MHT CET · Physics · Alternating Current
In a series LR circuit, \(X_L=R\), power factor is \(P_1\). If a capacitor of capacitance \(C\) with \(X_C=X_L\) is added to the circuit the power factor becomes \(P_2\). The ratio of \(P_1\) to \(P_2\) will be
- A \(1: 3\)
- B \(1: \sqrt{2}\)
- C \(1: 1\)
- D \(1: 2\)
Answer & Solution
Correct Answer
(B) \(1: \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \text { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}} \\ & \text { Given: } \mathrm{X}_{\mathrm{L}}=\mathrm{R} \\ \therefore \quad & \mathrm{P}_1=\frac{\mathrm{R}}{\sqrt{2 \mathrm{R}^2}}=\frac{1}{\sqrt{2}}\end{array}\)
When \(\mathrm{C}\) is connected, the circuit becomes a series LCR circuit.
\(\begin{array}{ll}
\therefore & \text { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}} \\
& \text { Given: } \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}} \\
\therefore & \mathrm{P}_2=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2}}=1 \\
\therefore & \mathrm{P}_1: \mathrm{P}_2=1: \sqrt{2}
\end{array}\)
When \(\mathrm{C}\) is connected, the circuit becomes a series LCR circuit.
\(\begin{array}{ll}
\therefore & \text { Power factor }=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}} \\
& \text { Given: } \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}} \\
\therefore & \mathrm{P}_2=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2}}=1 \\
\therefore & \mathrm{P}_1: \mathrm{P}_2=1: \sqrt{2}
\end{array}\)
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