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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2024 · 8 April · Evening Shift · Physics · Alternating Current

An alternating emf E=110 2 100 t volt is applied to a capacitor of...

An alternating emf E=1102sin⁡100t\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t} volt is applied to a capacitor of 2μF2 \mu \mathrm{F}, the rms value of current in the circuit is _____mA\mathrm{mA},

Correct answer

0

Step-by-step solution

C=2μf;E=1102sin⁡(100t)XC=1ωc=1100×2×10−6=100002=5000Ωio=11025000irms=110250002=1105 mA=22 mA\begin{aligned} C & =2 \mu f ; \quad E=110 \sqrt{2} \sin (100 \mathrm{t}) \\ \mathrm{X}_{\mathrm{C}} & =\frac{1}{\omega \mathrm{c}}=\frac{1}{100 \times 2 \times 10^{-6}} \\ = & \frac{10000}{2}=5000 \Omega \\ \mathrm{i}_{\mathrm{o}} & =\frac{110 \sqrt{2}}{5000} \\ \mathrm{i}_{\mathrm{rms}} & =\frac{110 \sqrt{2}}{5000 \sqrt{2}} \\ & =\frac{110}{5} \mathrm{~mA} \\ & =22 \mathrm{~mA}\end{aligned}
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