Alternating Current

JEE Main level, single correct, medium

162 of 190

Question 162 of 190

A series LRLR circuit connected with an ac source E=(25sin⁡1000t)VE=(25\sin 1000t)V has a power factor of 12\dfrac{1}{\sqrt{2}}. If the source of emf is changed to E=(20sin⁡2000t)VE=(20\sin 2000t)V, the new power factor of the circuit will be :

Show answer and solution

Answer: (C) 15\dfrac{1}{\sqrt{5}}

For LRLR circuit, cos⁡ϕ=R(XL)2+R2=12\cos \phi =\dfrac{R}{\sqrt{{\left({X}_{L}\right)}^{2}+{R}^{2}}}=\dfrac{1}{\sqrt{2}}
⇒R=XL\Rightarrow R={X}_{L}
Now, ω′=2ω{\omega }^{'}=2\omega
Therefore, (XL)′=ω′L=2ωL=2XL{\left({X}_{L}\right)}^{'}={\omega }^{'}L=2\omega L=2{X}_{L}
Hence, cos⁡ϕ′=R(2XL)2+R2=15\cos {\phi }^{'}=\dfrac{R}{\sqrt{{\left(2{X}_{L}\right)}^{2}+{R}^{2}}}=\dfrac{1}{\sqrt{5}}

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.