Question

In the adjacent circuit, the instantaneous current equation is

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

Answer 1

Answer: (B)

Let $\varphi$ is the phase difference between $V$ and $i$
$\text{tan}\varphi =\frac{X_L}{R}=\frac{\omega L}{R}=\frac{100\times 1}{100}=1$
$\Rightarrow \varphi =\frac{\pi }{4}$
Since this is an $LR$ circuit, $V$ will lead $i$
$\therefore$ The instantaneous current equation will be
$i_{}=i_0\text{sin}\left(100t-\varphi \right)$ where $\varphi =\frac{\pi }{4}$ and $i_0=\frac{V_0}{Z}$
$i_0=\frac{V_0}{\sqrt{R^2+{\left(\omega L\right)}^2}}=\frac{200}{\sqrt{{\left(100\right)}^2+{\left(100\times 1\right)}^2}}=\sqrt{2}$
$\therefore i=\sqrt{2}\text{sin}\left(100t-\frac{\pi }{4}\right)$