Alternating Current

JEE Main level, numerical answer, medium

113 of 190

Question 113 of 190

In an LCRLCR series circuit, an inductor 30mH30mH and a resistor 1Ω1\Omega are connected to an ACAC source of angular frequency 300rads−1.300rad{s}^{-1}. The value of capacitance for which the current leads the voltage by 45∘45^{\circ} is 1x×10−3F\dfrac{1}{x}\times {10}^{-3}F. Then the value of xx is

Show answer and solution

Answer: 3

tan⁡ϕ=xC−xLR\tan \phi =\dfrac{{x}_{C}-{x}_{L}}{R}
tan⁡45=xC−xLR\tan 45=\dfrac{{x}_{C}-{x}_{L}}{R}
xC−xL=R{x}_{C}-{x}_{L}=R
1ωC−ωL=R\dfrac{1}{\omega C}-\omega L=R
1ωC−300×0.03=1\dfrac{1}{\omega C}-300\times 0.03=1
1ωC=10\dfrac{1}{\omega C}=10
C=110ω=110×300C=\dfrac{1}{10\omega }=\dfrac{1}{10\times 300}
C=13×10−3C=\dfrac{1}{3}\times {10}^{-3}
X=3X=3

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 (AC Circuits), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.