R1 + R2 in Parallel with R3 + R4 + R5 in Series | Find Req

[arl146]

Homework Statement

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".

Homework Equations

parallel: R_eq = 1/((1/R1)+(1/R2)+...)
series: R_eq = R1 + R2 + R3 +...

The Attempt at a Solution

well I am thinking that R1 and R3 is in series and R2 and R4 is in series but from there I am not sure what to do. the R5 gets me confused

 

[ehild]

The Attempt at a Solution

well I am thinking that R1 and R3 is in series and R2 and R4 is in series but from there I am not sure what to do. the R5 gets me confused

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

 

[jayaganesh]

Use wheat stone bridge idea to solve this.

 

[arl146]

i don't really know it ..

but uhhh oh .. I am 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?

 

[ehild]

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

ehild

 

[arl146]

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

 

[ehild]

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

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

 

[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

 

[ehild]

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

ehild

 

[arl146]

nothing in my textbook about it .. professor briefly went over it today in class. there's 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 R4can'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?

 

[willem2]

nothing in my textbook about it .. professor briefly went over it today in class. there's 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

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', which you can solve.

 

[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 that's 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?

 

[ehild]

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'-[STRIKE]8[/STRIKE])/R4 ?
if that's 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?

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

ehild

 

[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 don't think that's right .. why isn't it what I had? what's wrong with the 8?

 

[arl146]

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

 

[azizlwl]

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

I2 passes 2 resistors
I4 passes 4 resistors

 

[arl146]

3? well i don't get what the equation would be still with that information
I_4=(V_P-V_P'-8)/R4 so is this wrong then too?

 

[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.

 

[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 don't understand the potential difference across R4

 

[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

 

[azizlwl]

Out of 5 resistors you can make only 3 circuits/equations out of it. Thus to solve the problem you have to have only 3 unknowns.

1. R1 & R2
2. R1 & R5 & R4
3 R3 & R4

Can you create only 3 unknowns current?

 

[ehild]

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

The terminals of R4 are connected to P' and to the negative terminal of the battery. The current I4 flows from P' to the negative terminal. We used the negative terminal as the zero point for the potential.
So the current flows from a point at potential V_P' to a point at potential 0. The potential drops along R4 by V_P'-0. You get I4 dividing that potential difference by the resistance.


ehild

 

[ehild]

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

It depends how you choose the direction of the current I4. If its flows from P' to the negative terminal of the battery you need the difference V_P' and zero. That means I4=V_P'/R4
By the way, [(8-V_P)-(V_P-V_P')]-8 =8-V_P-V_P+V_P'-8=V_P'.

ehild

 

[ehild]

azizlwl: You suggest the loop method, but the OP started the node method. His method is correct, no need to change it.

ehild

 

[arl146]

I think i may have thought that at one point. well my homework was already due so i didn't get it. but still, i have those 7 equations. I can substitute the 5 equations (1 for each I) into the 2 current equations and end up getting VP and VP' so what do I then do with it to find the Req? do i add the currents for I(total) and then use V=I(total)*Req? where V is just 8 or...

 

[ehild]

I think i may have thought that at one point. well my homework was already due so i didn't get it. but still, i have those 7 equations. I can substitute the 5 equations (1 for each I) into the 2 current equations and end up getting VP and VP' so what do I then do with it to find the Req? do i add the currents for I(total) and then use V=I(total)*Req? where V is just 8 or...

Yes, add the currents that flow into the battery, and divide the voltage of the battery with I total.


ehild

 

[arl146]

ok thanks. when the answers become available ill have to check to see if i would have been right

 

[azizlwl]

azizlwl: You suggest the loop method, but the OP started the node method. His method is correct, no need to change it.

ehild

Noted.
Maybe let me complete my method. It's really messy with linear algebra and solve it with online solver(can we use this in exam ). Hope there's easier approach.
Subt. I1=x, I2=y, I3=z

4x+6y=8 ...(1)

4x+4(x-y)+7(z+(x-y))=8
4x+4x-4y+7z+7x-7y=8
15x-11y+7z=8 ...(2)

2z+7(z+(x-y))=8
9z+7x-7y=8
7x-7y+9z=8 ...(3)

With the help of online solver,
http://easycalculation.com/matrix/matrix-algebra.php
Shows wrong assumption of current flow.

 

[ehild]

ok thanks. when the answers become available ill have to check to see if i would have been right

It is a rather ugly calculation... What have you got for the resultant resistance? Is it something near 4.6 ohm?

ehild

 

[arl146]

what happens if i get a negative value for V?
i got V_P' = 3.4413 V and V_P = -3.709497 V
may be my calculations but ...i ended up with 0.8477 for Req

 

[Laphanga]

Use Delta to Wye Transformation. It's another way to find equivalent resistance of this network.

 

[ehild]

what happens if i get a negative value for V?
i got V_P' = 3.4413 V and V_P = -3.709497 V
may be my calculations but ...i ended up with 0.8477 for Req

Sorry, I did not notice that your I2 was also wrong. I2 flows from potential V_P to zero, so I2=Vp/R2.

ehild