# 65W bulbs connected in parallel (hopefully easy)

1. Apr 13, 2008

### HyperSniper

I know that this has to be an easy problem, but my teacher left us in December and we got stuck with an idiot who hasn't taught us anything since then......

1. The problem statement, all variables and given/known data
How many 65-W light bulbs can be connected in parallel across a potential difference of 85V before the total current in the circuit exceeds 2.1 A?

2. Relevant equations
$$\Sigma$$1$$/$$Req=1$$/R1$$+1$$/R2$$+1$$/R3$$...
P=IV
V=IR

3. The attempt at a solution
I can't get an answer that makes any sense, but this is what I tried to do.

P=65-W
V=?
I=2.1A

P=IV
65=(2.1A)V
30.95V=V
R=14.738$$\Omega$$

X= number of bulbs

$$\epsilon$$= I((1/R)X)^-1
85V= (2.1A)(1/14$$\Omega$$)X)^-1
(14$$\Omega$$/X)(2.1A)=85V
29.4V/X=85V
29.4V=(85V)X
.34588=X

Last edited: Apr 13, 2008
2. Apr 13, 2008

### mikelepore

I assume "before circuit exceeds 2.1A" is a typing mistake that means "before the total current out of the voltage source exceeds 2.1A".

Each bulb has a power rating P=VI, so it draws current I = P/V = calculate. That's one bulb. How many bulbs to reach the allowed maximum current?

Last edited: Apr 13, 2008
3. Apr 13, 2008

### HyperSniper

Opps. I'm sorry, I'll go fix it.

So does that mean it's just...

65W/85V=.76A

2.1A=(X)(.76A)
X=2.76

Rounding down not to exceed the maximum...

2 bulbs?

4. Apr 13, 2008

### mikelepore

It looks good to me. Just one suggestion for future problems that say "no not exceed" something, your rounding to 0.76 instead of saying 0.764 might have changed the answer. It's okay in this case.

Last edited: Apr 13, 2008
5. Apr 13, 2008

Thank you.