Question 16: AC fundamentals

Question

In the circuit shown in figure, the AC source has angular frequency $\omega$ = 2000 rad s
$-$1
. The amplitude of the current will be nearest to

Accepted Answer

Given, $\omega$ = 2000 rad s
$-$1
L = 5 mH = 5 $	imes$ 10
$-$3
H
C = 50 $\mu$F = 50 $	imes$ 10
$-$6
F
V = 30 V
Inductive reactance,
X
L
= $\omega$L = 2000 $	imes$ 5 $	imes$ 10
$-$3
= 10 $\Omega$
and capacitive reactance,
${X_C} = {1 \over {\omega C}} = {1 \over {50 	imes {{10}^{ - 6}} 	imes 2000}} = {{100} \over {10}} = 10\Omega $
$\Rightarrow$ X
L
= X
C
The given circuit is resonant circuit.
Hence, impedance of circuit, Z = R = 4 + 6 + 0.5 = 10.5$\Omega$
Amplitude of current, ${I_0} = {V \over Z} = {{30} \over {10.5}} = 2.85$ A

Answer

2.85 A

Answer

3 A

Answer

0.5 A

Answer

3.625 A

AC fundamentals

In the circuit shown in figure, the AC source has angular frequency \omega = 2000 rad s -1 . The amplitude of the current will be nearest to

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