D.C. circuits, Kirchhoff’s laws and potential dividers

What the exam asks

• Kirchhoff’s first and second laws, and the conservation laws behind them.
• Series and parallel combinations, including networks that mix the two.
• Power and energy: $P = IV = I^2R = V^2/R$ and $W = VIt$.
• Potential dividers: calculating $V_{\text{out}}$, choosing resistor values, and sensor circuits.
• The effect of real meters and loads on a circuit (loading).
• Predicting and explaining changes when a switch closes, a component heats up or the light level changes.

Core ideas

Kirchhoff’s laws

First law: the sum of currents entering a junction equals the sum leaving it. This is conservation of charge: charge cannot build up at a point.

Second law: around any closed loop, the sum of the emfs equals the sum of the pds ($\sum \varepsilon = \sum IR$). This is conservation of energy: each coulomb gains as much energy from the sources as it transfers to the components.

To use the second law, choose a direction around the loop. An emf counts as positive if you pass through the cell from − to +. An $IR$ term counts as positive if you move through the resistor in the direction of the assumed current. A negative answer simply means the current flows the other way.

Series and parallel

Cambridge derivation: in series, the same $I$ flows, so $IR = IR_1 + IR_2$ and . In parallel, the same is across each branch, so .

Shortcuts: two resistors in parallel give $\dfrac{R_1R_2}{R_1 + R_2}$; identical resistors in parallel give . A parallel combination is always . Adding a branch in parallel always lowers the total resistance.

Power and energy

Choose the form that uses what is shared:

• In series (same $I$), use $P = I^2R$: the largest resistor dissipates the most power.
• In parallel (same $V$), use $P = V^2/R$: the resistor dissipates the most power.

Potential dividers

Two resistors in series across a supply $V_{\text{in}}$ share the pd in the ratio of their resistances. The output across $R_2$ is

A potentiometer (a variable resistor with a sliding contact) gives a continuously variable output from 0 to $V_{\text{in}}$. That is why a potential divider, not a series variable resistor, is used to supply a component when you want its pd to start from zero.

Sensor circuits. Replace one resistor with a thermistor or an LDR:

• The output is taken across one component. That component’s share of $V_{\text{in}}$ rises when its resistance rises relative to the other.
• For “output rises as it gets dark”, put the LDR across the output (its resistance rises in the dark).
• For “output rises as it gets hot”, put the thermistor in the other position; its resistance falls, so the fixed resistor’s share rises.
• The switching point occurs when the sensor’s resistance reaches a particular value. Replacing the fixed resistor with a variable resistor lets you adjust that point.

Loading. Anything connected across the output (a meter, a buzzer, a logic input) is in parallel with $R_2$. It lowers the combined resistance and so lowers $V_{\text{out}}$. Loading is negligible only if the load’s resistance is much larger than $R_2$.

Meters

An ideal ammeter has zero resistance and goes in series. An ideal voltmeter has infinite resistance and goes in parallel. A real voltmeter with a resistance comparable to the component it measures reads low, because it draws current and lowers the resistance of that part of the circuit. A voltmeter wrongly placed in series takes almost all the supply pd, so the current is tiny and a lamp in the same loop does not light.

Potentiometer and null methods (Cambridge)

A uniform resistance wire carrying a steady current has a pd that is proportional to length. A cell is connected through a galvanometer to a sliding contact. At the balance point the galvanometer reads zero, so no current is drawn from the cell being tested. The balance length is therefore proportional to its emf, not its terminal pd. For example, if 1.00 m of wire has 2.00 V across it and a cell balances at 0.745 m, its emf is $2.00 \times 0.745 = 1.49$ V.

Worked examples

Exam technique

• Redraw messy networks with the junctions clearly marked before calculating anything.
• “What happens when…” questions: follow the chain in order: resistance of the changed part, total resistance, total current, pds across fixed resistors, then the pd across the parallel part. Write each step as an arrow; examiners reward the chain.
• Resistor that dissipates the most power: decide first whether the resistors share $I$ or $V$.
• Potential divider design: find the current in the branch or use the ratio $V_1/V_2 = R_1/R_2$. It is usually quicker than rearranging the formula.

Common mistakes

Quick recap

• First law: charge is conserved at a junction. Second law: around any loop, sum of emfs = sum of $IR$ (energy is conserved).
• Series: $R = R_1 + R_2$, same current. Parallel: , same pd; the total is less than the smallest resistor.