Question

The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is
A. 0.4
B. 0.5
C. 0.8
D. 1.0

Answer 1

Hint: In this solution, the power factor is the ratio of the potential difference of resistance to the square root of sum of potential difference across resistor and potential difference across capacitor and inductor.

Complete step by step solution:
Given:
The L-C-R circuit has the potential differences across the resistance, capacitance and inductance; the given values are as follows.
The potential difference across the resistance is ${V_R} = 80\;{\rm{V}}$.
The potential difference across the capacitance is ${V_c} = 40\;{\rm{V}}$.
The potential difference across the inductance is ${V_L} = 100\;{\rm{V}}$.
The power factor is the ratio dissipated current and the product of the voltage and the current in the circuit, then
The equation of the power factor in the L-C-R circuit is,
${p_f} = \dfrac{{{V_R}}}{{\sqrt {{V_R}^2 + \left( {{V_L}^2 - {V_C}^2} \right)} }}$
Here, ${p_f}$ is the power factor of the L-C-R circuit.
Substitute the values in the above equation.
\[\begin{array}{l}
{p_f} = \dfrac{{{V_R}}}{{\sqrt {{V_R}^2 + \left( {{V_L}^2 - {V_C}^2} \right)} }}\\
{p_f} = \dfrac{{80}}{{\sqrt {{{80}^2} + {{\left( {100 - 40} \right)}^2}} }}\\
 = \dfrac{{80}}{{\sqrt {6400 + 3600} }}\\
 = \dfrac{{80}}{{100}}\\
 = 0.8\,
\end{array}\]

Therefore, the option is (C), the correct answer is 0.8.

Note: Make sure to use a negative sign to the potential difference of the capacitance and if should be subtracted from the Potential difference of the inductance and not with the resistance. Remember the potential difference of resistance always be in the numerator. Due to the ratio of the same units, there will be no units to the power factor.