Verified Solution • 2025–26 Session moderate

QJEEM21Ph34dc: JEE Main 2021 Physics - Class 12 JEEM21Ph34dc Solution: In the given circuit the AC source has = 100 rad s -1 .…

In the given circuit the AC source has ω\omega = 100 rad s-1. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?

JEE Main 2021 (Online) 26th August Evening Shift Physics - Alternating Current Question 93 English

Multiple Choice Options

  • A 5.9 A
  • B 3.16 A
  • C 0.94 A
  • D 6:00 AM

Step-by-Step Verified Solution

ZC=(1ωC)2+R2{Z_C} = \sqrt {{{\left( {{1 \over {\omega C}}} \right)}^2} + {R^2}}

=(1100×100×10−6)2+1002= \sqrt {{{\left( {{1 \over {100 \times 100 \times {{10}^{ - 6}}}}} \right)}^2} + {{100}^2}}

ZC=(100)2+(100)2{Z_C} = \sqrt {{{(100)}^2} + {{(100)}^2}}

=1002= 100\sqrt 2

ZL=(ωL)2+R2{Z_L} = \sqrt {{{(\omega L)}^2} + {R^2}}

(100×0.5)2+502\sqrt {{{(100 \times 0.5)}^2} + {{50}^2}}

=502= 50\sqrt 2

iC=200ZC=2001002=2{i_C} = {{200} \over {{Z_C}}} = {{200} \over {100\sqrt 2 }} = \sqrt 2

iL=200ZL=200502=22{i_L} = {{200} \over {{Z_L}}} = {{200} \over {50\sqrt 2 }} = 2\sqrt 2

cos⁡ϕ1=100102=12⇒ϕ1=45∘\cos {\phi _1} = {{100} \over {10\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _1} = 45^\circ

cos⁡ϕ2=50502=12⇒ϕ2=45∘\cos {\phi _2} = {{50} \over {50\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _2} = 45^\circ

JEE Main 2021 (Online) 26th August Evening Shift Physics - Alternating Current Question 93 English Explanation

I=IC2+IL2I = \sqrt {I_C^2 + I_L^2}

=2+8= \sqrt {2 + 8}

=10= \sqrt {10}

I = 3.16 A

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