Finding voltage across a resistor in a circuit

[nothing909]

Find the voltage across the 4Ω resistor (resistor 3).

This is all show in the pictures provided:

I have started by reducing all the resistors into one resistor, but please take a look over my working and tell me if it's all correct.

After reducing all the resistors into one, I used the resistance to calculate the supply current.

Then, with the supply current, I calculate the voltage over the first resistor. I then subtracted the first resistors voltage with the supply voltage to work out resistor 2's voltage.

At this point I'm stuck. I don't understand where to go from here to work out the voltage over the 4Ω resistor. I know I haven't got long to go

 

[CWatters]

Are you familiar with the potential divider rule ?

PS: I checked up to image 484 and I think your answer at that stage 2 + 2.968 is correct.

 

[nothing909]

Are you familiar with the potential divider rule ?

no, can you explain, please?

 

[CWatters]

Ok if not familiar with the potential divider rule (yet) I will suggest another method of proceeding...

With reference to image 484. You have calculated the voltage on the 2.968R resistor is 17.922V. So mark that voltage on the earlier drawing 483. In fact you can pretend/draw a new voltage source exists across the 2.968R resistor.

You will be able to ignore (erase) components to the left of the new voltage source and work out the voltages and currents on the right of the new voltage source.

If you don't understand let me know and I will try and mark up your drawing.

 

[nothing909]

Ok if not familiar with the potential divider rule (yet) I will suggest another method of proceeding...

With reference to image 484. You have calculated the voltage on the 2.968R resistor is 17.922V. So mark that voltage on the earlier drawing 483. In fact you can pretend/draw a new voltage source exists across the 2.968R resistor.

You will be able to ignore (erase) components to the left of the new voltage source and work out the voltages and currents on the right of the new voltage source.

If you don't understand let me know and I will try and mark up your drawing.

Can you explain a little more? Explain the potential difference maybe, see if I can understand. What do you mean "you can pretend/draw a new voltage source?"

 

[donpacino]

Hey nothing909,

I've you out with a very similar problem in one of your previous posts. You can look at that for more guidance

 

[CWatters]

Ok this is what I meant about marking up your drawing 483...

 

[CWatters]

Does that suggest a way you can work out the voltage across the 4R?

 

[nothing909]

Hey nothing909,

I've you out with a very similar problem in one of your previous posts. You can look at that for more guidance

I have looked at it. The only reason I'm getting confused with this is because there's so much extra resistors, it's just confusing me.

 

[nothing909]

Does that suggest a way you can work out the voltage across the 4R?

Umm, not really, I still don't understand.

 

[nothing909]

The voltage across the 4 ohm resistor is 12 (or around that). That's the answer, I just don't know how to get it.

 

[CWatters]

Ok you calculated that the voltage across the 2.968 resistor was 17.922V.

So then working backwards through your drawings..

Start at the bottom of image 483 (the one I modified)...
The 2.968R came from a 6R and a 5.875R in parallel. So the 17.922V is also the voltage you would measure on the 6R and the 5.875R.

Then look at the top of image 483..
The 5.875R came from the 4R and the 1.875R in series. So you will also have 17.922 across the series combination of the 4R and the 1.875R .

 

[CWatters]

The voltage across the 4 ohm resistor is 12 (or around that). That's the answer, I just don't know how to get it.

You have 17.922V across the 5.875R so the current through it is?

 

[nothing909]

You have 17.922V across the 5.875R so the current through it is?

3.051

 

[nothing909]

Ok you calculated that the voltage across the 2.968 resistor was 17.922V.

So then working backwards through your drawings..

Start at the bottom of image 483 (the one I modified)...
The 2.968R came from a 6R and a 5.875R in parallel. So the 17.922V is also the voltage you would measure on the 6R and the 5.875R.

Then look at the top of image 483..
The 5.875R came from the 4R and the 1.875R in series. So you will also have 17.922 across the series combination of the 4R and the 1.875R .

ohhh, do i just do voltage division between the 4 ohm resistor and the 1.875?

 

[CWatters]

The same 3.051A flows through both the 4R and the 1.875R as they are in series so the voltage drop across the 4R is?

 

[CWatters]

ohhh, do i just do voltage division between the 4 ohm resistor and the 1.875?

Yes. If you are familiar with the voltage divider :-)

 

[nothing909]

The same 3.051A flows through both the 4R and the 1.875R as they are in series so the voltage drop across the 4R is?

14.04?

 

[nothing909]

Yes. If you are familiar with the voltage divider :-)

so...

17.922x4/5.875 = 12.202v

ye, is that correct? that's the voltage across the 4 ohm resistor?

 

[CWatters]

Sorry we cross posted there.

 

[CWatters]

Correct. You can also do...

3.051 * 4 = 12.202

 

[nothing909]

thank you, i knew it was easy as that. took me long enough to try and figure out how to reduce all the resistors into one. xD thanks anyways.