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 :

The correct answer is 5 Ω.

Based on the text and the attached image, we have a series LCR circuit containing an inductor (L), a capacitor (C), and a resistor (R) connected in series with an AC voltage source. The given values across the components are:

- Potential difference across the inductor, $V_L=40\text{V}$
- Potential difference across the capacitor, $V_C=10\text{V}$
- Potential difference across the resistor, $V_R=40\text{V}$

In a series LCR circuit, the net RMS voltage ($V_{rms}$) is determined by calculating the phasor sum of the individual voltages across the resistor, inductor, and capacitor. The formula is:

$V_{rms}=\sqrt{V_R^2+(V_L-V_C{)}^2}$

Substitute the given potential differences into the formula:

$V_{rms}=\sqrt{(40{)}^2+(40-10{)}^2}$

$V_{rms}=\sqrt{1600+(30{)}^2}$

$V_{rms}=\sqrt{1600+900}=\sqrt{2500}=50\text{V}$

The problem states that the amplitude (peak value) of the current flowing through the circuit is $I_0=10\sqrt{2}\text{A}$. To find the impedance, we first need to determine the RMS current ($I_{rms}$), which is related to the peak current by:

$I_{rms}=\frac{I_0}{\sqrt{2}}$

$I_{rms}=\frac{10\sqrt{2}}{\sqrt{2}}=10\text{A}$

Finally, we calculate the impedance ($Z$) of the circuit using the AC version of Ohm's law:

$Z=\frac{V_{rms}}{I_{rms}}$

$Z=\frac{50}{10}=5\text{Ω}$

Therefore, the total impedance of the LCR series circuit is 5 Ω.