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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. - CBSE (Science) Class 12 - Physics

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Question

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.

Solution

Initial current, I1 = 5.0 A

Final current, I2 = 0.0 A

Change in current`dI=I_1-I_2=5A`,

Time taken for the change, t = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

e = L `"di"/"dt"`

`L=e/((di)/(dt))`

`=200/(5/0.1)=4H`

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

 

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 6: Electromagnetic Induction
Q: 8 | Page no. 230

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Solution 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. Concept: Inductance - Self-Inductance.
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