Question

In a series resonant LCR circuit, for the power dissipated to become half of the maximum power dissipated, the current amplitude is

• A. \(\frac{1}{\sqrt{2}}\) times its maximum value.
• B. \(\frac{1}{2}\) times its maximum value.
• C. twice its maximum value.
• D. \(\sqrt{2}\) times its maximum value.

Hint:

In an LCR circuit, the power dissipated is proportional to the square of the current amplitude. Maximum power occurs at resonance when current is maximum. The frequencies at which power dissipated is half of the maximum power are called half-power frequencies. At these frequencies, the current amplitude is \(1/\sqrt{2}\) times the maximum current amplitude. This is a crucial concept in understanding the bandwidth and Q-factor of resonant circuits.

Answer 1

The Correct Option is A

Step 1: Recall the formula for power dissipated in an LCR circuit.
In a series LCR circuit, the average power dissipated (\(P\)) is due to the resistance and is given by: \[ P = I_{\text{rms}}^2 R \] where \(I_{\text{rms}}\) is the root mean square current and \(R\) is the resistance.
Alternatively, in terms of peak current amplitude (\(I_0\)), where \(I_{\text{rms}} = \frac{I_0}{\sqrt{2}}\): \[ P = \left(\frac{I_0}{\sqrt{2}}\right)^2 R = \frac{I_0^2 R}{2} \] Step 2: Define maximum power dissipated.
In a series LCR circuit, maximum power dissipation occurs at resonance. At resonance, the impedance is minimum and equal to the resistance (\(Z = R\)), leading to maximum current amplitude. Let \(I_{0,\text{max}}\) be the maximum current amplitude at resonance. The maximum power dissipated (\(P_{\text{max}}\)) is: \[ P_{\text{max}} = \frac{I_{0,\text{max}}^2 R}{2} \] Step 3: Set up the condition for half maximum power and solve for current amplitude.
We are given that the power dissipated \(P\) is half of the maximum power dissipated \(P_{\text{max}}\): \[ P = \frac{1}{2} P_{\text{max}} \] Substitute the expressions for \(P\) and \(P_{\text{max}}\): \[ \frac{I_0^2 R}{2} = \frac{1}{2} \left(\frac{I_{0,\text{max}}^2 R}{2}\right) \] \[ \frac{I_0^2 R}{2} = \frac{I_{0,\text{max}}^2 R}{4} \] Cancel \(R\) from both sides and multiply by 2: \[ I_0^2 = \frac{I_{0,\text{max}}^2}{2} \] Take the square root of both sides to find \(I_0\): \[ I_0 = \sqrt{\frac{I_{0,\text{max}}^2}{2}} \] \[ I_0 = \frac{I_{0,\text{max}}}{\sqrt{2}} \] \[ I_0 = \frac{1}{\sqrt{2}} \times I_{0,\text{max}} \] This means the current amplitude at half maximum power is \(\frac{1}{\sqrt{2}}\) times its maximum value. 
Final Answer: \( \boxed{\frac{1}{\sqrt{2}} \times \text{its maximum value}} \).