Question

In the inductive circuit given in the figure, the currents rises after the switch is closed. At instant when the current is

$15\, mA,$ then potential difference across the inductor will be:

• A. zero
• B. 240 V
• C. 180 V
• D. 60 V

Answer 1

Answer: (C)

Here : Current in the circuit
$(i)=15\, mA =15 \times 10^{-3} A$
Resistance $R=4000$ Volt
Applied voltage in the circuit $=240\, V$
At any constant, the emf of the battery is equal to the sum of potential drop on the resistor and the emf developed in the induction coil.
Hence, $E=i R+L \frac{d i}{d t}$
$240=15 \times 10^{-3} \times 4000+L \frac{d i}{d t}$
Hence $L \frac{d i}{d t}=\Sigma=240-60=180\, V$