1. Feb 13, 2007

### BunDa4Th

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

Three 80.0 W, 135 V lightbulbs are connected across a V = 135 V power source, as shown in Figure P18.48. (Assume that the resistance of each bulb is constant even though, in reality, the resistance increases markedly with current.)

http://www.webassign.net/sercp/p18-48alt.gif

(a) Find the total power delivered to the three bulbs. W
(b) Find the potential difference across each light bulb.
deltaVR1 = V
deltaVR2 =
deltaVR3 =
2. Relevant equations

I^2R = P
DeltaV = IR
DeltaV/I = R
p/v = I

3. The attempt at a solution

The first thing I did was find the current which is 80/135 = .5925. From this I found the resistance which is 135/.5925 = 227.85 ohm.

From there I tried using the formula I^2R= P, which gave me the incorrect answer. Then I thought since R2 and R3 are parallel I can combine it by using the formula 1/R = 1/R1 + 1/R2. But that got me a bit confuse and still gave me an incorrect answer.

I am not sure where to start and what steps I am doing wrong or missing. Any help would be great.

2. Feb 13, 2007

### Integral

Staff Emeritus
You can use the fact that all bulbs have the same resistance. The parallel pair have a total resistance of $\frac R 2$ the total resistance is $\frac {3R} 2$.

Can you use that to determine the voltage drop across each bulb?

3. Feb 13, 2007

### BunDa4Th

Total Resistance is 341.775?

I am not sure how i would find the voltage of drop across each bulb, but im thinking its R/2 is used for R2 and R3 which give a voltage of 67.5 V.

i did 227.85/2 = 113.925 then multiply that by .5925 A to get 67.5 V

Most likely this is incorrect, I have tried asking the professor but she had another class right after and couldn't answer my question and i also had a lab which did not help explain this type of situation.

4. Feb 13, 2007

### Staff: Mentor

Your total resistance is correct. Now just use the voltage divider equation to figure out what the voltage across the two parallel resistors (light bulbs) is. Remember that in the voltage divider equation, the voltage is the input voltage multiplied by the lower resistance, all divided by the total resistance.

Then once you have the intermediate voltage across the top of the two parallel resistors, you have enough information to calculate the total power in the circuit.

5. Feb 14, 2007

### BunDa4Th

Okay i got lost in words just there but if i read correctly it is

to find R2 and R3 (parallel resistor) it would be

(135)(227.85/2)/341.775 giving a voltage of 45.

6. Feb 14, 2007

### BunDa4Th

Okay, I did the above correctly but I have a problem finding the total power deliver to the 3 bulbs.

I did P = VI on each bulb with the given voltage i just found and added up the power. which is incorrect.

or do i use the formula I^2R = P?

7. Feb 14, 2007

### Staff: Mentor

I would use the formula P = V^2/R for each of the two pieces (the single resistor at the top, and the parallel combination of the two resistors at the bottom).

8. Feb 14, 2007

### BunDa4Th

Okay I used the above formula to find each of the bulbs power and add it up but it gives me an incorrect answer. I tried using total resistance and 227.85 as the resistance and it still gave an incorrect answer.

I am now sure why this is not working or if i am using the correct number.

90^2/227.85 = 35.55 W
45^2/113.925 = 17.77 W
45^2/113.925 = 17.77 W

added those up and still incorrect.

9. Feb 14, 2007

### Staff: Mentor

You're double-counting the power of the lower leg. The power you calculated for the single upper resistor/bulb is correct, but remember that since the resistance of each bulb is 227.8 Ohms, the parallel combination of the two bottom resistors/bulbs gives you the single equivalent resistance of 113.9 Ohms.

Alternately, you could calculate the power dissipated in each of the bottom two bulbs, but you need to use their full resistance of 227.8 Ohms if you want to do it that way.

10. Feb 15, 2007

### BunDa4Th

Thanks a lot for the help. I finally understand it now.

again thanks for all the help berkeman.