A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R=3 Ω, L=25.48 mH , C=796 μ F. The impedance of the circuit is

Question

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R=3 Ω, L=25.48 mH , C=796 μ F. The impedance of the circuit is (A) 3 Ω (B) 4 Ω (C) 5 Ω (D) 6 Ω

Tardigrade Admin · Accepted Answer

XL=2 π f L=(2 π)(50)(25.48 × 10-3)=8 Ω XC=(1/2 π f ⋅ C)=(1/2 π(50)(796 × 10-6))=4 Ω Z=√R2+(XL-XC)2=√32+(8-4)2=5 Ω

Q. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which $R = 3 Ω, L = 25.48 m H, C = 796 μ F$ . The impedance of the circuit is

$3 Ω$

$4 Ω$

$5 Ω$

$6 Ω$

$X_{L} = 2 π f L = (2 π) (50) (25.48 \times 1 0^{- 3}) = 8 Ω$
$X_{C} = \frac{1}{2 π f \cdot C} = \frac{1}{2 π ( 50 ) ( 796 \times 1 0 ^{- 6} )} = 4 Ω$
$Z = R^{2} + (X_{L} - X_{C})^{2} = 3^{2} + (8 - 4)^{2} = 5 Ω$

Q. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R=3Ω,L=25.48mH,C=796μF. The impedance of the circuit is

3Ω

4Ω

5Ω

6Ω