Understanding Kirchhoff's Laws: Solving a Parallel Circuit Problem

[Mdhiggenz]

Homework Statement

my work

I know part A is incorrect because for my part B my quadratic contains an imaginary,which I doubt he would due. Some help would be awesome since it is due tomorrow, and will probably be on the exam as well.

 

[Saitama]

Part A looks correct to me. However, in part B, i guess you missed that there are two R_o resistors instead of 2R_o. Rewrite the equation.

 

[Mdhiggenz]

Hmm so what you are saying is.

Req= R0 + (2ReqR0/Req+2R0)+R0 ?

If that were the case couldn't I could the R0's and just have a single 2R0?

Thanks for the help btw

 

[Saitama]

Hmm so what you are saying is.

Req= R0 + (2ReqR0/Req+2R0)+R0 ?

Yes, i mean that.

If that were the case couldn't I could the R0's and just have a single 2R0?

Yes you can have single 2R_o, but you have included two 2R0 in the equation you posted.

 

[Mdhiggenz]

Excellent I got Req= 1+-√5, and I use the positive value

Now in part c.

I'm a bit confused as to what I use.

I tried further condensing the picture I came up with in part A but that made no sense, then I reread what they are asking and I suppose I use the original picture and using kirchhoffs rule around the loop from acbd as labeled in the picture. However looking at the picture acbd does not include i1 so should I include acebdf?

Thank you

 

[gneill]

Now in part c.

I'm a bit confused as to what I use.

I tried further condensing the picture I came up with in part A but that made no sense, then I reread what they are asking and I suppose I use the original picture and using kirchhoffs rule around the loop from acbd as labeled in the picture. However looking at the picture acbd does not include i1 so should I include acebdf?

Thank you

You might want to think in terms of KCL rather than KVL. Effectively the current entering the junction divides into two paths and subsequently rejoins at another junction. You happen to know the resistances of the two paths, so you can work out how the current divides between the paths.

 

[Mdhiggenz]

I'm still rather confused, because I think I can solve for I1 using simply junction C.

I know that R0 is going in junction C and R0 is going out of junction C

But it doesn't add up unless I reverse the direction of one of the junctions.

For example, R0+R0=2R0 Obeys KCL

In which i1 would be 2R0.

 

[gneill]

I'm still rather confused, because I think I can solve for I1 using simply junction C.

I know that R0 is going in junction C and R0 is going out of junction C

But it doesn't add up unless I reverse the direction of one of the junctions.

For example, R0+R0=2R0 Obeys KCL

In which i1 would be 2R0.

R₀ is a resistance, not a current.

The network after the junction has equivalent resistance Req, so the current i_o divides between the resistance 2R_o and Req.

 

[Mdhiggenz]

So I0=Ix+i1

That much I understand,

What I don't understand is how do I use the relationship of Req and 2Ro to solve for I1.

The way you drew the picture it seems that i1 is = to Req?

 

[gneill]

So I0=Ix+i1

That much I understand,

What I don't understand is how do I use the relationship of Req and 2Ro to solve for I1.

The way you drew the picture it seems that i1 is = to Req?

Resistance is not current.

You might want to look up "current divider". Like the voltage divider, it's one of the more handy circuits to know the math for.

 

[Mdhiggenz]

I looked one up online, and want to make sure I have it correct.

It would be I0(Req)=I1

Where I0 is = to R0

 

[gneill]

I looked one up online, and want to make sure I have it correct.

It would be I0(Req)=I1

Where I0 is = to R0

No. I don't understand why you insist on equating current to resistance. They have different units. Current is in Amps, while resistance is in Ohms. For an equation to be correct, units must always agree.

Two resistors in parallel must have the same potential (voltage) across them. If R1 and R2 are the resistances and the currents through them are I1 and I2 respectively, then by Ohm's law the relationship

I1*R1 = I2*R2 ...(equating the voltage drops across each resistor due to their currents)

must hold. If the total current is I0 = I1 + I2, where I0 is a given (known) value, then you have two equations in two unknowns I1 and I2 which you should be able to solve for.