Question

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 the figure. The 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 course is

• A. 5 Ω
• B. 4√2 Ω
• C. 5√2 Ω
• D. 4 Ω

Answer 1

The Correct Option is A

To find the impedance of the LCR series circuit, we need to analyze the given information and apply the concepts of AC circuits.

1. The potential differences across the inductor (L), capacitor (C), and resistor (R) are given as 40 V, 10 V, and 40 V respectively.
2. The amplitude of current flowing through the circuit is \(10\sqrt{2} \, \text{A}\).
3. Impedance (\(Z\)) in an LCR circuit is calculated using the formula: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\) where:
   • \(X_L\) is the inductive reactance,
   • \(X_C\) is the capacitive reactance.
4. Using the potential difference and the current, the impedance can also be calculated as: \(Z = \frac{V}{I}\) where \(V\) is the total potential difference and \(I\) is the current.
5. Since the potential difference across R is 40 V, we have: \(R = \frac{V_R}{I} = \frac{40}{10\sqrt{2}} \, \Omega = 2\sqrt{2} \, \Omega\). Thus, the value of \(R\) is used in the impedance calculation.
6. The net voltage across the series combination is given by: \(V = \sqrt{V_R^2 + (V_L - V_C)^2}\) Substituting the values: \(V = \sqrt{40^2 + (40 - 10)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{V}\)
7. Finally, the impedance \(Z\) is calculated as: \(Z = \frac{V}{I} = \frac{50}{10\sqrt{2}} = \frac{50}{10\sqrt{2}} \, \Omega = 5 \, \Omega\)

Thus, the impedance of the circuit is 5 Ω.