In the inductive circuit given in the figure, the current rises after the switch is closed. At instant when the current is $15\ mA$, then potential difference across the inductor will be [image:7_image7.emf]

Question

Dr. Abubacker · Accepted Answer

(c) Here: Current in the circuit
$(i) = 15\ mA = 15 	imes 10^{- 3}A$
Resistance $R = 4000\ Volt$
Applied voltage in the circuit $= 240\ V$
At any instant, the $emf$ of the battery is equal to the sum of potential drop on the resistor and the $emf$ developed in the induction coil
Hence, $E = iR + L\frac{di}{dt}$
$240 = 15 	imes 10^{- 3} 	imes 4000 + L\frac{di}{dt}$
Hence, $L\frac{di}{dt} = 240 - 60 = 180\ V$

Q. In the inductive circuit given in the figure, the current rises after the switch is closed. At instant when the current is $15\ mA$, then potential difference across the inductor will be

Explanation

Reviewed by Dr. Abubacker

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