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 : [figure]
- 5
- 4 2
- 5/ 2
- 4
Answer
(1) 5
Solution
The voltages across L, C and R are rms values (as read by voltmeters).
V = VR2 + (VL - VC)2
= 402 + (40 - 10)2
= 1600 + 900 = 50 V
rms current: Irms = I0 2 = 10 2 2 = 10 A
∴ Z = VIrms = 5010 = 5
Answer: (1)