Series LR Circuit - Practice Questions & MCQ

Series LR circuit-

The above figure shows that pure inductor of inductance $L$ connected in series with a resistor of resistance $R$ through sinusoidal voltage, which is given by - $\mathrm{V}=\mathrm{V}_0 \sin (\omega \mathrm{t}+\varphi)$.

The alternating current I , which is flowing in the circuit gives rise to voltage drop $\mathrm{V}_{\mathrm{R}}$ across the resistor and voltage drop $\mathrm{V}_{\mathrm{L}}$ across the coil. As we have studied in previous concept that the voltage drop $\mathrm{V}_{\mathrm{R}}$ across R would be in phase with current but voltage drop across the inductor will lead the current by a phase factor $\pi / 2$.

So, the voltage drop across $R$ is $-V_R=I R$
voltage drop across the inductor L is - $\mathrm{V}_{\mathrm{L}}=\mathrm{I}(\omega \mathrm{L})$
Where, I is the value of current in the circuit at a given instant of time
So, the voltage phasor diagram is -

In the above figure, we have taken current as a reference quantity because same amount of current flows through both the components. Thus from phasor diagram - 

$
\begin{aligned}
V & =\sqrt{V_R^2+V_L^2} \\
& =I \sqrt{R^2+\omega^2 L^2} \\
& =I Z
\end{aligned}
$

where,

$
Z=\left(R^2+\omega^2 L^2\right)^{1 / 2}
$

Here, Z is known as Impedence of the circuit.

By using the formula of impedence we can write that -

$
I=\frac{V_0 \sin (\omega t-\phi)}{Z}
$

This is current in steady state which lags behind applied voltage by an angle $\varphi$.
From here and the above figure, we can write that -

$
\tan \varphi=\frac{\omega L}{R}=\frac{X_L}{R}
$
 

Important term - 

1. Power factor - 

$
\begin{aligned}
& \cos \phi=\frac{R}{Z} \\
& R \rightarrow \text { resistance } \\
& Z \rightarrow \text { impedence }
\end{aligned}
$

2. Inductive susceptance ( $S_L$ ) -

It is the reciprocal of reactance.

$
S_L=\frac{1}{X_L}=\frac{1}{2 \pi \nu L}
$