A) zero
B) 240 V
C) 180 V
D) 60 V
Correct Answer: C
Solution :
At any instant, the emf of the battery is equal to the sum of potential drop across the resistor and emf developed-in the induction coil. Therefore, we have \[E=IR+L\frac{dI}{dt}\] or \[L\frac{dI}{dt}=E-IR\] Substituting\[E=240V,\,\,I=15mA\] \[=15\times {{10}^{-3}}A,\,\,R=4000\Omega \], we get \[L\frac{dI}{dt}=180V\] Hence, from Faradays law, we have the potential developed in the induction coil is \[E\,\,=L\frac{dI}{dt}=180V\]You need to login to perform this action.
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