Electrical question

Daniel S

I'm studying for an electrical engineering repeat module and I'm stuck on question one...

Anyway here's the question:



I know it's really feckin' simple but I'm struggling with it (mech engineer btw).

Am I right in thinking that:

Reffective of R2, R3 and R4 => (430)(220)/(430+220) => 145.54 Ohms

If I add a node "x" at the top right of the circuit that:
Vx=(10v)*(145.54+150)/(145.54+150+220)=5.733v

Basically I don't know what effect R4 and R5 have on the voltage/current at "A". All the material I have seems to be really really easy, then skips a fair bit and then I'm left with this.

brightspark

try redrawing it, so that it is more obvious what resistors are in parallel and what are in series

TheChizler

Try putting another node at the intersection of R3 and R4?

Daniel S

brightspark wrote: »

try redrawing it, so that it is more obvious what resistors are in parallel and what are in series

I know what resistors are in series and/or parallel, what I have a problem is the "Voltage divider", I don't get it.

TheChizler

Daniel S wrote: »

I know what resistors are in series and/or parallel, what I have a problem is the "Voltage divider", I don't get it.

It's just allocating the voltage to either resistor based on the ratio of the resisters to each other. Like allocating apples to John and Ann in the ratio 5:3.

brightspark

It's really a multistaged problem

work out the current flow though R1+(R2, R3, R4 = 145.54Ohms) and R5.

Ohms Law will then give you then give you the volt drop on R1 and R5.

This gives you the voltage remaining across R2, R3.

Calculate the current through R2, R3 and R4

Then work out the volt drop across R3 and add that to R5

pljudge321

I got an answer of 3.563 V if you want to check your answer.

TheChizler

pljudge321 wrote: »

I got an answer of 3.563 V if you want to check your answer.

I challenge your result with an answer of 3.567 V! Also 6.57 mA.

Daniel S

I get it now.


I still hate this bleedin' witchcraft people call electricity...

Daniel S

Turns out I made a balls of most of the questions I was doing. Can someone check my answers?

brightspark wrote: »

It's really a multistaged problem

brightspark wrote: »

work out the current flow though R1+(R2, R3, R4 = 145.54Ohms) and R5.

I got 145.54 Ohms also an 515.54 for Rt.

brightspark wrote: »

Ohms Law will then give you then give you the volt drop on R1 and R5.

Do I use the voltage divider here? For Vr1 I got 4.27v and for Vr5 I got 2.91v.

brightspark wrote: »

This gives you the voltage remaining across R2, R3.

So taking V(r2+r3)=10v-(4.27v+2.91v)=2.82v? Voltage drop is the same across R2 and R3 as it is across R4 as they're in parallel?

Do I use the voltage divider again?

Vr3=(2.82vx100ohms)/430ohms= 0.66v

Add this to Vr5 to get back up to node A:

Va=0.66v+2.91v=3.57v?

brightspark wrote: »

Calculate the current through R2, R3 and R4

The current will be the same at all points between R2 and R3 as they're in series right? So I can just treat that as one resistor? Using the current divider:

Iin=10v/515.54ohms=0.0194A

Ir3=0.0194(1/430)/(1/430+1/220)=0.00657A?

If I can do these questions, there's still some hope of me passing this exam!

pljudge321

Daniel S wrote: »

Turns out I made a balls of most of the questions I was doing. Can someone check my answers?


I got 145.54 Ohms also an 515.54 for Rt.

Do I use the voltage divider here? For Vr1 I got 4.27v and for Vr5 I got 2.91v.

So taking V(r2+r3)=10v-(4.27v+2.91v)=2.82v? Voltage drop is the same across R2 and R3 as it is across R4 as they're in parallel?

Do I use the voltage divider again?

Vr3=(2.82vx100ohms)/430ohms= 0.66v

Add this to Vr5 to get back up to node A:

Va=0.66v+2.91v=3.57v?



The current will be the same at all points between R2 and R3 as they're in series right? So I can just treat that as one resistor? Using the current divider:

Iin=10v/515.54ohms=0.0194A

Ir3=0.0194(1/430)/(1/430+1/220)=0.00657A?

If I can do these questions, there's still some hope of me passing this exam!

You got the same voltage answer as me and from memory that looks about right for the current. Do a couple more of them to build up your feeling for whats going on and it should click.

There are going to be numerous ways to solve each question so don't get bogged down about whether you should use voltage division or find the current flowing through each branch or the likes. After a while you'll start to recognize which method is fastest.

Daniel S

pljudge321 wrote: »

You got the same voltage answer as me and from memory that looks about right for the current. Do a couple more of them to build up your feeling for whats going on and it should click.

There are going to be numerous ways to solve each question so don't get bogged down about whether you should use voltage division or find the current flowing through each branch or the likes. After a while you'll start to recognize which method is fastest.

I've just done ten years of the question before lunch. I have the hang of them now. On to question 2,3 and 4. (I'll leave Q5-8 as the exam is capped at C3 (repeat) and I already have some marks from lab reports, attendance etc.).

Thanks again!