A sinusoidaI 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 = 3 Ω, L = 20 mH and C = 625 μF. The peak current in the circuit is

Question

A sinusoidaI 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 = 3 Ω, L = 20 mH and C = 625 μF. The peak current in the circuit is (A) 30 √ 2 A (B) 60 A (C) 100 A (D) 60 √ 2 A

Tardigrade Admin · Accepted Answer

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

Q. A sinusoidaI 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 = 3 $Ω$ , L = 20 mH and C = 625 $μ$ F. The peak current in the circuit is

30 $2$ A

60 A

100 A

60 $2$ A

The impedance of the circuit is
Z = $R^{2} + (X_{L} - X_{C})^{2}$
$X_{L} = ω L = 400 \times 20 \times 1 0^{- 3} = 8$ H
$X_{C} = \frac{1}{ω C} = \frac{1}{400 \times 625 \times 1 0 ^{- 6}} = 4 F$
Z = $(3)^{2} + (8 - 4)^{2} = 5$
i = $\frac{E}{Z} = \frac{300}{5} = 60$ A

Q. A sinusoidaI 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 = 3 Ω, L = 20 mH and C = 625 μF. The peak current in the circuit is

30 2​ A