# Voltage Drops In Series Circuits

1. Apr 6, 2005

### scorpa

Hello!

I am doing some physics questions and am having trouble with voltage drops. If anyone could help me understand it a bit better I would appreciate it.

The first question is this:

1) Three identical lamps are connected in series to a 6.0 V battery. What is the voltage drop across each lamp?

- All they give you is the voltage of the battery, so do you have to assume the other values are 1 or something like that. I know this one should be easy but I can't seem to get it. The lamps are all identical, so can you just assume that the voltage drop is 6V across each lamp? I really don't know what I am talking about here, as I am sure you can tell.

2) A string of eighteen identical Christmas tree lights are connected in series to a 120V source, the string dissipates 64.0 W.

a) What is the equivalent resistance of the light string?

P = IV
64W/120V
I = 0.533 A
R = V/I
120V/0.533A
R = 225 ohms

Equivalent resistance is 225 ohms?

b) What is the resistance of a single light?

As it is connected in series 225 ohms/18 lights = 12.5 ohms

c) What is the power dissipated by each lamp?

V = IR 120 V/12.5A = 9.6 A
P = IV 9.6A*120V = 1152 W which to me seems way to high

If one of the bulbs in the problem burn out and the lamp has a wire that shorts out the lamp filament when it burns out the resistance of the lamp is dropped to zero.

d) What is the resistance of the light string now?

e) Find the power dissipated by the string.

f) Did the power go up or down when the bulb burned out?

At this point I am pretty lost, and really need help. I know it should be something really easy but I can't think of what to do.

3) Amy needs 5.0V for some integrated circuit experiments. She uses a 6.0V battery, and two resistors to make a voltage divider. One resistor is 330 ohms. SHe decides to make the second resistor smaller. What value should it have?

Once again I have no clue about this one. For some reason voltage drop questions have been causing me a lot of grief today.

Thanks for any help you can give me, I really appreciate the help.

2. Apr 6, 2005

### Ouabache

Do you know Kirchoff's Voltage Law?
(applied to series circuits: the source voltage equals the sum of the voltage drops).

So analytically you would have:
V = iR1+iR2+iR3
V= i ( R1+R2+R3) <----- in this case R represents the impedance of each lamp and since the lamps are all the same, R1=R2=R3
so V = i (3R).

What is a voltage drop?? (what is formula for voltage ..... hint:V=iR),
The drop across each lamp would be iR..
From our derivation, V/3 = iR and you are given V.
So go ahead and calculate voltage drop across each lamp, iR = ??

3. Apr 6, 2005

### Ouabache

you need to review the reasoning of part (a)
Hint: remember something about voltage drops across each lamp?
V = iR1+iR2+iR3..+iR18 = i(R1+R2+R3... +R18) . So the current i does not go thru just one lamp but all 18. And the power is dissipated across all 18 lamps.
The reason your final power seems way to high is because your current is way to high.. Go back and check your current calculation.

Is this your analysis of what happens if a bulb burns out? If so, you may want to recheck the meaning of a short circuit versus and open circuit..

If we assume questions d, e and f pertain to after a bulb burns out;
I won't answer these directly but give you a hint by rewording your questions:

d) what is the resistance of an open circuit?
e) how much power is dissipated in an open circuit?
f) Is the answer to (e) larger or smaller than the dissipated power of the whole string of bulbs?

4. Apr 6, 2005

### Ouabache

I wouldn't worry about voltage drops here..
A better way to answer this question is to study up on voltage dividers
for example ---> http://www.aikenamps.com/VoltageDividerRule.htm
I'll give you some hints, the needed voltage is Vout, your source voltage is Vin,

Since you don't know which resistor is 330 ohms, try both cases:
(i) R1= 330 , solve for R2, (ii) R2 = 330, solve for R1.
The correct case will be the one with smaller 2nd resistor.

5. Apr 7, 2005

### scorpa

OK thanks for the help, I go give it another try.

6. Apr 7, 2005

### scorpa

Amy needs 5.0V for some integrated circuit experiments. She uses a 6.0V battery, and two resistors to make a voltage divider. One resistor is 330 ohms. She decides to make the second resistor smaller. What value should it have?

Ok for this question I got the other resistance to be 66 ohms, so I hope thats right. I used V2 = (vR2)/(R1+R2)

7. Apr 7, 2005

### scorpa

A string of eighteen identical Christmas tree lights are connected in series to a 120V source, the string dissipates 64.0 W. What is the power dissipated by each lamp?

Ok, if the entire string puts out 64 W will the power of put out by each lamp not just be 3.56 W?

8. Apr 7, 2005

### Delta

yep, 3.56W is correct. The reason for the incorrect 1152W is because you've taken the voltage across the whole string where is each lamp takes one eighteenth of the total potential.

9. Apr 7, 2005

### scorpa

Wow I feel like an absolute retard now

10. Apr 7, 2005

### Delta

Success involves making mistakes, least you shouldn't make that one again.

11. Apr 7, 2005

### scorpa

OK now here is another question, I am hoping that I get it now, but I want to make sure of myself.

1) A 75.0 W bulb is connected to a 120V source.

a) What is the current through the bulb?

P/V = I 75W/120V = 0.625A

b) What is the resistance of the bulb?

V/I = R 120V/ 0.625A = 192 ohms

c) A lamp dimmer puts a resistance in series with the bulb. What resistance would be needed to reduce the current to 0.300A?

120V/0.300A = 400 ohms-192ohms = 208 ohms

2) In the previous problem you found the resistance of the lamp and the dimmer resistor. This is were I am starting to screw up again. I'm not sure where to go now.

a) Assuming the resistances are constant find the voltage drops acrros the lamp and the resistor.

b) Find the power dissipated by the lamp.

c) Find the power dissipated by the dimmer resistor.

12. Apr 7, 2005

### whozum

The voltage drop over a resistor =

Drop(V) = IR

Your current is 0.3A, and you have tow resistors, the lamp (192o) and the added resistor (208)

The voltage drop through the entire circuit (the two resistors) must add up to the source voltage (120V)

13. Apr 7, 2005

### scorpa

alright thank you very much

14. Apr 7, 2005

### Ouabache

You don't need to hope it is right. Try your calculated resistor value back into the voltage-divider formula and see if it generates the expected output voltage.
Regarding your newest questions.. It looks like some other folks (Delta, whozum) have been helping.

Regarding the power dissipated by each of the eighteen identical bulbs in series;
I already told you what Delta said, but sometimes it helps to hear it explained another way.
I mentioned, "V = iR1+iR2+iR3...+iR18" meaning that each iR is a voltage drop across one light bulb. So how does that help?

Factoring out current i, you would get V = i(R1+R2+R3...+R18). Solving for i = V/(R1+R2+R3...+R18).
Since the bulbs are identical, then (R1=R2=R3...=R18) = 18R so i = V/(18R) = 120/(18*12.5) = 120/225 = 0.533333 amps.

The power dissipated by each bulb is $$P= i^2R$$ because the current remains constant through each bulb in a series circuit.
So you're correct, the power dissipated per bulb is $$P= (0.53333)^2*12.5 = 3.56W$$

Last edited: Apr 8, 2005
15. Apr 7, 2005

### scorpa

Thanks again, I wish I was half as smart as you guys were.