The current passing through a choke coil of 15 henry is decreasing at the rate of 0.2 ampere per second. The e.m.f. developed across the coil :

Question

The current passing through a choke coil of 15 henry is decreasing at the rate of 0.2 ampere per second. The e.m.f. developed across the coil : (A) 3 volt (B) 7.5 volt (C) 15 volt (D) 30 volt

Tardigrade Admin · Accepted Answer

E = L (dI/dt), Here L = 15 H and (dI/dt) = 0.2 As-1 this gives E = 3V

Q. The current passing through a choke coil of 15 henry is decreasing at the rate of 0.2 ampere per second. The e.m.f. developed across the coil :

3 volt

7.5 volt

15 volt

30 volt

$E = L \frac{d I}{d t}$ , Here L = 15 H and $\frac{d I}{d t}$ = 0.2 A $s^{- 1}$ this gives E = 3V

Q. The current passing through a choke coil of 15 henry is decreasing at the rate of 0.2 ampere per second. The e.m.f. developed across the coil :

3 volt

7.5 volt

15 volt

30 volt

E=LdtdI​, Here L = 15 H and dtdI​ = 0.2 As−1 this gives E = 3V

$E = L \frac{d I}{d t}$ , Here L = 15 H and $\frac{d I}{d t}$ = 0.2 A $s^{- 1}$ this gives E = 3V