Question

A capacitor and resistor are connected with an $AC$ source as shown in the figure. Reactance of capacitor is $X_c=3\Omega$ and resistance of resistor is $4\Omega$. Phase difference between current $I$ and $I_1$ is $(\text{Take}{\tan }^{-1}\frac{4}{3}={53}^{\circ })$

• A. ${90}^{\circ }$
• B. $0^{\circ }$
• C. ${53}^{\circ }$
• D. ${37}^{\circ }$

Answer 1

The correct option is C ${53}^{\circ }$

Here,

$I\to$current from source

and, $I_1\to$current in resistance

$RC$ is in parallel connection with respect to source and for parallel connection, voltage across both will be same as source $V=V_0\sin \omega t$

But, for current we make phasor taking into account that for capacitor, current leads voltage by $\frac{\pi }{2}$ whereas for resistance, current and voltage are in same phase.


So, $I_c=\frac{V_0}{X_c}$

and, $I_R=I_1=\frac{V_0}{R}$

$\therefore \tan \varphi =\frac{I_C}{I_1}=\frac{R}{X_c}=\frac{4}{3}$

$\Rightarrow \varphi ={\tan }^{-1}\frac{4}{3}={53}^{\circ }$

Hence, option $\left(c\right)$ is correct.