Finding current through resistor (Kirchhoff's Laws)

[BMcC]

Homework Statement

Consider the circuit shown in the diagram below, for R1 = 5 Ω, R2 = 8 Ω, R3 = 8 Ω, R4 = 8 Ω, and V0 = 8.0 V. Calculate the current through R4.

Homework Equations

Loop rule: The sum of all potential changes around a closed loop is zero

Junction rule: The currents entering and leaving a junction are equal

V = IR

The Attempt at a Solution

I guess I'm kind of lost with where to begin. The examples in class were a little more simplified but I'll say what my thought process is to solving it.

First I found the equivalent resistance through the whole circuit

1/R4 + 1/R3 = 1/Req₃₄
1/8 + 1/8 = 1/Req₃₄

Req₃₄ = 4 Ω

Then that resistance + R2 = the equivalent resistance on the far left loop

4 + 8 = 12 Ω

Once more, to find the equivalent resistance between that and R1, I went

1/12 + 1/R1 = 1/Req

Req for the circuit = 3.53 Ω

Then to find the current through R4, I've been using V = IR where V = 8.0 V and Req = 3.53

I = 2.27 A

This is incorrect, I'm just not quite sure how else to approach this problem. Am I on the right track?

Thanks!

 

[Hollumber]

Yes, you are definitely on the right track. Right now, it looks like you have calculated the circuit current (current through the battery). If this was a series circuit, then everything would also have the same current, but because there are some parallel components, you must calculate how much current goes through each "fork" (use junction rule).

Remember that the lower the resistance, the more current that's going to go through that wire when it splits, so you will have to calculate the Req of each wire coming from each fork. (You are doing very well, so I think this is enough guidance for now. If you are still stumped, let us know ^_^)

 

[Adithyan]

Potential difference across R₄ is not 8.0 V. Voltage of the battery divides itself between R_eq34 and R₂. Think how will 8.0V divide itself.

[Hint: Take the potential difference as V₁ and V ₂ for potential difference across R_eq34 and R₂ respectively. After that find out the ratio [itex]\frac{V₁ }{V₂ }[/itex] , using the relation V=IR]

 

[BMcC]

Yes, you are definitely on the right track. Right now, it looks like you have calculated the circuit current (current through the battery). If this was a series circuit, then everything would also have the same current, but because there are some parallel components, you must calculate how much current goes through each "fork" (use junction rule).

Remember that the lower the resistance, the more current that's going to go through that wire when it splits, so you will have to calculate the Req of each wire coming from each fork. (You are doing very well, so I think this is enough guidance for now. If you are still stumped, let us know ^_^)

Okay, so I have the total circuit current and all of the individual resistances and the battery's voltage. I'm a little confused on where to go from here to be honest!

 

[majormaaz]

Okay, so I have the total circuit current and all of the individual resistances and the battery's voltage. I'm a little confused on where to go from here to be honest!

I'd say that since R₄ is pretty much the last resistor you get to, you might as well take it leg by leg. For example, have you noticed that R₁ branch is parallel to the whole equivalent R₂₃₄ branch? What do you know about the voltage across parallel branches?

After that, it's just Junction Rule to find out how much current remains after the R₁ branch, and you're on your way.

 

[Rellek]

It looks like you have two branches of wire that share the same two nodes. This means they must be equipotential, right?

On the second branch, notice again that your third and fourth resistor share nodes, so these are also equipotential.

Perhaps if you could find the voltage drop over R₂, and subtract that from the total voltage drop over that length of wire, you could find the drop over your two parallel resistors, R₃ and R₄

 

[NascentOxygen]

Using a coloured pen, trace the complete circuit taken by the battery current passing through R4.