Question

Applied AC voltage is given as, $V=V_0\sin \left(\omega t\right)$. Corresponding to this voltage, match the following two columns.

Column $\text{I}$	Column $\text{II}$
$a$. $I=I_0\sin \left(\omega t\right)$ $b$. $I=-I_0\cos \left(\omega t\right)$ $c$. $I=I_0\sin (\omega t+\pi /6)$	$p$. only $R$ circuit $q$. only $L$ circuit $r$. may be $RC$ circuit

• A. $a\to p;b\to q;c\to r$
• B. $a\to q;b\to p;c\to r$
• C. $a\to p;b\to r;c\to q$
• D. $a\to r;b\to q;c\to p$

Answer 1

The correct option is A $a\to p;b\to q;c\to r$

Given:
$V=V_0\sin \left(\omega t\right)$

In pure resistive circuit, current and voltage remains in the phase.

$\Rightarrow I=I_0\sin \left(\omega t\right)$

$\therefore a\to p$

In pure inductive circuit, voltage leads ahead the current by $\pi /2$.

$\Rightarrow I=I_0\sin (\omega t-\pi /2)=-I_0\cos \left(\omega t\right)$

$\therefore b\to q$

In AC RC circuit, current lead the voltage by some phase.

$\Rightarrow I=I_0\sin (\omega t+\varphi )$

$\therefore c\to r$

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