Question

Two wires of same metal have the same length but their cross-sections area 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. $40/3\Omega$
• C. $5/2\Omega$
• D. $100\Omega$

Answer 1

Answer: (A)

$40\Omega$

(Let R' denote thicker and R denote thinner resistor)
We know,
$R=\frac{\rho \times L}{A}⟹\frac{R^{\prime }}{R}=\frac{A}{A^{\prime }}$
$⟹\frac{10}{R}=\frac{1}{3}⟹R=30\Omega$
Since they are connected in series,the equivalent resistance is given by,
$R_{eq}=R+R^{\prime }⟹R_{eq}=30+10=40\Omega$