Question

Two wires of the same metal have same length, but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of thicker wire is $10\, \Omega.$ The total resistance of the combination will be

• A. $40\, \Omega$
• B. $100\, \Omega$
• C. $(5/2)\, \Omega$
• D. $(40/3)\, \Omega$

Answer 1

Answer: (A)

$40\, \Omega$

Answer 2

Ratio of cross-sectional areas of the wires = 3 : 1 and resistance of thick wire $(R_1)=10\, \Omega.$
Resistance $(R)=\rho \frac{l}{A} \propto\frac{1}{A}.$
Therefore $\frac{R_1}{R_2}=\frac{A_2}{A_1}=\frac{1}{3}$ or $R_2 = 3R_1 = 3 \times 10 = 30\, \Omega$ and equivalent resistance of these two resistances in series combination
$= R_1+R_2 =30+10 =40\, \Omega.$