First slide

Phasors and Phasor Diagrams

Question

In the circuit shown in figure, the frequency of source voltage is $ω = 2000 rad / s$ . The amplitude of the current will be nearest to

Moderate

A

2 $\sqrt{2}$ A
B

3.3 A
C

$2 / \sqrt{5} A$
D

$\sqrt{5} A$

Solution

$Lω = 5 \times 10^{- 3} \times 2000 = 10 Ω$
$\frac{1}{Cω} = \frac{1}{50 \times 10^{- 6} \times 2000} = \frac{100}{10} Ω = 10 Ω$
Since $Lω = \frac{1}{Cω},$
$∴ Z = R = 0 .10 + 4 + 6 = 10 .1 Ω$
$I_{} = \frac{E}{Z} = \frac{20}{10 .1} A \approx 2 A \Rightarrow I_{0} = 2 \times \sqrt{2} A = 2 \sqrt{2} A$

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