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 :

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 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 : (A) 4 √2 Ω (B) 5 / √2 Ω (C) 4 Ω (D) 5 Ω

Tardigrade Admin · Accepted Answer

Given: VL=40 volt; VR=40 volt; VC=10 volt Now, V RMS =√ V R 2+( V L - V C )2 =√(40)2+(40-10)2=50 V I RMS =( I O /√2)=(10 √2/√2)=10 A ∵ V RMS = I RMS × Z ∴ Z=(V RMS / l RMS )=(50/10)=5 Ω

$42 Ω$

$5/ 2 Ω$

$4 Ω$

$5 Ω$

Given:
$V_{L} = 40$ volt; $V_{R} = 40$ volt; $V_{C} = 10$ volt Now,
$V_{RMS} = V_{R}^{2} + (V_{L} - V_{C})^{2}$
$= (40)^{2} + (40 - 10)^{2} = 50 V$
$I_{RMS} = \frac{I _{O}}{2} = \frac{10 2}{2} = 10 A$
$∵ V_{RMS} = I_{RMS} \times Z$
$∴ Z = \frac{V _{RMS}}{l _{RMS}} = \frac{50}{10} = 5Ω$

42​Ω

5/2​Ω

4Ω

5Ω

Given: VL​=40 volt; VR​=40 volt; VC​=10 volt Now, VRMS​=VR2​+(VL​−VC​)2​ =(40)2+(40−10)2​=50V IRMS​=2​IO​​=2​102​​=10A ∵VRMS​=IRMS​×Z ∴Z=lRMS​VRMS​​=1050​=5Ω

$42 Ω$

$5/ 2 Ω$

$4 Ω$

$5 Ω$

Given:
$V_{L} = 40$ volt; $V_{R} = 40$ volt; $V_{C} = 10$ volt Now,
$V_{RMS} = V_{R}^{2} + (V_{L} - V_{C})^{2}$
$= (40)^{2} + (40 - 10)^{2} = 50 V$
$I_{RMS} = \frac{I _{O}}{2} = \frac{10 2}{2} = 10 A$
$∵ V_{RMS} = I_{RMS} \times Z$
$∴ Z = \frac{V _{RMS}}{l _{RMS}} = \frac{50}{10} = 5Ω$