In a series R - C circuit shown in figure, applied voltage is 10, V and voltage across capacitor is

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7.Alternating Current

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In a series $R$ $-$ $C$ circuit shown in figure, the applied voltage is $10\, V$ and the voltage across capacitor is found to be $8 \,V$. Then, the voltage across $R$ and the phase difference between current and the applied voltage will respectively be

A

$6 V, \tan ^{-1}\left(\frac{4}{3}\right)$

B

$3 V, \tan ^{-1}\left(\frac{3}{4}\right)$

C

$6 V, \tan ^{-1}\left(\frac{5}{3}\right)$

D

None of these

(AIIMS-2018)

Solution

We know that, for series $R-C$ circuit

$V^{2}=V_{C}^{2}+V_{R}^{2}$

$100=64+V_{R}^{2}$

$\Rightarrow V_{R}^{2}=36 \Rightarrow V_{R}=\sqrt{36}=6 V$

Also, tan $\phi=\frac{V_{C}}{V_{R}} \Rightarrow \tan \phi=\frac{8}{6}$

$\tan \phi=\frac{4}{3}$

$\therefore \phi=\tan ^{-1}\left(\frac{4}{3}\right)$

Standard 12

Physics

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