Question

Two wires of the same metal have the same length but their cross-sectional areas are in the ratio of 4:1. They are joined in series. The resistance of the thicker wire is 20 Ω. The total resistance of the combination is:

• A. 25 Ω
• B. 100 Ω
• C. 16 Ω
• D. 340 Ω

Answer 1

The Correct Option is B

The resistance of a wire is given by \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Since the wires are made of the same metal and have the same length, the resistivity (\( \rho \)) and length (\( L \)) are constant. Therefore, the resistance is inversely proportional to the cross-sectional area.

Let \( R_1 \) be the resistance of the thicker wire and \( R_2 \) be the resistance of the thinner wire. The ratio of their cross-sectional areas is \( 4:1 \), so the ratio of their resistances is \( 1:4 \) (since resistance is inversely proportional to area).

Given that \( R_1 = 20\ \Omega \), we have:

\[ \frac{R_2}{R_1} = 4 \implies R_2 = 4R_1 = 4(20\ \Omega) = 80\ \Omega \]

When resistors are connected in series, their resistances add up:

\[ R_\text{total} = R_1 + R_2 = 20\ \Omega + 80\ \Omega = 100\ \Omega \]

The total resistance of the combination is \( 100\ \Omega \).