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. If the resistance of the thicker wire is $10\Omega$, then 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

The correct option is A $40\Omega$

As we know resistance $R$ is given by

$R=\rho \frac{l}{A}$

For same material and same length,

$\frac{R_2}{R_1}=\frac{A_1}{A_2}=\frac{3}{1}$

$\therefore R_2=3R_1$

Resistance of thicker wire is
$R_1=10\Omega$ (given)

$\therefore$ Resistance of thin wire

$R_2=3\times 10=30\Omega$

Total resistance of series combination is $R_{eq}$

$R_{eq}=R_1+R_2=10+30=40\Omega$

Hence, option $\left(a\right)$ is correct.