A capacitor of 10μF and an inductor of 1H are joined in series. An AC of 50Hz is applied to this combination. What is the impedance of the combination?
A
5(π2−5)πΩ
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B
100(10−π2)πΩ
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C
10(π2−5)πΩ
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D
10(10−π2)πΩ
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Solution
The correct option is B100(10−π2)πΩ From given,
Z=|XL−Xc| XL=ωL=2π×f×L=2π×50×1 XL=100πΩ and XC=1ωC=12π×50×10×10−6=1000πΩ Now XC=1000πΩ Z=|XL−XC| Z=∣∣100π−1000π∣∣Ω =100(10−π2)πΩ