Question

An inductor of reactance $X_L=4\Omega$ and resistor of resistance $R=3\Omega$ are connected in series with a voltage source of emf $\epsilon =\left(20V\right)\left[\sin \right(100\pi rad/s\left)t\right]$. The current in the circuit at any time t will be?

• A. $I=\left(4A\right)\left[\sin \right(100\pi rad/s)t+{37}^o]$
• B. $I=\left(4A\right)\left[\sin \right(100\pi rad/s)t-{37}^o]$
• C. $I=\left(4A\right)\left[\sin \right(100\pi rad/s)t+{53}^o]$
• D. $I=\left(4A\right)\left[\sin \right(100\pi rad/s)t-{53}^o]$

Answer 1

The correct option is A $I=\left(4A\right)\left[\sin \right(100\pi rad/s)t+{37}^o]$

The inductor and the resistor are in series. Given, $R_L=4\Omega ,R=3\Omega$

Resultant impedance R=$\sqrt{R_L^2+R^2}=\sqrt{4^2+3^2}=5\Omega$

In RL circuit, current due to inductor leads by resistance.

Therefore phase angle, $\varphi ={tan}^{-1}\frac{3}{4}$

= ${37}^0$

Peak voltage, $e_o=20V$

Current, I=$\frac{e_0}{R}=\frac{20}{5}=4A$

Since the current is leading. Therefore value of I is $4A\left[sin\right(100\pi )t+{37}^0]$