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A 0.30μFcapacitor is connected across an AC generator that produces a peak voltage of 10V. What is the peak current to and from the capacitor if the emf frequency is

(a) 100Hz?

(b)100kHz?

Short Answer

Expert verified

(a)Io=1.9×10-3A

role="math" localid="1649956592903" (b) Io=1.9A

Step by step solution

01

Part (a) Step 1: Given Information:

We have been given that

Capacitance (C)= 0.30μF

Peak voltage (Eo)= 10V

Emf frequency (n)= localid="1649958487243" 100Hz

02

Part (a) Step 2: Solving:

Peak current Iois given by :Io=Eo×Ӭ×C

Putting the value of(Ӭ)as2πnin the above equation:Io=Eo×2πn×C

Using the given values in the updated equation we get:Io=(10)×2π(100)×0.30×10-6

There the peak value of current for a frequency of 100HzisIo=1.9×10-3A

03

Part (b) Step 1: Given Information:

We have been given that

Capacitance (C)= 0.30uF

Peak voltage (Eo)= 10V

Emf frequency (n)=100kHz

04

Part (b) Step 2: Solving:

Using the expression of Io that we earlier used:Io=Eo×2πn×C

Putting the known values:Io=(10)×2π100×103×0.30×10-6

Io=1.9A

Therefore the peak current value for an emf frequency of100kHzis Io=1.9A

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