\Omega$ resistor in series. The right outside branch contains an ideal $6.0\,\mathrm{V}$ battery, also positive at the top, and a $3.0\,\Omega$ resistor in series. A $4.0\,\Omega$ resistor forms the common centre branch between the same top and bottom junctions. Define both mesh currents $I_1$ and $I_2$ clockwise. Find both mesh currents and the current through the centre resistor.

Draw a circuit with two rectangular loops side by side sharing one centre vertical branch, all three branches connected at a common top wire and common bottom wire. In the left outside vertical branch show an ideal 12 V battery with plus at top and minus at bottom and a 2.0 Ω resistor above it. In the middle vertical branch show a 4.0 Ω resistor. In the right outside vertical branch show an ideal 6.0 V battery with plus at top and minus at bottom and a 3.0 Ω resistor above it. Inside the left rectangle show a large clockwise circular arrow I_1, and inside the right rectangle show a large clockwise circular arrow I_2. Add short opposing current arrows next to the middle resistor, one downward for I_1 and one upward for I_2. Place junction labels A at the top and B at the bottom. Do not display solved currents or an answer. — Both clockwise mesh currents pass through the central resistor in opposite directions.

Show the solution

Step 1: Write the left loop equation. The left mesh goes upwards through its battery and downwards through the centre resistor:

12-2I_1-4(I_1-I_2)=0,\quad 6I_1-4I_2=12.

Step 2: Write the right loop equation. The right mesh goes downwards through its battery and upwards through the centre resistor:

-6-3I_2-4(I_2-I_1)=0,\quad -4I_1+7I_2=-6.

Step 3: Solve the simultaneous equations.

6I_1-4I_2&=12,\\ -4I_1+7I_2&=-6,\\ I_1&=\frac{30}{13}=2.30769\ldots\,\mathrm{A},\\ I_2&=\frac{6}{13}=0.461538\ldots\,\mathrm{A}. \end{aligned}$$ **Step 4: Find the centre-branch current.** The left mesh drives current down the shared resistor, while the right mesh drives it up:

I_{\mathrm{centre}}=I_1-I_2=\frac{24}{13}=1.84615\ldots,\mathrm{A}.

**Answer: $I_1=2.31\,\mathrm{A}$ clockwise, $I_2=0.462\,\mathrm{A}$ clockwise and $1.85\,\mathrm{A}$ downward through the centre resistor.**

Common mistakes

Reversing a battery sign without checking the terminals

Write $+\mathcal E$ when travelling from negative to positive and $-\mathcal E$ in the reverse direction.

Using $I_1+I_2$ for opposing mesh currents in a shared resistor

Draw the mesh arrows; in the common branch their contributions oppose, so use a signed difference.

Adding resistor drops without a consistent traversal direction

Choose the loop direction once and assign each $IR$ drop or rise from that direction.

Rejecting a negative calculated current

Keep the result and reverse the assumed arrow to interpret the actual direction.

Check your understanding

1. At a junction, currents of $3.0\,\mathrm{A}$ and $1.5\,\mathrm{A}$ enter. One outgoing current is $2.0\,\mathrm{A}$ . What is the other outgoing current?

2. A single loop contains an ideal $9.0\,\mathrm{V}$ source and resistors of $3.0\,\Omega$ and $6.0\,\Omega$ in series. What is the current?

3. Two clockwise mesh currents oppose each other through a shared $5.0\,\Omega$ resistor. In the downward direction, $I_1=2.0\,\mathrm{A}$ and $I_2=0.50\,\mathrm{A}$ acts upwards. What is the downward voltage drop across that resistor?

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