Question

Current in a circuit falls steadily from 2.0 A to 0.0 A in 10ms. If an average emf of 200 V is induced,then calculate the self inductance of the circuit

Answer 1

E = - L di/dt

E avg = - L (∆i / ∆t)

=> 200 = - L ( 0-2 ) / 0.01 = 2 L * 100
=> L = 1 H ...... ans..

Answer 2

The self inductance of the circuit is 1 H.

Explanation:

Given that,

Initial current, I = 2 A

Final current, I' = 0

Time, $t=10\ ms=0.01\ s$

Induced emf in the circuit, $\epsilon=200\ V$

To find,

The self inductance of the circuit.

Solution,

Due to changing current, an emf will be induced which is given by :

$\epsilon=-L\dfrac{dI}{dt}$

$L=\dfrac{-\epsilon}{(dI/dt)}$

$L=\dfrac{-\epsilon\times t}{(I'-I)}$

$L=\dfrac{-200\times 0.01}{(0-2)}$

L = 1 H

So, the self inductance of the circuit is 1 H.

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Self inductance

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