Question 12: AC fundamentals

Question

In the circuit shown in the figure, the AC source gives a voltage V = 20 cos (2000t) neglecting source resistance, the voltmeter and ammeter reading will be

Accepted Answer

Given, R
1
= 8 $$\Omega$$
R
2
= 2 $$\Omega$$
L = 5 mH
C = 50 $$\mu$$F
and V
0
= 20 cos (2000t)
Impedance, $$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} $$
As, here, $${X_L} = \omega L = 2000 	imes 5 	imes {10^{ - 3}} = 10\Omega $$
Similarly, $${X_L} = {1 \over {\omega C}} = {1 \over {2000 	imes 50 	imes {{10}^{ - 6}}}} = 10\Omega $$
$$\because$$ $${X_L} = {X_C}$$, Hence, $${i_{\max }} = {{{V_0}} \over Z} = {{20} \over {(8 + 2)}} = 2A$$
Hence, $${i_{rms}} = {{{i_{\max }}} \over {\sqrt 2 }} = {2 \over {\sqrt 2 }} = 1.41 \,A$$
and $$V = {R_2}\,{i_{rms}} = 1.41 	imes 2 = 2.82\,V$$

Answer

0 V, 0.47 A

Answer

2.82 V, 1.41 A

Answer

1.41 V, 0.47 A

Answer

1.5 V, 8.37 A

AC fundamentals

In the circuit shown in the figure, the AC source gives a voltage V = 20 cos (2000t) neglecting source resistance, the voltmeter and ammeter reading will be

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