Question

An e.m.f. \(e=E_{0}\cos \omega t\) is applied to a circuit containing L, C and R in series where \(X_{L}=4R\) and \(X_{C}=R\). The average power dissipated in the circuit is

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

Hint:

Find the impedance from the reactances, then use average power equal to the square of the rms emf times R divided by the square of the impedance.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a series LCR circuit the average power is dissipated only in the resistor.
It is given by \(P=\dfrac{E_{rms}^{2}R}{Z^{2}}\), where \(Z\) is the impedance.

Step 2: Impedance
\(Z=\sqrt{R^{2}+(X_L-X_C)^{2}}=\sqrt{R^{2}+(4R-R)^{2}}=\sqrt{10}\,R\)
So \(Z^{2}=10R^{2}\).

Step 3: RMS emf
The peak emf is \(E_0\), so \(E_{rms}=\dfrac{E_0}{\sqrt{2}}\) and \(E_{rms}^{2}=\dfrac{E_0^{2}}{2}\).

Step 4: Calculation
\(P=\dfrac{(E_0^{2}/2)\,R}{10R^{2}}=\dfrac{E_0^{2}}{20R}\)
This matches option (A).

Final Answer:
The average power dissipated in the circuit is E0 squared divided by 20R. \[ \boxed{\text{(A) }\dfrac{E_{0}^{2}}{20R}} \]