Alternating Current

JEE Main level, single correct, medium

168 of 190

Question 168 of 190

A series LR\mathrm{LR} circuit is connected to an ac source of frequency ω\omega and the inductive reactance is equal to 2R2 \mathrm{R}. A capacitance of capacitive reactance equal to R\mathrm{R} is added in series with L\mathrm{L} and R\mathrm{R}. The ratio of the new power factor to the old one is :

Show answer and solution

Answer: (D) 52\sqrt{\frac{5}{2}}

Power factor (old) { }_{\text {(old) }}
=RR2+XL2=RR2+(2R)2=R5R=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \mathrm{R})^2}}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}
Power factor (new) { }_{\text {(new) }}
=RR2+(XL−XC)2=RR2+(2R−R)2=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(X_L-X_C\right)^2}}=\frac{R}{\sqrt{R^2+(2 R-R)^2}}
=R2R=\frac{R}{\sqrt{2} R}
∴ New power factor  Old power factor =R2RR5R=52\therefore \frac{\text { New power factor }}{\text { Old power factor }}=\frac{\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}}{\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}}=\sqrt{\frac{5}{2}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: AC circuits, phasors and resonance simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Alternating Current (Quality and Power Factor), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.