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Question: If resistance of 100 Ω , inductance of 0.5 henry and capacitance of \(10 \times 10^{- 6}F\) are conn...

If resistance of 100 Ω , inductance of 0.5 henry and capacitance of 10×10−6F10 \times 10^{- 6}F are connected in series through 50 Hz ac supply, then impedance is

A

1.876

B

18.76

C

189.72

D

101.3

Answer

189.72

Explanation

Solution

Z=R2+(XL−XC)2Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

=1002+(0.5×100π−110×10−6×100π)2=189.72Ω= \sqrt{100^{2} + \left( 0.5 \times 100\pi - \frac{1}{10 \times 10^{- 6} \times 100\pi} \right)^{2}} = 189.72\Omega