A series LCR circuit is connected to an alternating source of emf \( E \). The current amplitude at resonance frequency is \( I_0 \). If the value of resistance \( R \) becomes twice of its initial value, then amplitude of current at resonance will be:
Show Hint
- \( \frac{I_0}{2} \)
- \( 2I_0 \)
- \( I_0 \)
- \( \frac{I_0}{\sqrt{2}} \)
The Correct Option is A
Approach Solution - 1
Approach Solution -2
Step 1: Recall the formula for current amplitude in a series LCR circuit.
The current amplitude is given by: \[ I = \frac{E}{Z}, \] where \( Z \) is the impedance of the circuit: \[ Z = \sqrt{R^2 + (X_L - X_C)^2}. \]
Step 2: At resonance condition.
At resonance, \( X_L = X_C \). Thus, impedance becomes minimum: \[ Z = R. \] Therefore, current amplitude at resonance is: \[ I_0 = \frac{E}{R}. \]
Step 3: When resistance is doubled.
If resistance becomes \( 2R \), then the new impedance at resonance is: \[ Z' = 2R. \] Hence, the new current amplitude: \[ I' = \frac{E}{Z'} = \frac{E}{2R}. \]
Step 4: Relate new current with the original.
\[ I_0 = \frac{E}{R} \quad \Rightarrow \quad I' = \frac{I_0}{2}. \]
Final Answer:
\[ \boxed{I' = \dfrac{I_0}{2}} \]
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