First slide

Phasors and Phasor Diagrams

Question

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

Moderate

A

$1.876 Ω$
B

$18.76 Ω$
C

$189.72 Ω$
D

$101.3 Ω$

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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