Question

An $\text{AC}$ source producing emf $E=E_0(\sin 2\omega t+\sin 3\omega t)$ is connected in series with a capacitor and a resistor. The current found in the circuit is $i=i_1\sin (2\omega t+{\varphi }_1)+i_2\sin (3\omega t+{\varphi }_2)$. Then

• A. $i_1=i_2$
• B. $i_1<i_2$
• C. $i_1>i_2$
• D. $i_1$ may be less than, equal to or greater than $i_2$.

Answer 1

The correct option is B $i_1<i_2$

In an $\text{AC}$ circuit, $i_1=\frac{E_0}{Z_1}$ and $i_2=\frac{E_0}{Z_2}$

From the data given in the question,

Here, $Z_1=\sqrt{{\left(R\right)}^2+{\left(\frac{1}{2\omega C}\right)}^2}$

and , $Z_2=\sqrt{{\left(R\right)}^2+{\left(\frac{1}{3\omega C}\right)}^2}$

Clearly, $Z_2<Z_1\Rightarrow i_1<i_2$

Hence, option (b) is the correct answer.