Physics · 2023
JEE Main · 6 April 2023, Shift 2 · Q47
A capacitor of capacitance 150.0 μ F is connected to an alternating source of emf given by E =36 sin (120 π t ) V. The maximum value of current in…
A capacitor of capacitance $\displaystyle 150.0 \mu \mathrm{~F}$ is connected to an alternating source of emf given by $\displaystyle \mathrm{E}=36 \sin (120 \pi \mathrm{t}) \mathrm{V}$. The maximum value of current in the circuit is approximately equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 2 A$
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JEE Main 2023 Physics question, with the answer from NTA’s final answer key. Where our answers come from.