Question 528: Electrostats & Capacitance

Question

A circuit having a self inductance of 1 henry carries a current of 1 A . To prevent the sparking when the circuit is broken, a capacitor which can withstand 500 V is connected across the switch. What is the minimum value of the capacitance of the capacitor?

Accepted Answer

The energy stored in an inductor is given by the expression:
$$ E = \frac{1}{2} L I^2 $$
where $ L $ is the inductance and $ I $ is the current. In this case, the energy is stored in a self-inductance $ L = 1 $ henry with a current $ I = 1 $ ampere. Substituting these values into the expression:
$$ E = \frac{1}{2} 	imes 1 	imes 1^2 = \frac{1}{2} \, 	ext{joule} $$
To prevent sparking, the energy from the inductor is transferred to a capacitor when the circuit is broken. The energy stored in a capacitor is given by:
$$ E = \frac{1}{2} C V^2 $$
where $ C $ is the capacitance and $ V $ is the voltage across the capacitor. The maximum voltage the capacitor can withstand is $ V = 500 $ volts. Equating the two expressions for energy:
$$ \frac{1}{2} L I^2 = \frac{1}{2} C V^2 $$
Simplifying, we find:
$$ L I^2 = C V^2 $$
Solving for $ C $, the minimum capacitance, we have:
$$ C = \frac{L I^2}{V^2} $$
Substitute the given values $ L = 1 $ henry, $ I = 1 $ ampere, and $ V = 500 $ volts:
$$ C = \frac{1 	imes 1^2}{500^2} = \frac{1}{250000} $$
Converting to microfarads:
$$ C = 4 	imes 10^{-6} \, 	ext{farads} = 4 \, \mu 	ext{F} $$
Thus, the minimum value of the capacitance of the capacitor is:
Option B: $ 4 \, \mu 	ext{F} $

Answer

$2 \mu \mathrm{~F}$

Answer

$4 \mu \mathrm{~F}$

Answer

$6 \mu \mathrm{~F}$

Answer

$8 \mu \mathrm{~F}$

Electrostats & Capacitance

A circuit having a self inductance of 1 henry carries a current of 1 A . To prevent the sparking when the circuit is broken, a capacitor which can withstand 500 V is connected across the switch. What is the minimum value of the capacitance of the capacitor?

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