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 will be

• A. $40\Omega$
• B. $\frac{40}{3}\Omega$
• C. $\frac{5}{2}\Omega$
• D. $100\Omega$

Answer 1

Answer: (A)

$40\Omega$

Answer 2

For same material and same length

$\frac{R_{2}}{R_{1}} = \frac{A_{1}}{A_{2}} = \frac{3}{2}$ ⇒ $R_{2} = 3R_{1}$

Resistance of thick wire $R_{1} = 10\Omega$

∴ Resistance of thin wire $R_{2} = 30\Omega$

Total resistance in series = 10 + 30 = 40 Ω