15 mA, then potential difference across the inductor will be:
A) zero
B) 240 V
C) 180V
D) 60V
Correct Answer: C
Solution :
Here: Current in the circuit \[(i)=15\,mA=15\times {{10}^{-3}}A\] Resistance \[R=4000\text{ }Volt\] Applied voltage in the circuit = 240 V At any constant, 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, \[E=iR+L\frac{di}{dt}\] \[240=15\times {{10}^{-3}}\times 4000+L\frac{di}{dt}\] Hence \[L\frac{di}{dt}=\Sigma =240-60=180\text{ }V\]
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