# Can anyone check the equation (Capacitance that is required)

#### 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

#### Delta Prime

Joined Nov 15, 2019
1,311
Z=V/I ???

#### ironasona

Joined Nov 12, 2020
53

#### wayneh

Joined Sep 9, 2010
17,406
Why did the current drop to 0.5A in line 2?

And: Rlamp=V/I=120/5=60 Ohms ??

#### ironasona

Joined Nov 12, 2020
53
Why did the current drop to 0.5A in line 2?
do i do 240/500 or 240/5?
Edi: or 240/4.16

#### wayneh

Joined Sep 9, 2010
17,406
do i do 240/500 or 240/5?
Edi: or 240/4.16
The unstated goal is probably to maintain the power to the filament. You know the voltages in both arrangements.

#### 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
2,082
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

#### wayneh

Joined Sep 9, 2010
17,406
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)
C = Zc•2π•ƒ =1/86.5Ω •2•3.1416•50Hz =1/ 27,187 = 36.8µF

#### 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)
C = Zc•2π•ƒ =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

#### wayneh

Joined Sep 9, 2010
17,406
ooo i see ,where i went wrong , ok thank you very much , i wasn't 100% i was doing it right
I recommend watching your units going forward. Don't be lured to take the shortcut of leaving them out of your notes. Naked numbers are much harder to follow and check.