A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $I_0$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

Given Data:

• A series LCR circuit is connected to an AC source.
• The current amplitude at resonance is $I_0$.
• The resistance $R$ is doubled.
• We need to determine the new current amplitude at resonance.

Step 1: Understanding Resonance in LCR Circuit

At resonance, the inductive reactance ($X_L$) and capacitive reactance ($X_C$) cancel each other:

$$X_L = X_C$$

Thus, the total impedance ($Z$) of the circuit is purely resistive, i.e.,

$$Z = R$$

The current amplitude at resonance is given by Ohm’s law:

$$I_0 = \frac{E_0}{R}$$

where:

• $E_0$ = Peak voltage of the AC source
• $R$ = Resistance of the circuit

Step 2: Effect of Doubling Resistance

When the resistance $R$ is doubled (i.e., $R' = 2R$), the new current amplitude at resonance becomes:

$$I' = \frac{E_0}{R'}$$

 $$I' = \frac{E_0}{2R}$$

Using the original equation for $I_0$,

$$I' = \frac{I_0}{2}$$