Homework Help: 5 resistor circuit

1. Jun 5, 2012

arl146

1. The problem statement, all variables and given/known data
Five resistors with resistance R1 = 4.00 Ω, R2 = 6.00 Ω, R3 = 2.00 Ω, R4 = 7.00 Ω, R5 = 4.00 Ω, are connected to a battery with emf = 8.00 Volts as shown in the figure. Find the equivalent resistance of the circuit. Give you answer with a unit of "Ohm".

2. Relevant equations
parallel: Req = 1/((1/R1)+(1/R2)+....)
series: Req = R1 + R2 + R3 +....

3. The attempt at a solution
well im thinking that R1 and R3 is in series and R2 and R4 is in series but from there im not sure what to do. the R5 gets me confused

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2. Jun 5, 2012

ehild

Neither R1 and R3 nor R2 and R4 are in series. Do you know what series connection means?

The resultant resistance seen by the battery is the voltage of the battery divided by the current through the battery. To find the current, apply Kirchhoff's laws.

ehild

Last edited: Jun 5, 2012
3. Jun 5, 2012

jayaganesh

Use wheat stone bridge idea to solve this.

4. Jun 5, 2012

arl146

i dont really know it ..

but uhhh oh .. im not sure if i can apply them here it seems too complicated my mind just gets all frazzled. can you give me one eqn to work off?

5. Jun 5, 2012

ehild

Do you need the equivalent resistance between points P and P'?

ehild

Last edited: Jun 5, 2012
6. Jun 5, 2012

arl146

uhhh i guess? idk! im confused already. isn't that just R5 though?

7. Jun 5, 2012

ehild

No, the other resistances are also connected to those points. Do you really need the equivalent resistance between P and P'? Show the original text of the problem, please.

ehild

8. Jun 5, 2012

arl146

oh yea true .. well then i guess not .. but that is the original text, all i have in the original post is the whole problem

9. Jun 5, 2012

ehild

Equivalent resistance refers to two points. If you do not know those points I can not help.
Have you learnt something about Wheatstone Bridge? Look after in your textbook. Is there anything about it?

ehild

10. Jun 5, 2012

arl146

nothing in my textbook about it .. professor briefly went over it today in class. theres like 4 formulas that he got.
I1R1=Va-Vb
I1R1=Vb-Vc
I2R2=Va-Vd
I2R=Vd-Vc

where c is the left corner of the 'diamond' , a is the right corner, b is P in our case, d is P' in our case .. R is our R3 and R3 in his example is my R4

can't you just do from the left corner of the diamond that we'll call c like he did all the way to the right corner which is a?

11. Jun 5, 2012

willem2

Those can't be right. Your professor might have done something like the following,
but you seem to have made some errors in copying them.

Assuming that the equivalent resistance is wanted between the points where the battery is connected.

There are 2 unknown voltages, in the points P and P'. The points connected to the battery are at 8V and 0V.

If you know these unknown voltages, you can compute all the currents in the 5 resistances, The current between two points is the potential difference between
those two points divided by the resistance between them.
for example, the point connected to the battery is at 8V and the point P is
at V_P, so the potential differenct between them is (8-V_P) and the current
through it: I_1 = (8 - V_P)/R_1

When you've computed all 5 currents as a funtion of V_P and V_P', you can than use Kirchhoffs current law in the points P and P' to get 2 simultaneous equations for V_P and V_P', wich you can solve.

12. Jun 5, 2012

arl146

ok well I1 is easy... i assume I3 is the same thing? I_3=(8-V_P')/R3 right?
so when you want to do R2 and R4 how do you do it .. is it like I_2=(V_P-8)/R2?
and then the same thing for R4 .. I_4=(V_P'-8)/R4 ?
if thats the case then I_5=(V_P-V_P')/R5 ?

and then the current laws:

for P: I1-I2-I5=0
for P': I3+I5-I4=0

this is all right?

13. Jun 6, 2012

ehild

Everything is right except the equation for I4. Check the PD across R4.

ehild

14. Jun 6, 2012

arl146

ummm I_4=(V_P'-V_P+8)/R4? since the potential difference at the right corner just past R2 and R4 is V_P-8? no i dont think thats right .. why isn't it what I had? what's wrong with the 8?

Last edited: Jun 6, 2012
15. Jun 6, 2012

arl146

i don't understand why I4 is not right. Isn't it similar to I2?

16. Jun 6, 2012

azizlwl

I2 passes 2 resistors
I4 passes 4 resistors

17. Jun 6, 2012

arl146

3? well i dont get what the equation would be still with that information
I_4=(V_P-V_P'-8)/R4 so is this wrong then too?

Last edited: Jun 6, 2012
18. Jun 6, 2012

azizlwl

First equation/circuit remove R3, R4 and R5. You just got R1 and R2. Noted the I11 and I2.
Second equation/circuit, remove R2. Noted the I12 I3,I4 &I5

Combine them and you get the current and voltage on each resistor.

19. Jun 6, 2012

arl146

what youre saying makes no sense. well i got the first part, but the second part made absolutely no sense to me? i don't understand how to get the I4 equation i dont understand the potential difference across R4

20. Jun 6, 2012

arl146

would the potential difference across R4 be -V_P?
if across R3 it is (8-V_P) and across R5 it is (V_P-V_P') then [(8-V_P)-(V_P-V_P')] is the 'initial' and the final is 8. so [(8-V_P)-(V_P-V_P')]-8 is just -V_P?
or is it I4=(V_P'-2*V_P)/R4