Question 4: AC fundamentals

Question

In a series R-C 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
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m3o2m625/3fab84de-c9ec-4e4a-8e29-61a3e7e3bc6e/180fbdd0-a63f-11ef-a300-b723d4598e73/file-jaoe38c1m3o2m626.png?format=png)

Accepted Answer

We know that, for series R-C circuit
$V^2=V_C^2+V_R^2$
$\begin{aligned}
100 & =64+V_R^2 \\
\Rightarrow & V_R^2=36 \Rightarrow V_R=\sqrt{36}=6 \mathrm{~V}
\end{aligned}$
$\begin{aligned}
& 	ext { Also, } 	an \phi=\frac{V_C}{V_R} \Rightarrow 	an \phi=\frac{8}{6} \\
& 	an \phi=\frac{4}{3} \\
& 	herefore \phi=	an ^{-1}\left(\frac{4}{3}ight)
\end{aligned}$
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m3o2nyqa/fae1ba1b-01bd-4663-b8d2-bff41f618612/4a049720-a63f-11ef-a300-b723d4598e73/file-jaoe38c1m3o2nyqb.png?format=png)

Answer

$6 \mathrm{~V}, 	an ^{-1}\left(\frac{4}{3}ight)$

Answer

$3 V, 	an ^{-1}\left(\frac{3}{4}ight)$

Answer

$6 \mathrm{~V}, 	an ^{-1}\left(\frac{5}{3}ight)$

Answer

None of these

AC fundamentals

In a series R-C 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

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