Resistance and resistor arrangements

When a variable resistor is connected in series with a lamp, increasing its resistance increases the circuit’s total resistance. With the supply potential difference unchanged, the current falls and the lamp becomes dimmer. Reducing the resistance allows a larger current and makes the lamp brighter.

Linking potential difference, current and resistance

The quantities are related by:

$$V = I \times R$$

Here, $V$ is potential difference in volts (V), $I$ is current in amperes (A), and $R$ is resistance in ohms (Ω). A resistance of $1\,\Omega$ means that a potential difference of $1\,\text{V}$ produces a current of $1\,\text{A}$.

Rearrange the equation according to the quantity you need:

$$I = \frac{V}{R} \qquad R = \frac{V}{I}$$

Calculating current

For example, a resistor has a potential difference of 6 V across it and a resistance of 3 Ω. Calculate the current through it.

$$I = \frac{V}{R}$$ $$\frac{6}{3} = 2\,\text{A}$$

The result tells us how much charge flows each second. If the resistance were doubled while the potential difference stayed the same, the current would halve.

Calculating potential difference

For example, a 10 Ω resistor carries a current of 0.4 A. Calculate the potential difference across it.

$$V = IR$$ $$0.4 \times 10 = 4\,\text{V}$$

This is the potential difference needed across that resistor to produce the stated current.

Calculating resistance

For example, a resistor has a potential difference of 6 V across it and carries a current of 0.5 A. Calculate its resistance.

$$R = \frac{V}{I}$$ $$\frac{6}{0.5} = 12\,\text{Ω}$$

This resistance describes how strongly the component opposes current at those values of potential difference and current. For a fixed resistor at constant temperature, resistance stays constant, so current is directly proportional to potential difference. Other components do not necessarily have constant resistance.

Why resistors in series increase net resistance

In series, there is only one path: charge must pass through both resistors, one after the other. Adding a second resistor adds further opposition to the flow of charge, so the net resistance increases.

The resistances add:

$$R_{\text{total}} = R_1 + R_2$$

The same current passes through both resistors, while the supply potential difference is shared between them. With an unchanged supply, the greater net resistance means a smaller current.

Why resistors in parallel decrease net resistance

In parallel, each resistor is on a separate branch. Both branches have the same potential difference across them, and each provides a path for charge.

Adding the second branch allows an additional current to flow. The total current from the supply is the sum of the branch currents, so it increases even though the supply potential difference has not changed. Since $R_{\text{total}} = V/I_{\text{total}}$, a larger total current at the same potential difference means a smaller net resistance.

The combination has a resistance lower than either resistor alone. The individual resistors have not changed: it is the extra path that reduces the resistance of the combination.

$V$: volts (V); $I$: amperes (A); $R$: ohms (Ω).

Resistor arrangements

Series: charge passes through both resistors. Resistances add, so net resistance increases:

$$R_{\text{total}} = R_1 + R_2$$

Parallel: an extra branch provides another path for charge. At the same supply potential difference, total current increases, so net resistance decreases below either individual resistance.