A sinusoidal voltage of peak value 300 V and an angular frequency ω =400 textrad/s is applied to series L-C-R circuit, in which R=3Ω , L=20 mH and C=652μ F. The peak current in the circuit is

Question

A sinusoidal voltage of peak value 300 V and an angular frequency ω =400 textrad/s is applied to series L-C-R circuit, in which R=3Ω , L=20 mH and C=652μ F. The peak current in the circuit is (A)  30√2A  (B)  60 A  (C)  100 A  (D)  60√2 A

Tardigrade Admin · Accepted Answer

The impedance of the circuit is Z=√R2+(XL-CC)2 XL=ω L=400× 20× 10-2=8H XC=(1/ω C)=(1/400× 625× 10-6)=4F X=√(3)2+(8-4)2=5 The peak current in the circuit i=(E/Z)=(300/5)=60A

Q. A sinusoidal voltage of peak value 300 V and an angular frequency $ω = 400 rad/s$ is applied to series L-C-R circuit, in which $R = 3Ω, L = 20 m H$ and $C = 652 μ F .$ The peak current in the circuit is

$302 A$

$60 A$

$100 A$

$602 A$

The impedance of the circuit is
$Z = R^{2} + (X_{L} - C_{C})^{2}$
$X_{L} = ω L = 400 \times 20 \times 10^{- 2} = 8 H$
$X_{C} = \frac{1}{ω C} = \frac{1}{400 \times 625 \times 10 ^{- 6}} = 4 F$
$X = (3)^{2} + (8 - 4)^{2} = 5$
The peak current in the circuit
$i = \frac{E}{Z} = \frac{300}{5} = 60 A$

Q. A sinusoidal voltage of peak value 300 V and an angular frequency ω=400rad/s is applied to series L-C-R circuit, in which R=3Ω,L=20mH and C=652μF. The peak current in the circuit is

302​A

60A

100A

602​A