Question

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

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

Answer 1

Answer: (C)

For the same length and same material,
$\frac{R_{2}}{R_{1}}=\frac{A_{1}}{A_{2}}=\frac{3}{1}$ or $R_{2}=3\, R_{1}$
The resistance of thick wire, $R_{1}=10\, \Omega$
The resistance of thin wire $=3\, R_{1}=3 \times 10=30\, \Omega$
Total resistance $=10+30=40\, \Omega$