Question

An e.m.f. \(e=E_0 \cos \omega t\) is applied to a circuit containing \(L, C\) and \(R\) in series where \(X_L=3 R\) and \(\mathrm{X}_{\mathrm{C}}=\mathrm{R}\). The average power dissipated in the circuit is

• A. \(\frac{E_{0}^{2}}{5R}\)
• B. \(\frac{E_{0}^{2}}{10R}\)
• C. \(\frac{\mathrm{E}_{0}^{2}}{15\mathrm{R}}\)
• D. \(\frac{E_{0}^{2}}{20R}\)

Hint:

Find the impedance from the net reactance, then use average power equal to rms voltage squared times R over impedance squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(Z=\sqrt{R^2+(X_L-X_C)^2}\) and \(P_{avg}=\dfrac{V_{rms}^2R}{Z^2}\).

Step 2: Impedance
\(X_L-X_C=3R-R=2R\), so \(Z^2=R^2+4R^2=5R^2\).

Step 3: Average power
\(V_{rms}^2=\dfrac{E_0^2}{2}\), so \(P=\dfrac{E_0^2}{2}\cdot\dfrac{R}{5R^2}=\dfrac{E_0^2}{10R}\).

Final Answer:
The average power is E0 squared over 10R. \[ \boxed{\text{(B) }\dfrac{E_{0}^{2}}{10R}} \]