30,A 10 μ fcapacitor is connected across a200V, 50 HzAC supply. The p

1.	30,A 10 μ fcapacitor is connected across a200V, 50 HzAC supply. The peak currentthrough the circuit is:(A) 0.6 /2 A(B) 0.6A(C) 0.3 t AD) 0.6 v24A
Answer» We know, Capacitive reactance(Xc) =1÷(23.14fC) =1÷(23.14501010^-6) {~1 uF=10^-6 F~} =318 ohms(approx.) Again voltage across capacitor =200V So, Current across capacitor =200÷318=0.7 A The calculation was done assuming peak voltage =200 V. If 200V is the R.M.S value then peak voltage would be 200*root over of 2=283 V So current =283÷318= 0.89A = 0.6√2 A it is not in option and solution not understandable

30,A 10 μ fcapacitor is connected across a200V, 50 HzAC supply. The peak currentthrough the circuit is:(A) 0.6 /2 A(B) 0.6A(C) 0.3 t AD) 0.6 v24A

Answer»

We know, Capacitive reactance(Xc) =1÷(2*3.14*f*C)

=1÷(2*3.14*50*10*10^-6) {~1 uF=10^-6 F~}

=318 ohms(approx.)

Again voltage across capacitor =200V

So, Current across capacitor =200÷318=0.7 A

*The calculation was done assuming peak voltage =200 V.

If 200V is the R.M.S value then peak voltage would be 200*root over of 2=283 V

So current =283÷318= 0.89A = 0.6√2 A

it is not in option and solution not understandable

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30,A 10 μ fcapacitor is connected across a200V, 50 HzAC supply. The peak currentthrough the circuit is:(A) 0.6 /2 A(B) 0.6A(C) 0.3 t AD) 0.6 v24A

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