Question:
To have dissipative power in an LCR series circuit to be half
To have dissipative power in an LCR series circuit to be half
Show Hint
Whenever power is proportional to square of current:
\[
P\propto I^2
\]
Half power corresponds to:
\[
I=\frac{I_{\max}}{\sqrt{2}}
\]
This is called half-power condition.
Updated On: Jun 17, 2026
- current amplitude \(=2\times\) maximum current amplitude
- current amplitude \(=\dfrac{\text{Maximum current amplitude}}{2}\)
- current amplitude \(=(\text{Maximum current amplitude})^{1/2}\)
- current amplitude \(=\dfrac{\text{Maximum current amplitude}}{\sqrt{2}}\)
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The Correct Option is D
Solution and Explanation
Concept:
Power dissipated in an AC circuit is proportional to square of current:
\[
P\propto I^2
\]
Step 1: Write relation between powers. Suppose: \[ P_{\max}\propto I_{\max}^2 \] Given: \[ P=\frac{P_{\max}}{2} \] Therefore: \[ I^2=\frac{I_{\max}^2}{2} \]
Step 2: Take square root. \[ I=\frac{I_{\max}}{\sqrt{2}} \] Hence: \[ \boxed{\frac{\text{Maximum current amplitude}}{\sqrt{2}}} \]
Step 1: Write relation between powers. Suppose: \[ P_{\max}\propto I_{\max}^2 \] Given: \[ P=\frac{P_{\max}}{2} \] Therefore: \[ I^2=\frac{I_{\max}^2}{2} \]
Step 2: Take square root. \[ I=\frac{I_{\max}}{\sqrt{2}} \] Hence: \[ \boxed{\frac{\text{Maximum current amplitude}}{\sqrt{2}}} \]
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