Question

For the shown circuit, the phase difference between current $I$ and $I_1$ is -



Take ${\tan }^{-1}\left(\frac{3}{4}\right)={37}^{\circ }$ and ${\tan }^{-1}\left(\frac{4}{3}\right)={53}^{\circ }$

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

Answer 1

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


From figure,

$I_2=\frac{V}{X_C}=\frac{V}{3}$

$I_1=\frac{V}{R}=\frac{V}{4}$

$I_2$ is ${90}^{\circ }$ ahead of applied voltage function, and $I_1$ is in phase with applied voltage function.


$\tan \varphi =\frac{V/3}{V/4}=\frac{4}{3}$

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

Hence, option $\left(C\right)$ is the correct answer.