In a series CR circuit shown in figure, applied voltage is 10, V and voltage across capacitor is fou

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7.Alternating Current

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

A

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

B

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

C

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

D

none

Solution

$V_{R}=\sqrt{V^{2}-V_{C}^{2}}=\sqrt{(10)^{2}-(8)^{2}}=6 V$

$\tan \phi=\frac{X_{C}}{X_{R}}=\frac{V_{C}}{V_{R}}=\frac{8}{6}=\frac{4}{3}$

Standard 12

Physics

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(AIIMS-2019)