Alternating Current

JEE Main level, numerical answer, medium

165 of 190

Question 165 of 190

A series LCR circuit is connected to an AC source of 220V220V, 50Hz50Hz. The circuit contains a resistance R=80ΩR=80\Omega, an inductor of inductive reactance XL=70Ω{X}_{L}=70\Omega, and a capacitor of capacitive reactance XC=130Ω{X}_{C}=130\Omega. The power factor of circuit is x10\dfrac{x}{10}. The value of xx is:

Show answer and solution

Answer: 8

Power factor in LCR circuit is given by,
cos⁡ϕ=RZ\cos \phi =\dfrac{R}{Z}, where, ZZ is the total impedance of the circuit.
So, cos⁡ϕ=RR2+(XC−XL)2\cos \phi =\dfrac{R}{\sqrt{{R}^{2}+{\left({X}_{C}-{X}_{L}\right)}^{2}}}
Putting the values, we have
⇒cos⁡ϕ=80(80)2+(60)2\Rightarrow \cos \phi =\dfrac{80}{\sqrt{(80{)}^{2}+(60{)}^{2}}}
⇒cos⁡ϕ=80100⇒810\Rightarrow \cos \phi =\dfrac{80}{100}\Rightarrow \dfrac{8}{10}
Thus, the value of x=8x=8.

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.