Can anyone check the equation (Capacitance that is required)

Thread Starter

ironasona

Joined Nov 12, 2020
53
Hello, could anyone have a quick look and check if i have done it right .

Thank you

I=P/V=500/120=4.16A
Z=V/I=240/0.5=480 Ohms
Rlamp=V/I=120/5=60 Ohms

480^2-60^2=476 Ohms after rooting

Xc=1/2pie*f*C

1/2pie*50*476=6.6uF

Q
A metal-filament lamp is rated at 120V, 500W. It is to be connected in series with a capacitor to a power supply of 240V 50Hz
 

Thread Starter

ironasona

Joined Nov 12, 2020
53
The unstated goal is probably to maintain the power to the filament. You know the voltages in both arrangements.
i need to find the capacitance that is required:
so i did the new equation:
I=P/V=500/120=4.16A
Z=V/I=240/4.16=58 Ohms
Rlamp=V/I=120/4.16=29 Ohms
/58^2-29^2=50 Ohms

1/2Pie*(50)(50)=6uF is that right ??
 

BobaMosfet

Joined Jul 1, 2009
1,848
i need to find the capacitance that is required:
so i did the new equation:
I=P/V=500/120=4.16A
Z=V/I=240/4.16=58 Ohms
Rlamp=V/I=120/4.16=29 Ohms
/58^2-29^2=50 Ohms

1/2Pie*(50)(50)=6uF is that right ??
Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3
 

Thread Starter

ironasona

Joined Nov 12, 2020
53
Line 1 & 2. What should the current be to maintain 500W? How about
Power = 500W = 240V • 2.08A
Total impedance Z = V/I = 240V / 2.08A = 115.4Ω
Lamp impedance ZL = V/I = 120V / 4.16A = 28.8Ω
Capacitor impedance Zc = Z - ZL = 86.5Ω (series, impedances are additive)
Zc = 1/(2πƒC) https://www.allaboutcircuits.com/tools/capacitor-impedance-calculator/
C = Zc•2𕃠=1/86.5Ω •2•3.1416•50Hz =1/ 27,187 = 36.8µF
ooo i see ,where i went wrong , ok thank you very much , i wasn't 100% i was doing it right
 
Top