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

Answer 1

The correct option is C $40\Omega$

Resistance of a wire $R=\frac{\rho l}{A}$ where $\rho =$ resistivity, $l=$ length, $A=$ cross section of the wire.
As both have same material and length so $R\propto \frac{1}{A}$
Thus, $\frac{R_2}{R_1}=\frac{A_1}{A_2}=\frac{3}{1}\Rightarrow R_2=3R_1$.
here $R_1$ is the resistance of thicker wire so its resistance $R_1=10\Omega$ (given)
so, $R_2=3\left(10\right)=30\Omega$
As they are connected in series so the equivalent resistance is
$R_{eq}=R_1+R_2=10+30=40\Omega$