Question

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which\[R=3\Omega \],\[L=25.48\,mH\] and. \[C=786\mu F\]
Find:
a) Impedance of the circuit.
b) The phase difference between the voltage across the source and the current.
c) The power factor.

Answer 1

Hint: The impedance of the circuit is the root of the sum of the square of the resistor and the difference between the inductive and the capacitive reactance. The phase difference is the ratio of the difference between the inductive and the capacitive reactance by the resistance. The power factor is the cosine angle of the phase difference.
Formula used:
\[\begin{align}
  & Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}} \\
 & \tan \phi =\dfrac{{{X}_{L}}-{{X}_{C}}}{R} \\
 & PF=\cos \phi \\
\end{align}\]

Complete answer:
From the given information we have the data as follows.
\[\begin{align}
  & V=283\,V \\
 & f=50\,Hz \\
 & R=3\Omega \\
 & L=25.48\,mH=25.48\times {{10}^{-3}}\,H \\
 & C=786\mu F=786\times {{10}^{-6}}F \\
\end{align}\]
The circuit diagram representing the LCR circuit is,

a) Impedance of the circuit.
\[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\]
Firstly, we will compute the inductive reactance and the capacitive reactance.
The inductive reactance is given by the formula as follows.
\[\begin{align}
  & {{X}_{L}}=\omega L \\
 & \Rightarrow {{X}_{L}}=2\pi f\times L \\
 & \Rightarrow {{X}_{L}}=2\times 3.14\times 50\times 25.48\times {{10}^{-3}} \\
 & \therefore {{X}_{L}}=8\Omega \\
\end{align}\]
The capacitive reactance is given by the formula as follows.
\[\begin{align}
  & {{X}_{C}}=\dfrac{1}{\omega C} \\
 & \Rightarrow {{X}_{C}}=\dfrac{1}{2\pi f\times C} \\
 & \Rightarrow {{X}_{C}}=\dfrac{1}{2\times 3.14\times 50\times 796\times {{10}^{-6}}} \\
 & \therefore {{X}_{C}}=4\Omega \\
\end{align}\]
The impedance of the circuit is given by the formula as follows.
\[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\]
Substitute the values in the above formula
\[\begin{align}
  & Z=\sqrt{{{3}^{2}}+{{(8-4)}^{2}}} \\
 & \therefore Z=5\,\Omega \\
\end{align}\]
Therefore, the impedance value of the circuit is \[5\Omega \].
b) The phase difference between the voltage across the source and the current.
\[\tan \phi =\dfrac{{{X}_{L}}-{{X}_{C}}}{R}\]
Substitute the values in the above formula
\[\tan \phi =\dfrac{8-4}{3}\]
Continue further computation.
\[\begin{align}
  & \tan \phi =\dfrac{4}{3} \\
 & \Rightarrow \phi ={{\tan }^{-1}}\dfrac{4}{3} \\
\end{align}\]
Therefore, the value of the phase difference between the voltage across the source and the current is,
\[\phi =53.1{}^\circ \]
c) The power factor.
\[PF=\cos \phi \]
Substitute the values in the above formula
\[\begin{align}
  & PF=\cos 53.1{}^\circ \\
 & \therefore PF=0.6 \\
\end{align}\]
Therefore, the power factor is 0.6.
\[\therefore \] a) Impedance of the circuit is\[5\Omega \]. The phase difference between the voltage across the source and the current is\[53.1{}^\circ \]. The power factor is 0.6.

Note:
The power factor can also be calculated as the resistance value by the impedance value. As the circuit involves all the 3 components, that is, the resistor, inductor and the capacitor, thus, the resistance and both the reactants should be considered.