# Homework Help: Really Hard Circuit Question

1. Apr 13, 2005

### scorpa

Really Hard Circuit Question- With LINK

Hello there!!!

I am doing a really tough physics question on circuits :yuck: . I am attaching the picture of the circuit so you can actually see what is going on

These are the questions:

a) What is the current through R1?

This one was easy I just found the total resistance of the circuit and used the equation V= IR to get a current of 2.4 A which is the answer/

b) What is the current through R4?

I have no clue how to do this one, no matter what I do I get an answer either to high to to low. The answer should be 0.8 A.

c) What is the power dissipated in R2?

I have a feeling that I can't get this one because I can't get the one before it. The answer should be 5.1W.

d) What is the total power dissipated in the circuit?

I could do this one. It's just p = iv (12.0V)(2.4A) = 29W

Oh and one more really stupid thing. When you are drawing a graph and it says to plot resistance as a function of the cross-sectional area the resistance should go on the x-axis right?

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Last edited: Apr 13, 2005
2. Apr 13, 2005

### scorpa

There we go this one has the link

3. Apr 13, 2005

### scorpa

Oh and just in case you find it hard to read R1 = 3.0 ohms, R2= 2.0 ohms, R3= 1.0 ohms, R4 = 4.0 ohms, R5 = 2.0 ohms

4. Apr 13, 2005

### scorpa

plz someone help me...I'm desperate!!!

5. Apr 14, 2005

### Chi Meson

Sorry guy; we're still waiting to see the picture. Can you describe the circuit?

6. Apr 14, 2005

### Delta

a) good.

b) You know that the voltage across the R2,3,4,5 network is the same across the R2-R3 branch, and also the R4-R5 branch. so... if you know the total current going into this network and the resistance of the network R2,3,4,5 you can find the voltage across it. Then you can find the current going through R4-R5.

c) you now know the voltage across the R2-R3 branch from above, therefore you know the current. Take the R2 resistance ....

d) good.

7. Apr 14, 2005

### scorpa

I put the picture on as an attachment, it shows up on my computer. I can't really describe the circuit very well without getting everything mixed up. Maybe I'll try attaching it again or something.

8. Apr 14, 2005

### scorpa

Thanks for the help Delta, but I still don't really get what to do. When I went back to the question I redid it and got the answers but I know it's not the right way to do it.
For B)

I went Current total is 2.4 A
I total = 2Ir4 + Ir4
2.4 A = 3 Ir4
0.80 A = Ir4
There for the current moving through R4 is 0.80 A.

C)

Ir2 = 2XIr4 there for the current moving through R2 is 1.6 A
V = IR 1.6A x 2.0 ohms = 3.2 V
P= IV 1.6A x 3.2 V = 5.1 W

Therefore the power moving through R2 is 5.1W

9. Apr 14, 2005

### roger

Hi Scorpa,

for part b, this is the procedure to calculate the current across r 4

you must calculate the pd at the junction of the R (3, 2 and 4) and the positive potential.

you can see that resistors 2,3,4, and 5 is equivalent to 2 ohms by calculating the resistors in parallel.
It follows immediately that the 2 ohm and the r1 forms a potential divider,and the pd across the 2 ohm is : 2/5 * 12v

the pd across the r4 is (another potential divider) 4/6 * 4.8 v

which gives you 3.2 v

then the current through r4 is 3.2v / 4 ohm = 0.8 A

10. Apr 14, 2005

### scorpa

Ok so what I did was totally wrong..lol???? Thanks for the help I'm going to go try it your way now. I did get that the resistance in total across 2.3.4.5 was 2 ohms, but I never tried using a potential divider equation.

11. Apr 14, 2005

### scorpa

Could you explain what you have done in more detail? I thought that I understood it but I guess I don't.

12. Apr 15, 2005

### roger

Scorpa

if we suppose that the r2 and r4 do not exist, then we simply have the 1ohm , 2ohm, and 3 ohm in series totaling 6ohms.

and if you were to find the pd across (r2+r3) you would have 1+2/6 * 12volt = 6 volt
(Using the potential divider , you calculate r2+r3/r1+r2+r3 * 12volts)

But recall that this circuit has r2 and r4 (2ohm and 4ohm) in series across the r2 and r3(also in series).
so you calculated the combined parallel resistance of r2 to r5 to be 2 ohm, which you would expect to be lower the 3ohm.(r2+r3)

now calculate the pd across the same points as above but this time its 2/5 * 12volts = 4.8volt

And finally, we calculate r4/r4+r5 * 4.8volt (using the potential divider) which gives the pd across r4, from which we calculate the current through r4 using V/R=I to give 0.8 Amps