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