Calculating voltage drp in simple circuits

[thecat]

Ok I'm doing a tafe course, teacher is hard to understand and I missed a couple of lessons and am seriously struggling to gt my head around this...

I'm Ok with resistance.. series you just add them up, parallel I always throught you divided the product by the sum. Old mate says 1/r=1/r1=1/r2... which I think works out as the same thing

But I'm struggling to work out voltage drops in mixed circuits

can anyone talk me through this. I know it should be easy but my brain just isn't working
-------------------------- l--------------R3------------l
______l ------R1----------l--------------300------------l
______l------140_____l___l__________________________l-----------l
______l_____________l____l------R4--------r5-----------l__________l
_____60v___________R2 800____ 200_____ 400___________________l
______l_____________l________________________________________l
______l_____________l________________________________________l
______l---------------l----------------------------------------------l

1 Find equivalent resistance for circuit
2 Find total current taken for circuit
3 What is the voltage drop across R3
4 Calculate power taken by R5

 

[tommy gun]

Volt Calc?

Hi Mate
When you get the answer could you let me know
My son is also doing Tafe and trying to help him
Ian

 

[Dales Cannon]

If I read your circuit correctly...

Replace R3, R4 and R5 with an equivalent resister Rt: 1/Rt=1/R3+1/(R4+R5)

Then you have a simple series parallel circuit with R1, R2 (parallel) and Rt. So then calculate total R and find the current flow and work back with i over each resistor?

yes?

 

[Steve-0]

What dales said. Find the current through the circuit and work it back.

 

[Case king]

Current will split through the parallel part involving r3, r4 &r5. once you work out the voltage drop across it you can find the current through r3 and the remaining current will go through r4 &r5 and then you can work out the power used in r5

 

[3viltoast3r]

I'm Ok with resistance.. series you just add them up, parallel I always throught you divided the product by the sum. Old mate says 1/r=1/r1=1/r2... which I think works out as the same thing Your way is right.


1 Find equivalent resistance for circuit
Rx = R3||(R4+R5)
Then Ry = Rx||R2
Then Req = Ry + R1

So basically, add series, then simplify parallel and repeat.
2 Find total current taken for circuit
I = V/R=60/Req
3 What is the voltage drop across R3
Find the voltage drop across R1,The connection where r3, r4 and r2 is all at the same voltage. Find the drop across R1.
Vr1 = I*R1. Then voltage across r3 is Vr3 = 60-vr1
4 Calculate power taken by R5
The voltage Vr3 is over the branch R= R4 + R5. So you have I = Vr3/(R4+R5). Then Voltage drop across R5 is easy to find, Then P = VI

Hope this helps!

 

[thecat]

Champions.

Thanks for that

 

[RYDA]

This thread just brings back horrible memories of highschool physics and it's electricity. I don't envy you guys.