TS EAMCET · Physics · Electromagnetic Induction
Consider a current in a circuit falls from \(6.0 \mathrm{~A}\) to \(1.0 \mathrm{~A}\) in \(0.2 \mathrm{~s}\). If an average emf of \(150 \mathrm{~V}\) is induced by the circuit, then the self inductance of the circuit is
- A \(2 \mathrm{H}\)
- B \(6 \mathrm{H}\)
- C \(4 \mathrm{H}\)
- D \(8 \mathrm{H}\)
Answer & Solution
Correct Answer
(B) \(6 \mathrm{H}\)
Step-by-step Solution
Detailed explanation
Given, \(I_1=6 \mathrm{~A}, I_2=1 \mathrm{~A}, \Delta t=0.2 \mathrm{~s}\) and \(e=150 \mathrm{~V}\) Average emf induced in a inductor, \[ e=L \frac{\left(Y_1-Y_2\right)}{\Delta t} \] Self inductance,…
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