An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ‘R’ are connected in series to an ac source of potential difference ‘V’ volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10 2 A. The impedance of the circuit is :

Correct Answer: 5 Ω

To find the impedance of the series LCR circuit, we can follow these steps:

Step 1: Calculate the total rms potential difference (V) of the source
In a series LCR circuit, the potential differences across the resistor ($V_R$), inductor ($V_L$), and capacitor ($V_C$) are not in phase. The net rms voltage $V$ is given by the relation:
$V=\sqrt{V_R^2+{(V_L-V_C)}^2}$
Given parameters in the question:
Potential difference across L, $V_L=40\text{V}$
Potential difference across C, $V_C=10\text{V}$
Potential difference across R, $V_R=40\text{V}$

Substituting these values into the voltage formula:
$V=\sqrt{{40}^2+{(40-10)}^2}$
$V=\sqrt{{40}^2+{30}^2}$
$V=\sqrt{1600+900}$
$V=\sqrt{2500}=50\text{V}$

Step 2: Calculate the rms current ($I_{\text{rms}}$)
The given amplitude (peak value) of the current flowing through the circuit is:
$I_0=10\sqrt{2}\text{A}$
The relationship between peak current and rms current is:
$I_{\text{rms}}=\frac{I_0}{\sqrt{2}}$
Substituting the value of $I_0$:
$I_{\text{rms}}=\frac{10\sqrt{2}}{\sqrt{2}}=10\text{A}$

Step 3: Calculate the impedance (Z) of the circuit
The impedance of the circuit is defined as the ratio of total rms potential difference to rms current:
$Z=\frac{V}{I_{\text{rms}}}$
Substituting the calculated values:
$Z=\frac{50}{10}=5\Omega$

Thus, the impedance of the circuit is 5 Ω.