Verified Solution • 2025–26 Session moderate

QJEEM21Phadbb: JEE Main 2021 Physics - Class 12 JEEM21Phadbb Solution: Find the peak current and resonant frequency of the following…

Find the peak current and resonant frequency of the following circuit (as shown in figure).

JEE Main 2021 (Online) 26th February Evening Shift Physics - Alternating Current Question 116 English

Multiple Choice Options

  • A 2 A and 100 Hz
  • B 2 A and 50 Hz
  • C 0.2 A and 100 Hz
  • D 0.2 A and 50 Hz

Step-by-Step Verified Solution

We know, z = (xL−xC)2+R2\sqrt {{{({x_L} - {x_C})}^2} + {R^2}}

xL=ωL=100×100×10−3=10Ω{x_L} = {\omega _L} = 100 \times 100 \times {10^{ - 3}} = 10\Omega

xC=1ωC=1100×100×10−6=10Ω{x_C} = {1 \over {{\omega _C}}} = {1 \over {100 \times 100 \times {{10}^{ - 6}}}} = 10\Omega

∴\therefore z=(10−100)2+R2=902+1202z = \sqrt {{{(10 - 100)}^2} + {R^2}} = \sqrt {{{90}^2} + {{120}^2}}

=30×5=150Ω= 30 \times 5 = 150\Omega

ipeak=Δvz=30150=15amp=0.2{i_{peak}} = {{\Delta v} \over z} = {{30} \over {150}} = {1 \over 5} amp = 0.2 amp

For resonant frequency,

ωL=1ωC⇒ω2=1LC⇒ω=1LC\omega L = {1 \over {\omega C}} \Rightarrow {\omega ^2} = {1 \over {LC}} \Rightarrow \omega = {1 \over {\sqrt {LC} }}

& f=12πLC=12π100×10−3×100×10−6f = {1 \over {2\pi \sqrt {LC} }} = {1 \over {2\pi \sqrt {100 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 6}}} }}

=100102π=100π2π=50= {{100\sqrt {10} } \over {2\pi }} = {{100\pi } \over {2\pi }} = 50 Hz

as 10≈π\sqrt {10} \approx \pi

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