question_answer

For an L-R circuit, the inductive reactance is equal to the resistance R of the circuit. An emf \[E={{E}_{0}}\cos \,\,(\omega \,t)\] is applied to the circuit. Then, the power consumed in the circuit is

A)  \[\frac{{{E}_{0}}}{R}\]

B)  \[\frac{E_{0}^{2}}{4R}\]

C)  \[\frac{4R}{{{E}_{0}}}\]                  

D)  \[\frac{R}{{{E}_{0}}}\]

Correct Answer: B

Solution :

For L-R combination,                 Given,                 \[{{X}_{L}}=R\] Impedance \[Z=\sqrt{{{R}^{2}}+{{R}^{2}}}\]                 \[=\sqrt{2}R\] Power consumed \[={{\left( \frac{{{V}_{rms}}}{Z} \right)}^{2}}R\] \[={{\left( \frac{{{E}_{0}}/\sqrt{2}}{\sqrt{2}R} \right)}^{2}}R=\frac{E_{0}^{2}}{4R}\]