Question: 28. 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 \mathrm{L}, \mathrm{C} and R is 40 V , 10 V and 40 V , respectively. The amplitude of current flowing through LCR series circuit is 10\sqrt{2}\,\mathrm{A}. The impedance of the circuit is :
(1) 4\sqrt{2}\,\Omega
(2) \frac{5}{\sqrt{2}}\,\Omega
(3) 4\,\Omega
(4) 5\,\Omega
Answer: Option (4)
Explanation:
In a series LCR circuit, the voltages across the inductor, capacitor and resistor are not in phase with each other.
The voltage across the resistor V_R is in phase with current,
while the voltages across the inductor V_L and capacitor V_C
are 90^\circ out of phase with the current and opposite to each other.
The net reactive voltage is given by
V_X = |V_L - V_C|.
Substituting the given values,
V_X = |40 - 10| = 30\,\text{V}.
The resultant voltage across the circuit is obtained using vector addition:
V = \sqrt{V_R^2 + V_X^2}.
Substituting the values,
V = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\,\text{V}.
The peak current in the circuit is given as I_0 = 10\sqrt{2}\,\text{A}.
The impedance of the circuit is defined as
Z = \frac{V_0}{I_0},
where V_0 is the peak value of the applied voltage.
Here, the given voltage 50\,\text{V} is the peak voltage.
Thus,
Z = \frac{50}{10\sqrt{2}} = \frac{5}{\sqrt{2}} \times \sqrt{2} = 5\,\Omega.
Therefore, the impedance of the circuit is 5\,\Omega,
and the correct answer is Option (4).