Resistors in Series & Parallel

Resistors in Series

When resistors are connected in series, they are arranged one after another in a single path for the current to flow.

• Total resistance in a series circuit is the sum of the individual resistances:

$$R_\text{total} = R_1 + R_2 + R_3 + \dots$$

• Current is the same through each resistor because there is only one path for the charge to flow.
• Potential difference (voltage) divides across the resistors. The sum of the voltages across each resistor equals the total voltage supplied by the battery.

This means the voltage across each resistor depends on its resistance. A resistor with a higher resistance will have a larger share of the total voltage.

For instance, if two resistors of 4 9 and 6 9 are connected in series, the total resistance is:

$$R_\text{total} = 4\,\Omega + 6\,\Omega = 10\,\Omega$$

If the total voltage supplied is 12 V, the current through the circuit is:

$$I = \frac{V}{R_\text{total}} = \frac{12\,\text{V}}{10\,\Omega} = 1.2\,\text{A}$$

The voltage across each resistor is then:

$$V_1 = IR_1 = 1.2 \times 4 = 4.8\,\text{V}$$

$$V_2 = IR_2 = 1.2 \times 6 = 7.2\,\text{V}$$

Resistors in Parallel

When resistors are connected in parallel, they are connected across the same two points, creating separate branches for the current to flow.

• Potential difference (voltage) across each resistor is the same because each resistor is connected directly across the power supply.
• Current divides between the branches. The total current supplied by the battery is the sum of the currents through each resistor.
• Total resistance in a parallel circuit is less than the smallest individual resistor. This is because adding more branches provides extra paths for the current.

The total resistance in parallel is found using the reciprocal formula:

$$\frac{1}{R_\text{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$

For example, if two resistors of 6 9 and 3 9 are connected in parallel, the total resistance is:

$$\frac{1}{R_\text{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$$

So,

$$R_\text{total} = 2\,\Omega$$

If the voltage across the circuit is 12 V, the current through each resistor is:

$$I_1 = \frac{V}{R_1} = \frac{12}{6} = 2\,\text{A}$$

$$I_2 = \frac{12}{3} = 4\,\text{A}$$

The total current from the battery is:

$$I_\text{total} = I_1 + I_2 = 2 + 4 = 6\,\text{A}$$

Calculating Total Resistance

To find the total resistance in circuits with resistors in series or parallel, use the following methods:

• Series: Add the resistance values directly:

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

• Parallel: Use the reciprocal sum formula:

$$\frac{1}{R_\text{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$$

For two resistors in parallel, you can also use the shortcut formula:

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

This is useful for quick calculations but only works for two resistors.

When circuits combine series and parallel parts, calculate the total resistance step-by-step:

• Calculate total resistance of parallel sections first.
• Add series resistances to the result.

For example, consider a circuit with a 2 9 and 4 9 resistor in parallel, connected in series with a 3 9 resistor:

First, find the parallel resistance:

$$\frac{1}{R_\text{parallel}} = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$

$$R_\text{parallel} = \frac{4}{3} \approx 1.33\,\Omega$$

Then add the series resistor:

$$R_\text{total} = 1.33 + 3 = 4.33\,\Omega$$