Question

In the adjacent circuit, the instantaneous current equation is

• A. $2sin \left(100 t - \frac{\pi }{4}\right)$
• B. $\sqrt{2}sin \left(100 t - \frac{\pi }{4}\right)$
• C. $\sqrt{2}sin \left(200 t - \frac{\pi }{4}\right)$
• D. $\sqrt{2}\left(100 t + \frac{\pi }{4}\right)$

Answer 1

Answer: (B)

Let $\phi$ is the phase difference between $V$ and $i$
$\text{tan} \phi=\frac{X_{L}}{R}=\frac{\omega L}{R}=\frac{100 \times 1}{100}=1$
$\Rightarrow \phi=\frac{\pi }{4}$
Since this is an $LR$ circuit, $V$ will lead $i$
$\therefore $ The instantaneous current equation will be
$i_{}=i_{0 }\text{sin} \left(100 \, t - \phi\right)$ where $\phi=\frac{\pi }{4}$ and $i_{0}=\frac{V_{0}}{Z}$
$i_{0}=\frac{V_{0}}{\sqrt{R^{2} + \left(\omega L\right)^{2}}}=\frac{200}{\sqrt{\left(100\right)^{2} + \left(100 \times 1\right)^{2}}}=\sqrt{2}$
$\therefore i=\sqrt{2}\text{sin} \left(100 \, t - \frac{\pi }{4}\right)$