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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:
1024310_902682918e374a9a831d8a79cdcf815f.png


A
0 V,0.47 A
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B
1.68 V,0.47 A
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C
0 V,1.4 A
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D
5.6 V,1.4 A
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Solution

The correct option is D $$5.6\ V, 1.4\ A$$
We know
$$Z=\sqrt{(R)^2+(X_L-X_C)^2}$$
Here $$R=10\Omega,\ X_L=\omega L=2000\times5\times10^{-3}=10\ \Omega$$
$$X_C=\dfrac{1}{\omega C}=\dfrac{1}{2000\times50\times10^{-6}}=10\ \Omega$$
So,
$$Z=\sqrt{(10)^2}$$
$$Z=10\ \Omega$$
Maximum current $$i_)=\dfrac{V_0}{Z}=\dfrac{20}{10}=2\ A$$
Hence,
$$I_{rms}=\dfrac{2}{\sqrt2}=1.4\ A$$
$$V_{rms}=4\times1.4=5.6\ V$$

Physics
NCERT
Standard XII

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