Question

2. A coil of inductance 0.5 H is point
connected to a 18 V battery.
Calculate the rate of growth of
current.
3.6 A/S
36 A/s
/
0 0.36 A/S
O 360 A/S​

Answer 1

Answer:

36A/S

Explanation:

Formula for the rate of growth of current in inductor is

$v = l \frac{di}{dt} \\ \frac{di}{dt} = v \div l = 36$

36A/S