Question

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

• A. $10\Omega$
• B. $20\Omega$
• C. $40\Omega$
• D. $100\Omega$

Answer 1

The correct option is C $40\Omega$

Resistance of the wire $R=\frac{\rho L}{A}$

where $\rho$ is the resistivity and $L$ and $A$ is the length and cross-section area of the wire, respectively.

$⟹$ $R\propto \frac{1}{A}$ (for same $L$ and $\rho$)

Ratio of cross-sectional area of thick wire to thin wire $\frac{A_1}{A_2}=\frac{3}{1}$

Thus ratio of resistance of thin wire to thick wire $\frac{R_2}{R_1}=\frac{A_1}{A_2}=\frac{3}{1}$

Or $\frac{R_2}{10}=\frac{3}{1}$

$⟹$ $R_2=30\Omega$

Total resistance of the (series) combination $R_{eq}=R_1+R_2$

$⟹$ $R_{eq}=10+30=40\Omega$