In an electrical circuit, the three resistors of 3 Ω, 6 Ω and 9 Ω are connected in series. The voltage drop across 3 Ω is 5 V. The voltage drop across the combination is

30 V

The correct answer is Option (2) → 30 V

Given:

Resistances in series: $R_1 = 3 \, \Omega$, $R_2 = 6 \, \Omega$, $R_3 = 9 \, \Omega$

Voltage drop across $R_1$: $V_1 = 5 \, \text{V}$

Current in series is same through all resistors:

$I = \frac{V_1}{R_1} = \frac{5}{3} \, \text{A}$

Voltage drop across 6 Ω:

$V_2 = I \cdot R_2 = \frac{5}{3} \cdot 6 = 10 \, \text{V}$

Voltage drop across 9 Ω:

$V_3 = I \cdot R_3 = \frac{5}{3} \cdot 9 = 15 \, \text{V}$

Total voltage drop:

$V = V_1 + V_2 + V_3 = 5 + 10 + 15 = 30 \, \text{V}$