Question

In the AC network shown in the figure, the rms current flowing through the inductor and capacitor are $0.6\text{A}$ and $0.8\text{A}$, respectively. Then, the current coming out of the source is -

• A. $1.0\text{A}$
• B. $1.4\text{A}$
• C. $0.2\text{A}$
• D. $1.8\text{A}$

Answer 1

The correct option is C $0.2\text{A}$


Suppose, AC voltage source is $E=E_0\sin \left(\omega t\right)$.

Then,

$I_1=I_{1_0}\sin (\omega t+\pi /2)$, and

$I_2=I_{2_0}\sin (\omega t-\pi /2)$

From figure,

$I=\sqrt{I_1^2+I_2^2+2I_1{I}_2\cos \varphi }$

$\Rightarrow I=\sqrt{{0.8}^2+{0.6}^2+2\left(0.8\right)\left(0.6\right)\cos \varphi }$

$\Rightarrow I=\sqrt{1+0.96\cos \varphi }$

Here, phase difference,

$\varphi =\pi \Rightarrow \cos \varphi =-1$

$\therefore I=0.2\text{A}$

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