Calculating current from diagram.

[rcmango]

Homework Statement

from the diagram i drew I want to calculate the current in the battery for the circuit, can someone help me with this. Want to use symbols too. R1 and R2 are in series, they are in parallel with R3, and R4 is in series with R1, R2, and R3. The diagram is attached.

Homework Equations

The Attempt at a Solution

Im not sure where to start with this problem, can you get me jump started. The problem is that I realize that the diagram I have drawn has 3 resistors that are in parallel, then one that they will always be in series with. So I'm confused as to how to calculate the current when I have both parallel and series connections? I can't just use 1/r1 + 1/r2 + 1/r3/ ... etc, or series as well just series = r1+r2+r3.. etc. Please give me some help and stick with me with this today.

 

[gneill]

You want to work step by step, replacing sets of resistors that you can combine (series or parallel) with single resistors, until you are left with just a single resistor representing the whole resistor network.

To begin with, you've already identified R1 and R2 as being in series, so what single resistance value can replace them?

 

[rcmango]

Okay could we call R1 and R2 10 Ohms?

 

[gneill]

Okay could we call R1 and R2 10 Ohms?

Yes. So replace R1 and R2 with a single resistor of 10 Ohms. What's next? Do you see another series or parallel opportunity to pursue?

 

[rcmango]

Okay well 10 Ohms is in parellel with R3 could we add those together?

 

[cupid.callin]

yes we can ... and it'll give ___ .

 

[rcmango]

25 Ohms, so now I have 25 Ohms of parallel circuitry with an 8 Ohm series.

Whats next.

 

[cupid.callin]

25 Ohms

how will it give 25Ω?

what formula do we use for parallel resistor combination?

 

[rcmango]

Oh I see for the parallel connects, we must use this: (1/10 + 1/15) = 1/66_7 = 5.9999 Ohms? Now what about the last series, just add that together?

 

[gneill]

Oh I see for the parallel connects, we must use this: (1/10 + 1/15) = 1/66_7 = 5.9999 Ohms? Now what about the last series, just add that together?

Call it 6 Ohms: (1/10 + 1/15)^-1 = 10*15/(10 + 15) = 6

If resistors are in series you add them...

 

[rcmango]

Oh okay, now its just two resistors in series, so 6 Ohms + 8 Ohms

So the total Ohms of the circuit: 14 Ohms

 

[gneill]

Oh okay, now its just two resistors in series, so 6 Ohms + 8 Ohms

So the total Ohms of the circuit: 14 Ohms

Right. So there's a battery (of some unspecified voltage V) connected across 14 Ohms. What's the current?

 

[rcmango]

V=i*r

So current of this circuit would be (Battery Voltage/ 14 Ohms)= current (Amps)

 

[cupid.callin]

Yes that's correct.

 

[rcmango]

Thankyou!