Question

To have dissipative power in an LCR series circuit to be half

• A. current amplitude \(=2\times\) maximum current amplitude
• B. current amplitude \(=\dfrac{\text{Maximum current amplitude}}{2}\)
• C. current amplitude \(=(\text{Maximum current amplitude})^{1/2}\)
• D. current amplitude \(=\dfrac{\text{Maximum current amplitude}}{\sqrt{2}}\)

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.

Answer 1

The Correct Option is D

Step 1: Link power to current.
In an AC circuit the power used up depends on the square of the current. \[ P \propto I^2 \] So if the current changes, the power changes as its square.

Step 2: Write the maximum power.
At resonance the current is largest, giving the most power. \[ P_{max} \propto I_{max}^2 \]
Step 3: Use the half power condition.
We want the power to drop to half of the maximum. \[ P = \frac{P_{max}}{2} \]
Step 4: Turn this into a current relation.
Since power follows current squared, \[ I^2 = \frac{I_{max}^2}{2} \]
Step 5: Take the square root.
\[ I = \frac{I_{max}}{\sqrt2} \]
Step 6: State the result.
So the current amplitude must be the maximum divided by root two. \[ \boxed{I = \dfrac{I_{max}}{\sqrt2}} \]