Q: A capacitor and resistor are connected with an AC source as shown in figure. Reactance of capacitor is X_{C }= 3Ω and resistance of resistor is 4Ω . Phase difference between current I and I_{1} is [ tan^{-1} (3/4)= 37° ]

(a) 90°

(b) zero

(c) 53°

(d) 37°

Ans: (c)

Sol: $ \displaystyle I_1 = \frac{V}{R} $

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

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

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

As I_{1} is in Phase with applied voltage but I_{2} is lead by 90° .

$\displaystyle tan\phi = \frac{V/3}{V/4} $

$ \displaystyle tan\phi = \frac{4}{3} $

$ \displaystyle \phi = tan^{-1}(\frac{4}{3}) $

φ = 53°