# Charge entering and leaving a light bulb

1. May 6, 2013

### iriver4

1. The problem statement, all variables and given/known data

A light bulb is rated at 30 w when operated at 120 v. How much charge enters (and leaves) the light bulb in 4.0 min?

ans is 17 C

2. Relevant equations

I = P/V
i = dq/dt

3. The attempt at a solution

I = P/V = 30/120 = 0.25A

q = I*t = (0.25)(4*60) = 60 C

thinking maybe the answer given is a typo but need confirmation please

2. May 6, 2013

### Staff: Mentor

I thought a bit about if the difference could be that the 0.25A is RMS, and maybe there needs to be some correction for that. But the RMS value is what is used in the power calculation, and is equivalent to the DC value that the current would have if the light bulb were run on DC instead of AC.

Does anybody else see where the 17C came from?

3. May 7, 2013

4. May 7, 2013

### rude man

Maybe it's my daily allotment of Shiraz, but my vote is zero:
q = integral i dt = 0 over 4*60*60 integer cycles.

5. May 7, 2013

### haruspex

You raise the key question of whether this is DC or AC. However, even if it is AC we can get a nonzero result by noting that it doesn't matter which way the charge flows in at any time, so it's ∫|I|.dt. OTOH, this gives me 2PT/(πV) = 2*30*240/(π*120) ≈ 38C.

6. May 7, 2013

### rude man

???
A semantic issue. I'll stick with zero C.

7. May 7, 2013

### mukundpa

If it is an AC than the current calculated is rms current. Peak value of this current will be Imax = 0.25*√2 A

To calculate charge flowing in one direction only we have to integrate instantaneous current for half cycle and thus the average current for half cycle is 2 Imax/π.

In 4 min charge is flowing in one direction for 2 min and thus the net charge flow in one direction during 4 min (for 2 min) will be
(2√2*0.25/π)*2*60 = 27 C

8. May 7, 2013

### SammyS

Staff Emeritus
I like it !