# Almost Solved my ele. hmwk, need someone to check it!

#### shemer77

1. The problem statement, all variables and given/known data
1) a 12m length of wire with uniform cross sectional area 2.75x10^-7 has a potential difference of 15v applied across it, producing a current of 20A
a) find the resistivity of the wire B) what is the power dissipated in the wire?

2) 2 circuits are made. In circuit 1, 4 resistors (1,2,3,6 ohms) are connected in series. in circuit 2, the same 4 resistors are connected in parallel. The battery emf is 24 volts in both cases. Calculate the equivalent resistance of the each circuit and the current through each resistor.

3)http://img25.imageshack.us/my.php?image=46010928.png
determine the equivalent resistance and the total power deliverd by the battery for this ciruit.
http://img25.imageshack.us/my.php?image=46010928.png

3. The attempt at a solution
1) a) r=I/V r=15/20 r=.75
b) p=IV p=20*15 p=300

2) Equivalent resistance for the series would be 12 because i just added the resistors up and i solved for current doing 24=I(12) V=IR so then the current through each resistor would be 2,4,6,12 respectively
for the parallel i did 1/rp=1/r1+1/r2+1/r3+1/r4 and got 1/3=rp, i solved current using v*r=I
24(1/3)=8 and then for each resistor the current was 24,12,8,4 respectively.

3) I am totally confused on this one?

Last edited:

#### Kruum

1 a) is correct answer, but you have symbols for current and voltage wrong.

b) There is an equation for resistivity, that includes lenght and cross sectional area.

2) Req for series is correct, but the currents aren't. The current of one loop is constant inside that loop. So the currents through each resistor would be?

Req for parralleled is wrong. Your method of finding the total current is right, but for each resistor it's wrong. Kirchof's law states that if you divide one wire into multiple, the sum of the currents is the same as it was in one wire. But the voltage won't be altered in "cross roads".

3) You need to find the Req for the circuit and then you can use the equation for power.

#### shemer77

your right there is an equation for resistivity, that includes lenght and cross sectional area.
but i dont know what the restistivy of my matiral is.

2) for the series circuit would the current through each resistor be 2?
B)i dont get you

3)thanks for nothing on that one, lol you just read the question too me.

#### Kruum

your right there is an equation for resistivity, that includes lenght and cross sectional area.
but i dont know what the restistivy of my matiral is.
That's what your trying to figure out. The equation should be $$R= \rho \frac{l}{A}$$ you know everything else but $$\rho$$.

2) for the series circuit would the current through each resistor be 2?
Yes.

B)i dont get you
Then you should read about Kirchoff's law. But to get you going, the voltage through each resistor is the same, current isn't.

3)thanks for nothing on that one, lol you just read the question too me.
Your welcome! :tongue2: http://en.wikipedia.org/wiki/Resistor#Series_and_parallel_resistors Check that out. You should see those type of connections in the circuit, which you then can use to determine Req for the circuit.

#### shemer77

lol thanks man, didnt mean too sound like a loser.

#### shemer77

i still dont see how my curent through each resistor is wrong for the parallel cirucit.

#### mplayer

i still dont see how my curent through each resistor is wrong for the parallel cirucit.
Try drawing the circuit out with the 4 resistances in parallel connected to the voltage source. You should be able to see that the voltage across each resistor is the same as the voltage source. So if you know the voltage across each resistor, and the resistance of each resistor, you should be able to solve for the current through each resistor.

#### mplayer

3)
determine the equivalent resistance and the total power deliverd by the battery for this ciruit.
Start collapsing the circuit from the outside, moving towards the voltage source. Use parallel combination where you need it, and use series combination where you need it. It may help to redraw the circuit each time you make a combination, in order to easily see subsequent combinations. Once the circuit is down to just a voltage source and an equivalent resistor, it should be easy to find the power. There are multiple ways to find it.

#### shemer77

Try drawing the circuit out with the 4 resistances in parallel connected to the voltage source. You should be able to see that the voltage across each resistor is the same as the voltage source. So if you know the voltage across each resistor, and the resistance of each resistor, you should be able to solve for the current through each resistor.
isnt the formula i=V/R1 , if thats true then for resisotr one it would be 24/1

#### mplayer

isnt the formula i=V/R1 , if that's true then for resistor one it would be 24/1
Yep, that's one branch current, just get the other 3 for that parallel circuit and you should be set for that part. After you've calculated those currents, you will need to re-figure the equivalent parallel resistance REQ, it is not 1/3 ohms but it's close.

Once you get the correct REQ, calculate the current through it. You will notice that the current through the REQ is the sum of the individual branch currents. In other words: IREQ = IR1 + IR2 + IR3 + IR4

#### shemer77

is it .5?

for number 3 is the equivalent resistance 6?

#### mplayer

is it .5?

for number 3 is the equivalent resistance 6?
Yep, both are right, good job

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving