Ohm's law and resistor networks

By the end of this topic you will be able to

State Ohm's law and use it to relate voltage, current and resistance.
Find the equivalent resistance of series, parallel and series–parallel combinations.
Calculate the current, voltage and power in each resistor.

Voltage, current and resistance

Voltage $V$ (volts) is the energy per unit charge that pushes current round a circuit. Current $I$ (amperes) is the flow of charge. Resistance $R$ (ohms) is how strongly a component opposes the current. For a resistor, they are related by Ohm's law:

V = IR

The power turned into heat in a resistor is

P = VI = I^2 R = \frac{V^2}{R}

Series circuits

Components in series share one current; their voltages add up to the supply:

R_{\text{eq}} = R_1 + R_2 + R_3, \qquad V = V_1 + V_2 + V_3

Series resistors act as a voltage divider: the larger the resistor, the larger its share of the voltage, $V_k = V\,R_k/R_{\text{eq}}$ .

Parallel circuits

Components in parallel share one voltage; their currents add up:

\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}, \qquad I = I_1 + I_2 + I_3

The equivalent resistance is always less than the smallest branch: adding a path makes it easier for current to flow. Parallel resistors form a current divider: more current takes the smaller resistance.

Mixed networks

Reduce step by step: combine the parallel pair into one resistor, add it to the series resistor, then work backward to find each voltage and current.

Interactive modelSwitch between series, parallel and mixed. Change the resistances and read the currents.

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Worked example

A 12 V supply is connected to $R_1 = 100\ \Omega$ in series with $R_2 = 200\ \Omega$ and $R_3 = 300\ \Omega$ . Find the current and the voltage across each resistor.

Series. $R_{\text{eq}} = 100 + 200 + 300 = 600\ \Omega$ and $I = 12/600 = 20$ mA.

Voltages. $V_1 = 0.020\times100 = 2$ V, $V_2 = 4$ V, $V_3 = 6$ V; these add up to 12 V ✓.

Power. $P = 12\times0.020 = 0.24$ W in total.

Now parallel. The same resistors in parallel on 12 V: $R_{\text{eq}} = 54.5\ \Omega$ , total current 220 mA, with branch currents 120, 60 and 40 mA, which add up to 220 mA ✓.

Practice problems

A 9 V battery drives a 450 Ω resistor. Find the current and power.

Show answer

$I = 9/450 = 20$ mA; $P = 9\times0.02 = 0.18$ W.
Find the equivalent resistance of 100 Ω and 100 Ω in parallel.

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$R_{\text{eq}} = 50\ \Omega$ .
With $R_1 = 100$ , $R_2 = 200$ , $R_3 = 300\ \Omega$ in the "R1 in series with R2 ∥ R3" arrangement on 12 V, find the current drawn from the supply.

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$R_2 \parallel R_3 = 120\ \Omega$ ; $R_{\text{eq}} = 220\ \Omega$ ; $I = 12/220 = 54.5$ mA.
Why is the voltage across R2 and R3 the same in the mixed circuit?

Show answer

They are connected between the same two nodes (in parallel), and parallel branches share the same voltage.

Key formulae

Ohm's law: $V = IR$
Power: $P = VI = I^2 R = \frac{V^2}{R}$
Series resistance: $R_{\text{eq}} = R_1 + R_2 + R_3$
Parallel resistance: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (40 min). 5 min: water-pipe analogy; 10 min: Ohm's law and power; 10 min: series and parallel; 15 min: model, switching among the three arrangements.

Activity. In pairs: pick R values so that the mixed circuit draws exactly 50 mA from 12 V.

Discussion. Why are house lights wired in parallel and not in series?

Assessment idea. A voltage divider that must output a given voltage at a given load.