Question

Current is a circuit falls steadily from 2.0A to 0.0A in 10ms. If an average emf of 22V is induced, calculate its

self-inductance.​

Answer 1

Solution :

As per the given data ,

• Initial current ( I ') = 2 A
• Final current ( I " )= 0 A

• EMF induced = 22 V
• Time = 10 ms = 0.01 s

The EMF is induced in the circuit due to the change in flow of current .

$\leadsto\boxed{E = - L\dfrac{\Delta I }{t}}$

$\leadsto E = - L\bigg(\dfrac{I" - I' }{t}\bigg )$

Now let's substitute the given values in the above equation ,

$\leadsto 22 = - L\bigg(\dfrac{0 - 2 }{0.01 }\bigg )\\ \\ \leadsto 22 = - L \times - 200 \\ \\ \leadsto L = \dfrac{22}{200} \\ \\ \leadsto \boxed{ L = 0.1 H}$

The self inductance of the circuit is 0.1 H .