MHT CET · Physics · Alternating Current
An e.m.f. \(E=E_0 \cos \omega t\) is applied to circuit containing \(L\) and \(R\) in series. If \(X_L=2 R\), then the power dissipated in the circuit is
- A \(\frac{E_o{ }^2}{12 R}\)
- B \(\frac{\mathrm{E}_0{ }^2}{10 \mathrm{R}}\)
- C \(\frac{\mathrm{E}_0{ }^2}{8 \mathrm{R}}\)
- D \(\frac{E_0{ }^2}{6 R}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{E}_0{ }^2}{10 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
Impedance in the circuit is given by,
\(Z^2= R^2+X_L^2=R^2+(2 R)^2=5 R^2\)
\(P= \frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \cos \theta\)
\(=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{\sqrt{2} Z} \times \frac{R}{Z}\)
\(=\frac{E_0^2 R}{2 Z^2}\quad\ldots\left(\because Z^2=5 R^2\right)\)
\(=\frac{E_0^2 R}{10 R^2}\)
\(=\frac{E_0^2}{10 R}\)
\(Z^2= R^2+X_L^2=R^2+(2 R)^2=5 R^2\)
\(P= \frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \cos \theta\)
\(=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{\sqrt{2} Z} \times \frac{R}{Z}\)
\(=\frac{E_0^2 R}{2 Z^2}\quad\ldots\left(\because Z^2=5 R^2\right)\)
\(=\frac{E_0^2 R}{10 R^2}\)
\(=\frac{E_0^2}{10 R}\)
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