If resistance of 100 Ω, inductance of 0.5 H and capacitance of 10×10−6 F are connected in series through 50 Hz AC supply, then impedance is

Z=R2+XL−XC2=1002+0.5×100π−110×10−6×100π2=189.72 Ω

If resistance of 100 Ω, inductance of 0.5 H and capacitance of 10×10−6 F are connected in series through 50 Hz AC supply, then impedance is

Z=R2+XL−XC2=1002+0.5×100π−110×10−6×100π2=189.72 Ω

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If resistance of $100 Ω,$ inductance of 0.5 H and capacitance of $10 \times 10^{- 6} F$ are connected in series through 50 Hz AC supply, then impedance is

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$1.876 Ω$

$101.3 Ω$

$18.76 Ω$

$189.72 Ω$

answer is C.

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Detailed Solution

$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$

$= \sqrt{100^{2} + {(0 .5 \times 100 π - \frac{1}{10 \times 10^{- 6} \times 100 π})}^{2}}$

$= 189 .72 Ω$

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If resistance of 100 Ω, inductance of 0.5 H and capacitance of 10×10−6 F are connected in series through 50 Hz AC supply, then impedance is

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If resistance of $100 Ω,$ inductance of 0.5 H and capacitance of $10 \times 10^{- 6} F$ are connected in series through 50 Hz AC supply, then impedance is