Question

A circuit has self-inductance of 1H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400V is used. The least capacitance of capacitor connected across the switch must be equal to
$(A)50\mu F$
$(B)25\mu F$
$(C)100\mu F$
$(D)12.5\mu F$

Answer 1

To prevent sparking, what we have to consider is No current or Voltage should be exceeded with the limit of instruments capacity. Here the value of current and voltage are fixed, and what we can adjust is the value of capacitance. And we can consider conservation of energy as well.
Formula used:
Energy stored in the capacitor, ${E_c} = \dfrac{1}{2}C{V^2}$
And energy stored in the inductor, ${E_L} = \dfrac{1}{2}L{I^2}$
Where, $C$ is the value of capacitance, $V$ is the potential difference across plates of capacitor, $L$ is the value of inductance, and $I$ is the current flowing through the inductor.

Complete step by step answer:
Given, Self inductance $L = 1H$
Current in the circuit $I = 2A$
The voltage to which the capacitor can withstand $V = 400V$
Spark happens if more current flows through the capacitor than its capacity. Here we can adjust its capacity by changing the value of capacitance according to the current in the circuit.
Whenever the switch is on, the capacitor begins to charge and whenever we turn off the switch, the capacitor discharges through the inductor in the same circuit.
For having No spark situation, Energy stored in the capacitor must be equal to energy stored in the inductor. If this condition violates, then spark will happen in the circuit.
Now recalling our prior knowledge in capacitor and inductor.
We know, energy stored in the capacitor, ${E_c} = \dfrac{1}{2}C{V^2}$
And energy stored in the inductor, ${E_L} = \dfrac{1}{2}L{I^2}$
Where, $C$ is the value of capacitance (that we have to find out), $V$ is the potential difference across plates of capacitor (its maximum value is given$400V$ ), $L$ is the value of inductance($1H$) and $I$ is the current flowing through the inductor ($1A$ ).
Energy stored in the capacitor $=$ Energy stored in the Inductor
$\dfrac{1}{2}C{V^2} = \dfrac{1}{2}L{I^2}$
$C{V^2} = L{I^2}$
The value of capacitance $C = \dfrac{{L{I^2}}}{{{V^2}}}$
Now substituting the values, $C = \dfrac{{1 \times {2^2}}}{{{{400}^2}}}$
$C = \dfrac{4}{{16000}} = 2.5 \times {10^{ - 5}} = 25 \times {10^{ - 6}}F$
Hence, answer is $25\mu F$, option(B).

Note:
Self induction is that phenomenon in which a change in electric current in a coil produces an induced emf in the coil itself. ... Self inductance of a coil is defined as the ratio of self-induced emf to the rate of change of current in the coil.