Find the Value of Resistor R Please

[Andre36]

So Im hear doing some homework and I cannot figure how to find the value of resistance r in both of these diagrams. Can someone show me the formula pls. You can click to enlarge them.

Thanks.

Diagram 1




Diagram 2

 

[Knobhead]

Show what you've worked out so far.

I'm sorry but I will not do your home work.

 

[spark 68]

Resistors in series = R total = (R1 + R2 + R3........Rn.)

Resistors in parallel R total = 1/(1/R1 + 1/R2 + 1/R3...1/Rn)


Hint: Ohms Law

V=IR

R=V/I

I=V/R

You will need to work out the series and parallel parts out seperately.

I hope this helps you

 

[Knobhead]

I take it you've not even attempted to solve it.

OP legged it after being asked what he’d done.

 

[Andre36]

Show what you've worked out so far.

I'm sorry but I will not do your home work.

I have yet to work out anything so far because the whole thing is confusing me. I am electrical installations student and I am not as experienced as you are.

I understand OHMs Law but I'm not sure how to apply it in these examples to get the answer.

 

[Andre36]

Resistors in series = R total = (R1 + R2 + R3........Rn.)

Resistors in parallel R total = 1/(1/R1 + 1/R2 + 1/R3...1/Rn)


Hint: Ohms Law

V=IR

R=V/I

I=V/R

You will need to work out the series and parallel parts out seperately.

I hope this helps you

Thanks for your help. It's pretty late now and Im tired so I will go through this post again in the morning. Thank you.

 

[Knobhead]

OK

A:
Work out the volt drop across the parallel resistors, you have the current flow through one of them and it’s value. Subtract the voltage drop across the parallel resistors from the supply voltage, using this work out the total current flow through the single resistor. With the total current flow now known along with the current flow in one leg of the parallel network you can then work out the current flow and resistance of the 2[SUP]nd[/SUP] leg.

B:
You have the supply voltage and total current flow from which you can work out the total resistance of the whole circuit. Work out the value of the parallel network and subtract from the total.

It’s taken me ten times longer to write this than do the calculation, but if I just give you the answer you learn nothing.
One thing you have to learn is to split a circuit/question down in to constitute parts.

Let me know the answers you get

 

[Deleted member 9648]

Andre36 there has been a number of posts where students are simply using the forum to get their homework done for free...members have got wise to this hence the suspicion.....this may or may not apply to you,I suspect not as you have not simply asked for the answer. But surely this homework has not simply been dumped on you with no previous?...have you not covered this in class or do you not have a theory textbook to refer to?

 

[Smudge]

at the risk of being hounded by the mass's, the OP states he is doing homework and is openly asking for help with the formulas, not for the answers. Granted its in the wrong section of the forum, but i thought the point of this forum was to help each other on areas of weakness?

Andre36, if you need help with the theory, pm me and i will do my best to help guide you to solving it for yourself. I'm no genius by any means, but where i can help, i will.

 

[Andre36]

Andre36 there has been a number of posts where students are simply using the forum to get their homework done for free...members have got wise to this hence the suspicion.....this may or may not apply to you,I suspect not as you have not simply asked for the answer. But surely this homework has not simply been dumped on you with no previous?...have you not covered this in class or do you not have a theory textbook to refer to?

I understand this. It's not like I was asking him to complete the whole sheet, It was ONE question! Thanks for your understanding. No, the homework was not just dumped on us, your right...BUT we didn't have much time to go through it either and the teacher's time was taken up with an other group so I didn't have time to go over it with him. I am on my half term break now and we were told to finish it over our half term. I have one text book but it's no use as it's mainly about health and safety. I will go to my local library to see what they have.

Thanks.

 

[parker56]

Electrical Installation Calculations: Basic: For technical certificate Level 2: For Technical Certificate
I found this book a great help,just seen you can download to kindle.

 

[Andre36]

OK

A:
Work out the volt drop across the parallel resistors, you have the current flow through one of them and it’s value. Subtract the voltage drop across the parallel resistors from the supply voltage, using this work out the total current flow through the single resistor. With the total current flow now known along with the current flow in one leg of the parallel network you can then work out the current flow and resistance of the 2[SUP]nd[/SUP] leg.

I have worked it out this way:

6/15 = 0.4a
6/5 = 1.2a
=1.6a combined
6/1.6 = 3.75r


B:
You have the supply voltage and total current flow from which you can work out the total resistance of the whole circuit. Work out the value of the parallel network and subtract from the total.

30/0.5 = 60Rt

It’s taken me ten times longer to write this than do the calculation, but if I just give you the answer you learn nothing.
One thing you have to learn is to split a circuit/question down in to constitute parts.

Let me know the answers you get

Have I got them right?

 

[JorgeC]

I people!
At first excuse my bad english.
Now the problem, i agree with Tony at first diagram, but about the second diagram i only see the current flow passing trough the 20 ohm resistor, and not the totally flow current circuit, so the current is divided to the two parallel resistors. It is necessary to calculate the current flow at other branch.
I hope to be usefull for you Andre36.

 

[derek preece]

For the first question Iwould have worked out the total resistance of the circuit using ohms law R=V/I

Then work out the value of the resistances in parallel using 1/R = 1/R1+ 1/R2

Once you have these values you can subtract the total value of the resistances in parallel from the total resistance of the circuit giving you the resistance value required.
If you look at Tony's post you'll see that he has put that as 'B'

 

[tony mc]

Andre did you work out the answers ?

If so I would like to see what you put for picture 2 as I think you may have copied down wrong or am I missing something?

You need 55 ohms in the parallel bit but with a 20 omhs resistor already there you reading will be under 20 ohms.

I must be missing something !!!!

 

[spikester]

Andre

For diagram 1 work out the parallel network resistance and using the 1A calculate the voltage across the network, take this voltage away from the supply and that is the voltage across r, then it is ohm law again to work out r.

For diagram 2 uses ohm's law for the 20 ohms and 0.5A then this voltage can by taken away from supply to work out voltage across 5 ohms. For the 5 ohms calculate I then take 0.5A away from it and you now have I and V across r, simple ohms law again to work out r.

Hope this helps.

 

[JorgeC]

Hi! people!
Spikester is right and had answered the question. We know the resistance value ( 20 ohm and is cross with 0,5A),so now using the law ohm we get the voltage applied at parallel resistance. knowing this value is easy to get the voltage on 5 ohm resistance and then the current.

Now the rest is easy!

 

[tony mc]

Just had another look and can see what I was missing and the position of the 0.5A. I took it as 0.5 A for the entire cct at first must use my Mr Magoo glasses more often !


Thanks