Ncert Exercises & Solutions

Q 8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Current at the initial point, null

Current at the final pint, null

Therefore, change in current is, $d I = I_{1} - I_{2} = 5 A$

Total time taken, t = 0.1 s

Average emf, e = 200 V

We have the relation, for self – inductance (L) and average emf of the coil : $e = \frac{d i}{d t}$

$\begin{array}{l} L = \frac{e}{(\frac{d t}{d t})} \\ = \frac{200}{\frac{5}{0.1}} = 4 H \end{array}$

Hence, the self – induction of the coil is 4 H.

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