(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Hi, I found the following example in a high school physics book about electrical circuits. There's something about the example that I don't get, so I hope you can help me out. I give you the example text and the solution, and then I explain what I don't understand.

Example text:

"An alternating emf. of 200V and 50Hz is applied to a capacitor in series with 20V, 5W lamp. Find the capacitance required to run the lamp."

Solution:

Lamp:

V = 20V, P = I*V, i.e. I = P/V = 5W/20V = 0.25A

Circuit:

I = 200V/X (where X is the capacitive reactance)

X = 200V/I = 200V/0.25A = 800 ohm

X = 1/wC (where w is the angular frequency)

X = 1/wC = 800 ohm, i.e. C = 1/(w*800 ohm) = 1/(2*pi*50Hz*800 ohm) = 3.797 uF

What I don't understand:

Why is the reactance calculated as 200V/0.25A? I would expect the lamp to "take" 20V from the circuit, thus leaving 180V for the capacitor, i.e. X = 180V/0.25 = 720 ohm. Can someone please tell me why this is not the case?

2. Relevant equations

U = I*R (ohm's law)

P = I*V (power = current * voltage)

X = 1/wC (capacitive reactance)

w = 2*pi*f (angular frequency and frequency)

3. The attempt at a solution

I looked at an exercise and its solution in the book and it gave the same answer as the example from above. I looked for an answer on the internet but couldn't find a similar case. I read about lamps in electrical circuits, but what I found supported me in my expectation that the lamp should "take up" some voltage.

**Physics Forums - The Fusion of Science and Community**

# AC emf applied to capacitor and lamp in series

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: AC emf applied to capacitor and lamp in series

Loading...

**Physics Forums - The Fusion of Science and Community**