A 20V 5watt lamp is used on AC mains of 200 volts, 50√11/pi c.p.s. C

1.	A 20V 5watt lamp is used on AC mains of 200 volts, 50√11/pi c.p.s. Calculate the capacitor
Answer» EXPLANATION: Since the current through the lamp would be: I = P/V=50/100=0.5A and since this is a SERIES circuit we know the same amount of current flows throughout the circuit therefore we can say that the total impedance is: Z=V/I=200/0.5=400Ω by Ohm’s law and lamp’s RESISTANCE is: Rlamp=V/I=100/0.5=200Ω by Ohm’s law Through the impedance triangle Xc would be: Z²=Rlamp²+Xc² Xc=√(Z²-Xc²) Xc=√(400²-200²) Xc=√(120000) Xc=346.41Ω and now we can determine C from: Xc=1/2πfC therefore C=1/2πfXc C=1/(2π(50)(346.6)) C=0.000009F or 9µF Hopes this helps.

A 20V 5watt lamp is used on AC mains of 200 volts, 50√11/pi c.p.s. Calculate the capacitor

Answer»

Since the current through the lamp would be:

I = P/V=50/100=0.5A

and since this is a SERIES circuit we know the same amount of current flows throughout the circuit therefore we can say that the total impedance is:

Z=V/I=200/0.5=400Ω by Ohm’s law

and lamp’s RESISTANCE is:

Rlamp=V/I=100/0.5=200Ω by Ohm’s law

Through the impedance triangle Xc would be:

Z²=Rlamp²+Xc²

Xc=√(Z²-Xc²)

Xc=√(400²-200²)

Xc=√(120000)

Xc=346.41Ω

and now we can determine C from:

Xc=1/2πfC

therefore

C=1/2πfXc

C=1/(2π(50)(346.6))

C=0.000009F

or 9µF

Hopes this helps.

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