Current Electricity - Result Question 32
34. Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is $10 \Omega$. The total resistance of the combination will be
[1995]
(a) $10 \Omega$
(b) $20 \Omega$
(c) $40 \Omega$
(d) $100 \Omega$
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Answer:
Correct Answer: 34. (c)
Solution:
- (c) Length of each wire $=\ell$; Area of thick wire $(A_1)=3 A$; Area of thin wire $(A_2)=A$ and resistance of thick wire $(R_1)=10 \Omega$. Resistance
$(R)=\rho \frac{\ell}{A} \propto \frac{1}{A}$ (if $\ell$ is constant)
$\therefore \frac{R_1}{R_2}=\frac{A_2}{A_1}=\frac{A}{3 A}=\frac{1}{3}$
or, $R_2=3 R_1=3 \times 10=30 \Omega$
The equivalent resistance of these two resistors in series
$=R_1+R_2=30+10=40 \Omega$.
