In a series L-C circuit, if L=10-3 H and C=3 × 10-7 F is connected to a 100 V -50 Hz a.c. source, the impedance of the circuit is

Question

In a series L-C circuit, if L=10-3 H and C=3 × 10-7 F is connected to a 100 V -50 Hz a.c. source, the impedance of the circuit is (A) (105/3 π)-10 π (B) 0.1 π-3 × 10-5 π (C) (105/3 π)-(π/10) (D) None of these

Tardigrade Admin · Accepted Answer

L=10-3 H and C=3 × 10-7 F f=50 Hz , V =100 V Z=√(XL-xc)2 =(XL-XC) =ω L-(1/ω C) Putting ω, L C Z=((105/3 π)-(π/10))

Q. In a series $L - C$ circuit, if $L = 1 0^{- 3} H$ and $C = 3 \times 1 0^{- 7} F$ is connected to a $100 V - 50 Hz$ a.c. source, the impedance of the circuit is

$\frac{1 0 ^{5}}{3 π} - 10 π$

$0.1 π - 3 \times 1 0^{- 5} π$

$\frac{1 0 ^{5}}{3 π} - \frac{π}{10}$

None of these

$L = 1 0^{- 3} H$ and $C = 3 \times 1 0^{- 7} F$
$f = 50 Hz, V = 100 V$
$Z = (X_{L} - x_{c})^{2}$
$= (X_{L} - X_{C})$
$= ω L - \frac{1}{ω C}$
Putting $ω, L & C$
$Z = (\frac{1 0 ^{5}}{3 π} - \frac{π}{10})$

Q. In a series L−C circuit, if L=10−3H and C=3×10−7F is connected to a 100V−50Hz a.c. source, the impedance of the circuit is

3π105​−10π

0.1π−3×10−5π

3π105​−10π​