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 :

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

Tardigrade Admin · Accepted Answer

Here: Current in the circuit (i)=15 mA =15 × 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 (d i/d t) 240=15 × 10-3 × 4000+L (d i/d t) Hence L (d i/d t)= Sigma=240-60=180 V

Q. In the inductive circuit given in the figure, the currents rises after the switch is closed. At instant when the current is 15mA, then potential difference across the inductor will be :

zero

240 V

180 V

60 V

Q. In the inductive circuit given in the figure, the currents rises after the switch is closed. At instant when the current is $15 m A$ , then potential difference across the inductor will be :