Kirchhoff's Laws

PHYCells & Circuits

Kirchhoff's laws extend circuit analysis beyond simple series-parallel reduction to multi-loop networks with several sources. The junction rule expresses charge conservation and the loop rule expresses energy conservation, giving enough equations to solve for every branch current.

ISC examines two batteries feeding a common load, where you assign current directions, write the junction and loop equations, and solve.

Junction (current) rule

\sum I_{in} = \sum I_{out}

At any node the total current entering equals the total leaving. A statement of charge conservation; e.g.

I_1 + I_2 = I_3

at a node.

Loop (voltage) rule

\sum E - \sum I R = 0

Around any closed loop the algebraic sum of EMFs equals the sum of

IR

drops. A statement of energy conservation.

Sign convention for a loop

\Delta V = -IR \;\; (\text{along current}), \qquad \Delta V = +E \;\; (- \to + \text{ terminal})

Traversing a resistor in the current's direction gives a drop

-IR

; crossing a cell from

-

+

gives a rise

+E

. Reverse the sign if traversed the other way.

Two parallel batteries feeding a load

I_3 = I_1 + I_2, \qquad E_1 = I_1 r_1 + I_3 R, \qquad E_2 = I_2 r_2 + I_3 R

I_1, I_2

branch currents from the two cells,

I_3

through the common load

R

. Solve the three simultaneous equations for the currents.

Derivation

Assign branch currents $I_1$ (through cell 1), $I_2$ (through cell 2) and $I$ through $R$ , all directed into the node feeding $R$ . The junction rule gives:
$I = I_1 + I_2$
The loop rule around each cell's loop (cell $\to$ node $\to$ $R$ $\to$ back to the cell) gives two independent equations:
$E_1 = I_1 r_1 + IR, \qquad E_2 = I_2 r_2 + IR$
Solve each loop equation for its branch current and substitute into the junction rule:
$I = \dfrac{E_1-IR}{r_1} + \dfrac{E_2-IR}{r_2}$
Collect the terms in $I$ and solve for the current through $R$ :
$I\left(1+\dfrac{R}{r_1}+\dfrac{R}{r_2}\right) = \dfrac{E_1}{r_1}+\dfrac{E_2}{r_2} \;\Rightarrow\; I = \dfrac{E_1 r_2 + E_2 r_1}{r_1 r_2 + R(r_1+r_2)}$

The junction rule is conservation of charge; the loop rule is conservation of energy — together they fully determine the unknown currents.
Assume a direction for each branch current first; a negative answer simply means the real current flows opposite to your guess.
Across a resistor traversed along the assumed current direction, take the potential change as $-IR$ ; against it, $+IR$ .
Crossing a cell from negative to positive terminal is a rise $+E$ ; from positive to negative it is $-E$ , independent of current direction.
You need as many independent equations as unknown currents: one fewer than the number of junctions plus enough independent loops.
Include each branch's internal resistance $r$ as a series resistor in that branch's loop equations.
Check the solution by substituting the currents back into an unused loop or into the junction rule.

Where the marks go

Inconsistent sign convention — mixing up the direction of $IR$ drops or EMF rises within a single loop.
Treating a negative current as an error instead of reading it as a reversed direction.
Forgetting to include internal resistances $r_1, r_2$ in the loop equations.
Writing too few independent equations (e.g. two dependent loops) and being unable to solve for all currents.

How the board asks it

Numericaltwo parallel batteries feeding a common load
Two cells of emf $E_1 = 6\,\text{V}$ and $E_2 = 4\,\text{V}$ , with internal resistances $r_1 = 1\,\Omega$ and $r_2 = 2\,\Omega$ , are connected in parallel across an external resistance $R = 4\,\Omega$ . Using Kirchhoff's laws, calculate the current through each cell and the current through $R$ .
Define / statejunction (charge) rule and loop (energy) rule
State Kirchhoff's junction rule and loop rule, and name the conservation principle that each one expresses.
Applicationsetting up the junction and loop equations
Apply Kirchhoff's laws to the two-loop circuit shown, assign a direction to each branch current, and write the three independent equations needed to find $I_1$ , $I_2$ and $I_3$ (you need not solve them).
Numericalsolving branch currents in a two-loop network
In the network shown, two cells of emf $10\,\text{V}$ and $4\,\text{V}$ drive currents through resistors of $2\,\Omega$ , $4\,\Omega$ and $6\,\Omega$ . Using Kirchhoff's rules, calculate the current in each branch.
Derive / proveloop rule with sign convention
Two cells of emf $E_1$ and $E_2$ with internal resistances $r_1$ and $r_2$ are connected in parallel across a resistance $R$ . Using Kirchhoff's rules, derive an expression for the current $I$ through $R$ .
Give reasonsa negative current means a reversed assumed direction
On solving a circuit by Kirchhoff's laws a student obtains a branch current $I = -0.5\,\text{A}$ . Explain what the negative sign signifies about the actual direction of this current.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.