Question

A 1.5 microfarad capacitor is connected to a 220V,50Hz source.FIND the reactance and Irms,Ipeak in the circuit.??​

Topic; Alternating current class 12

Answer 1

Answer:

Given : C=15×10

−6

F

Rms value of voltage V

rms

=220 V

Frequency of source f=50 Hz

Capacitive reactance X

C

=

2πfC

1

=

2π×50×15×10

−6

1

=212.3Ω

Rms value of current I

rms

=

X

C

V

rms

=

212.3

220

=1.04 A

Peak value of current I

o

=

2

I

rms

=

2

×1.04=1.465 A

Since X

C

∝

f

1

So capacitive reactance gets halved if the frequency is doubled.

Also, I

rms

∝

X

C

1

Answer 2

Answer:

this is your Solution

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