A parallel combination of three resistors takes a current of 7.5 A from a 30 V battery. If the resistances of two resistors are 12 Ω and 10 Ω, then the resistance of the third resistor is

15 Ω

The correct answer is Option (3) → 15 Ω

Given:

Total current: $I_\text{total} = 7.5\ \text{A}$

Voltage: $V = 30\ \text{V}$

Resistors: $R_1 = 12\ \Omega$, $R_2 = 10\ \Omega$, $R_3 = ?$

Parallel combination: $I_\text{total} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$

Substitute values:

$7.5 = \frac{30}{12} + \frac{30}{10} + \frac{30}{R_3} = 2.5 + 3 + \frac{30}{R_3}$

$7.5 = 5.5 + \frac{30}{R_3} \Rightarrow \frac{30}{R_3} = 2 \Rightarrow R_3 = 15\ \Omega$

∴ Resistance of the third resistor = 15 Ω