Question

An electric current of $5\, A$ is passing through a circuit containing three wires arranged in parallel. If the lengths and radii of the wires are in the ratio 2: 3: 4 and 3: 4: 5 , then the ratio of currents passing through wires would be

• A. 3: 6: 10
• B. 4: 9: 16
• C. 9: 16: 25
• D. 54: 64: 75

Answer 1

Answer: (D)

Given $, l_{1}: l_{2}: l_{3}=2: 3: 4=2 l: 3 l: 4 l$
$r_{1}: r_{2}: r_{3}=3: 4: 5=3 r: 4 r: 5 r$
Resistance of a wire, $R=\frac{\rho l}{\pi r^{2}}$
$R_{1}=\frac{\rho 2 l}{\pi(3 r)^{2}}=\frac{2}{9} \frac{\rho l}{\pi r^{2}}=\frac{2}{9} R,$
$R_{2}=\frac{\rho 3 l}{\pi(4 r)^{2}}=\frac{3}{16} \frac{\rho l}{\pi r^{2}}=\frac{3}{16} R$
$R_{3}=\frac{\rho 4 l}{\pi(5 r)^{2}}=\frac{4}{25} \frac{\rho l}{\pi r^{2}}=\frac{4}{25} R$
$I_{1}=\frac{V}{R_{1}} ; I_{2}=\frac{V}{R_{2}}$ and $I_{3}=\frac{V}{R_{3}}$
$\therefore I_{1}: I_{2}: I_{3}=\frac{1}{R_{1}}: \frac{1}{R_{2}}: \frac{1}{R_{3}}$
$=\frac{9}{2 R}: \frac{16}{3 R}: \frac{25}{4 R}=\frac{9}{2}: \frac{16}{3}: \frac{25}{4}$
$=54: 64: 75$