Kirchhoff rules for branched circuits

Kirchhoff's rules extend Ohm's law for a complete circuit, $I = \frac{\varepsilon}{R + r}$, to networks that cannot be reduced to a single series–parallel chain.

Junction rule: at any node the algebraic sum of the currents is zero, $\sum_k I_k = 0$ — charge does not accumulate, so the current entering a node equals the current leaving it.

Loop rule: around any closed contour the algebraic sum of the potential differences is zero, $\sum_k \Delta V_k = 0$ — this is energy conservation per unit charge.

To use the loop rule, choose a traversal direction: a resistor crossed along the current contributes $-IR$, against the current $+IR$; a source crossed from $-$ to $+$ contributes $+\varepsilon$, from $+$ to $-$ gives $-\varepsilon$.

A negative answer is not an error: it means the real current flows opposite to the arrow you drew, and its magnitude is correct.

With $n$ nodes write $n-1$ independent junction equations and complete the system with the loop equations of the independent contours, so the number of equations equals the number of unknown branch currents.

Four wires meet at a node. Currents $I_1 = 3$ A and $I_2 = 2$ A flow into the node, and $I_3 = 1$ A flows out of it. Use the junction rule to find the current $I_4$ in the fourth wire and state its direction.

Nodes $A$ and $B$ are joined by three parallel branches.

• Branch 1: source $\varepsilon_1 = 12$ V with internal resistance $r_1 = 1\ \Omega$, positive terminal at $A$.
• Branch 2: source $\varepsilon_2 = 4$ V with internal resistance $r_2 = 1\ \Omega$, positive terminal at $A$.
• Branch 3: resistor $R = 2\ \Omega$.

Find the current in each branch.

Nodes $A$ and $B$ are joined by three branches.

• Branch 1: source $\varepsilon_1 = 10$ V and resistor $R_1 = 1\ \Omega$ in series, positive terminal at $A$.
• Branch 2: source $\varepsilon_2 = 5$ V and resistor $R_2 = 1\ \Omega$ in series, positive terminal at $A$.
• Branch 3: an unknown resistor $R_3$; the measured current through it is $3$ A flowing from $A$ to $B$.

Find $R_3$ and the currents in branches 1 and 2, then check the loop rule for the contour through branches 1 and 3.

Test both rules in the PhET simulation Circuit Construction Kit: DC.

1. Build a circuit with one battery ($\varepsilon \approx 9$ V) and two resistors connected between the same pair of nodes, so that the circuit has three branches.
2. Insert an ammeter in series in each branch and record the current together with its direction.
3. Check the junction rule: add the currents algebraically at each node; the sum must be zero within the reading error.
4. Connect a voltmeter across each branch and check the loop rule: walking around any closed contour, the algebraic sum of the measured potential differences must be zero.
5. Change one resistance, repeat the measurements, and explain how the currents redistribute.

If you reproduce this with real parts, use only a low-voltage battery (never mains) and keep the ammeter in series.

To make the sign conventions of the loop rule and the node labelling explicit before any algebra: the sketch shows which end of each element is the positive terminal and which way each assumed current flows.

A labelled sketch of a two-node, three-branch circuit: node $A$ on the left, node $B$ on the right; in each branch the source $\varepsilon_k$, its internal resistance $r_k$ and the current arrow $I_k$; one dashed contour with a traversal arrow marked $\sum_k \Delta V_k = 0$.

A car battery has EMF $\varepsilon = 12{,}0$ V and internal resistance $r = 0{,}05$ $\Omega$. Two headlights are connected in parallel, each filament having resistance $R = 2{,}4$ $\Omega$. Current flows from the positive terminal along a common wire to the branching node, where it splits into two branches.

The branching node yields the junction rule: $I_0 = I_1 + I_2$. The parallel branches have a combined resistance $R_{\text{par}} = 1{,}2$ $\Omega$, so the current in the common wire is $I_0 = \dfrac{\varepsilon}{R_{\text{par}} + r} = \dfrac{12{,}0}{1{,}25} = 9{,}6$ A. The loop rule for the loop containing each headlight reads $\varepsilon - I_0 r - I_k R = 0$, giving the voltage across the lamps $U = 12{,}0 - 9{,}6\cdot 0{,}05 = 11{,}52$ V and $I_1 = I_2 = U/R = 4{,}8$ A, consistent with $I_0 = I_1 + I_2$. If one filament burns out, its branch opens, but the other lamp keeps glowing: the current in it becomes $12{,}0/(2{,}4+0{,}05) = 4{,}9$ A.

A power bank with EMF $\varepsilon = 5{,}00$ V and internal resistance $r = 0{,}056$ $\Omega$ is connected through a splitter cable to a phone and a tablet. An ammeter reads $I_1 = 1{,}20$ A in the phone branch and $I_2 = 2{,}40$ A in the tablet branch, while a voltmeter at the output reads $U = 4{,}80$ V.

The common lead carries $I_0 = I_1 + I_2 = 3{,}60$ A — the junction rule, verifiable with a single ammeter at the output. The loop containing the source gives the loop rule $\varepsilon - I_0 r - U = 0$, hence $I_0 r = 5{,}00 - 4{,}80 = 0{,}20$ V and $r = 0{,}056$ $\Omega$. Each device branch is its own loop: $U - I_k R_k = 0$, so the phone resistance is $R_1 = 4{,}80/1{,}20 = 4{,}0$ $\Omega$ and the tablet resistance is $R_2 = 4{,}80/2{,}40 = 2{,}0$ $\Omega$. If the tablet is unplugged, the current drops to $5{,}00/(4{,}0+0{,}056) = 1{,}23$ A.

In a kitchen, an electric kettle rated $2000$ W and a toaster rated $1000$ W are plugged into one socket at $U = 230$ V through a splitter. The socket is protected by a $13$ A fuse.

The appliances form parallel branches, and the node in the splitter gives the junction rule: $I_0 = I_1 + I_2 = \dfrac{2000}{230} + \dfrac{1000}{230} = 8{,}70 + 4{,}35 = 13{,}05$ A. Each branch is a loop sharing the same mains lead: $U - I_k R_k = 0$, so the heater resistances are $R_1 = 230/8{,}70 = 26{,}4$ $\Omega$ and $R_2 = 230/4{,}35 = 52{,}9$ $\Omega$. The current in the common wire, $13{,}05$ A, exceeds the fuse rating of $13$ A, so the fuse opens the circuit; with the toaster switched off the current is $8{,}7$ A, below the limit, and the circuit keeps working.