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PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2024

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An alternating voltage $V (t) = 220 \sin 100 π t$ volt is applied to a purely resistive load of $50 Ω$ . The time taken for the current to rise from half of the peak value to the peak value is:

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Answer:

Solution:

Time taken to change current value from half of the peak value to the peak value is,
$t = \frac{T}{6}$
$\Rightarrow t = \frac{2 π}{6 ω} = \frac{π}{3 ω} = \frac{π}{300 π} = \frac{1}{300} = 3.33 ms$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2024

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