In a series $C R$ 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

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

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

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

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Zero, Zero

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Solution

The correct option is A $6 V, {tan}^{- 1} (\frac{4}{3})$
In $C R$ circuit
$V_{R} = \sqrt{V^{2} - V_{C}^{2}}$
$V_{R} = \sqrt{10^{2} - 8^{2}} = 6 V$

Phase difference
$tan ϕ = \frac{V_{C}}{V_{R}} = \frac{4}{3}$

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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 8 V. Then the voltage across R, and the phase difference between current and the applied voltage will respectively be

The correct option is A 6 V, tan−1(43) In CR circuit VR=√V2−V2C VR=√102−82=6 V Phase difference tan ϕ=VCVR=43

In a series $C R$ 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

The correct option is A $6 V, {tan}^{- 1} (\frac{4}{3})$
In $C R$ circuit
$V_{R} = \sqrt{V^{2} - V_{C}^{2}}$
$V_{R} = \sqrt{10^{2} - 8^{2}} = 6 V$

Phase difference
$tan ϕ = \frac{V_{C}}{V_{R}} = \frac{4}{3}$