Question

In a series LCR what will be the phase difference between voltage drop across inductor and capacitor.

Answer 1

Step 1: Establish the Equation

Here, we have an alternating current.

Therefore we have emf as $v=V_0\cos \omega t$ is applied to a circuit containing an inductance $L$, a capacitance $C$ and a resistance $R$ in series.

if $i$ is the instantaneous current then the instantaneous emf equation can be given as,

$Ri+L\frac{di}{dt}+\frac{1}{C}\int idt=v$

since, $v_R=Ri$, $v_L=L\frac{di}{dt}$ and $v_C=\frac{1}{C}\int idt$

Step 2: Derive the Impedance

$R\overrightarrow{I}+j\omega L\overrightarrow{I}-\frac{j}{\omega C}\overrightarrow{I}=\overrightarrow{V}$ $\left(1\right)$

where $j$ is a complex quantity.

$\Rightarrow \overrightarrow{I}=\frac{\overrightarrow{V}}{R+j\left(\omega L-{\displaystyle \frac{1}{\omega C}}\right)}=\frac{\overrightarrow{V}}{Z}$

where $Z$ is the impedance.

$\therefore Z=R+j\left(\omega L-\frac{1}{\omega C}\right)$

Step 3: Derive the voltages

Equation $\left(1\right)$ can also be written as,

$\overrightarrow{V}={\overrightarrow{V}}_R+\overrightarrow{V_L}+\overrightarrow{V_C}$

where ${\overrightarrow{V}}_R=\overrightarrow{I}R,{\overrightarrow{V}}_L=j\omega L\overrightarrow{I}$ and ${\overrightarrow{V}}_C=-\frac{j}{\omega C}\overrightarrow{I}$ are the phasors for the voltages across $R,L$ and $C$ respectively.

We know that ${\overrightarrow{V}}_R$ is always in phase with the current phasor $\overrightarrow{I}$, ${\overrightarrow{V}}_L$ leads over $\overrightarrow{I}$ by $90°$ and ${\overrightarrow{V}}_C$ lags behind $\overrightarrow{I}$ by $90°$.

The sum of ${\overrightarrow{V}}_R$, ${\overrightarrow{V}}_L$ and ${\overrightarrow{V}}_C$ gives the phasor $\overrightarrow{V}$ for the applied emf.

Step 4: Calculate the phase difference

Let us represent it by a vector diagram which is known as a phasor diagram.

Hence, from this diagram, we can see that the phase difference between voltage drop across inductor and capacitor is $180°$