Question

A 5.89 µF capacitor is connected to an emf given by $v_0 \sin(\omega t)$ where $V_0 = 120$ Volts and $\omega$ is $120 \pi$ rad/s. What is the magnitude of the current in Amperes through the capacitor at t = 6.98 seconds?

Name: A 5.89 µF capacitor is connected to an emf given by v0 sin(ω t) where V0 = 120 Volts and ω is 120 π rad/s. What is the magnitude of the current in Amperes through the capacitor at t = 6.98 seconds?
Uploaded: 2022-04-13T11:21:10-08:00
Duration: 1 min 43 s
Channel: Madhur L
Description: A 5.89 µF capacitor is connected to an emf given by v0 sin(ω t) where V0 = 120 Volts and ω is 120 π rad/s. What is the magnitude of the current in Amperes through the capacitor at t = 6.98 seconds?

          A 5.89 µF capacitor is connected to an emf given by $v_0 \sin(\omega t)$ where $V_0 = 120$ Volts and $\omega$ is $120 \pi$ rad/s. What is the magnitude of the current in Amperes through the capacitor at t = 6.98 seconds?

A 5.89 µF capacitor is connected to an emf given by v0 sin(ω t) where V0 = 120 Volts and ω is 120 π rad/s. What is the magnitude of the current in Amperes through the capacitor at t = 6.98 seconds?

Added by Carla K.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

04/13/2022

Step 1

Step 1: The current through a capacitor is given by: $$I = C \frac{dV}{dt}$$ where C is the capacitance and V is the voltage across the capacitor. Show more…

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A 5.89 µF capacitor is connected to an emf given by $v_0$ sin(ωt) where $V_0$ = 120 Volts and ω is 120 π rad/s. What is the magnitude of the current in Amperes through the capacitor at t = 6.98 seconds?