Question

Define self Inductance. Current in a circuit steadily from 2.0 A to 0.0 A in 10

milliseconds. If an average e.m.f. of 200V is induced, calculate the self-

inductance of the circuit​

Answer 1

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$\bold{Explanation:}$

The self inductance of the circuit is 1 H.

Explanation:

Given that,

Initial current, I = 2 A

Final current, I' = 0

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

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Self inductance

Answer 2

$\huge\underbrace\mathfrak \red{ANSWER }$

The self inductance of the circuit is 1 H.

Explanation:

Given that,

Initial current, I = 2 A

Final current, I' = 0

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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