Verified Solution • 2025–26 Session moderate

QJEEM19Ph417e: JEE Main 2019 Physics - Class 12 JEEM19Ph417e Solution: A circuit connected to an ac source of emf e = e 0 sin(100t)…

A circuit connected to an ac source of emf e = e0sin(100t) with t in seconds, gives a phase difference of π\pi /4 between the emf e and current i. Which of the following circuits will exhibit this ?

Multiple Choice Options

  • A RC circuit with R = 1 k Ω\Omega and C = 1μF
  • B RL circuit with R = 1k Ω\Omega and L = 1mH
  • C RC circuit with R = 1k Ω\Omega and C = 10 μF
  • D RL circuit with R = 1 k Ω\Omega and L = 10 mH

Step-by-Step Verified Solution

Given phase difference = π4{\pi \over 4} and ω\omega = 100 rad/s

⇒\Rightarrow Reactance (X) = Resistance (R) Now by checking option.

Option (A)
R = 1000 Ω\Omega and Xc = 110−6×100=104Ω{1 \over {{{10}^{ - 6}} \times 100}} = {10^4}\Omega

Option (B)
R = 103 Ω\Omega and XL = 10−3×100=10−1Ω{10^{ - 3}} \times 100 = 10^{-1} \Omega

Option (C)
R = 103 Ω\Omega and Xc = 110×10−6×100=103Ω{1 \over {{10 \times {10}^{ - 6}} \times 100}} = {10^3}\Omega

Option (D)
R = 103 Ω\Omega and XL = 10×10−3×100=1Ω10 \times {10^{ - 3}} \times 100 = 1\Omega

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