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.

Step-by-step Explanation:

From the provided circuit diagram, we can observe an LCR series circuit connected to an AC source of potential difference _V. The potential differences across each component are labeled as:
- 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, $V_R=40\text{V}$

1. Find the RMS potential difference of the AC source:
In a series LCR circuit, the net RMS voltage $V_{\text{rms}}$ is given by the formula:

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

Substituting the given values into the formula:

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

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

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

2. Find the RMS current flowing through the circuit:
We are given the peak current (amplitude of current) flowing through the circuit as $I_0=10\sqrt{2}\text{A}$.
The relation between RMS current $I_{\text{rms}}$ and peak current $I_0$ is:

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

Substituting the value of $I_0$:

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

3. Calculate the impedance of the circuit:
The impedance $Z$ of the series LCR circuit is defined as the ratio of the RMS voltage to the RMS current:

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

Substituting the values of $V_{\text{rms}}$ and $I_{\text{rms}}$:

$Z=\frac{50}{10}=5\text{Ω}$