Question

An inductance of 2H carries a current of 2A. To prevent sparking when the circuit is broken, a capacitor of 4$\mu$F is connected across the inductance. The voltage rating of the capacitor is of the order of :

• A. 1000 V
• B. 10 V
• C. 100 V
• D. $10^4$ V

Answer 1

Answer: (A)

1000 V

Answer 2

When a circuit containing an inductor is suddenly broken, the current through the inductor tries to drop to zero instantaneously. This rapid change in current (dI/dt) induces a very large electromotive force (EMF) across the inductor, given by $V = -L \frac{dI}{dt}$. This high induced voltage can cause sparking across the breaking contacts.

To prevent this sparking, a capacitor is connected across the inductance. When the circuit is broken, the energy stored in the inductor is transferred to the capacitor. By the principle of conservation of energy, the maximum energy stored in the inductor will be transferred to the capacitor, leading to a maximum voltage across the capacitor.

The energy stored in an inductor is given by: $E_L = \frac{1}{2}LI^2$

The energy stored in a capacitor is given by: $E_C = \frac{1}{2}CV^2$

Where: $L$ = inductance $I$ = current $C$ = capacitance $V$ = voltage across the capacitor

To prevent sparking, the energy stored in the inductor is completely transferred to the capacitor: $E_L = E_C$ $\frac{1}{2}LI^2 = \frac{1}{2}CV^2$ $LI^2 = CV^2$

We need to find the voltage rating of the capacitor ($V$). So, we rearrange the formula: $V^2 = \frac{LI^2}{C}$ $V = \sqrt{\frac{LI^2}{C}} = I\sqrt{\frac{L}{C}}$

Given values: Inductance, $L = 2$ H Current, $I = 2$ A Capacitance, $C = 4 \mu F = 4 \times 10^{-6}$ F

Substitute the values into the equation: $V = 2 \times \sqrt{\frac{2}{4 \times 10^{-6}}}$ $V = 2 \times \sqrt{\frac{1}{2 \times 10^{-6}}}$ $V = 2 \times \sqrt{0.5 \times 10^6}$ $V = 2 \times \sqrt{50 \times 10^4}$ $V = 2 \times \sqrt{50} \times \sqrt{10^4}$ $V = 2 \times \sqrt{50} \times 10^2$

We know that $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. $V = 2 \times 5\sqrt{2} \times 100$ $V = 10\sqrt{2} \times 100$ $V = 1000\sqrt{2}$

Using the approximate value $\sqrt{2} \approx 1.414$: $V \approx 1000 \times 1.414$ $V \approx 1414$ V

The voltage rating of the capacitor should be able to withstand this peak voltage. The calculated voltage is approximately 1414 V, which is of the order of 1000 V.