Could someone possibly give me tge

Jony130

Joined Feb 17, 2009
5,487
Well your Rth looks good, but Vth for sure is not equal to 3V. In fact Vth value is very strange, are you sure you do not have any error on the diagram?
 

Thread Starter

serkefti

Joined Dec 16, 2014
15
Well your Rth looks good, but Vth for sure is not equal to 3V. In fact Vth value is very strange, are you sure you do not have any error on the diagram?
thanks for ur respond?
which diagram u are talking about? the actual question paper ?
mine I did that with some help so don't really know what im doing.:confused:
 

Jony130

Joined Feb 17, 2009
5,487
I was talking about original question paper. For me the voltage source should be swap in places. Because now Vth = 0V .
 

Jony130

Joined Feb 17, 2009
5,487
But for this circuit Vth is 0V. So what is the point of doing part B if Vth = 0V. If Vth = 0V no current will flow through load resistor and if so no power transfer will occur.
And this is why I vote that the voltage source must be swapped
 

WBahn

Joined Mar 31, 2012
29,976
As given it's 0V, but that's just what it is. Keep in mind that the next part of the problem forces a change in one of the resistances, which will result in a different Vth.
 

Thread Starter

serkefti

Joined Dec 16, 2014
15
As given it's 0V, but that's just what it is. Keep in mind that the next part of the problem forces a change in one of the resistances, which will result in a different Vth.
so what would be the answer of part B as I really don't know what to do ?o_O
 

WBahn

Joined Mar 31, 2012
29,976
what do u mean?
Ive already done the Part A.
My apologies. I had another thread in mind.

So let's get your Vth correct. You aren't showing how you came up with 3V for Vth, you just state it. Show how you got that result so that we can try to figure out where you went wrong.
 

Thread Starter

serkefti

Joined Dec 16, 2014
15
My apologies. I had another thread in mind.

So let's get your Vth correct. You aren't showing how you came up with 3V for Vth, you just state it. Show how you got that result so that we can try to figure out where you went wrong.
to be honest I guessed that bit.
so not really sure what I was doing:(
 

Thread Starter

serkefti

Joined Dec 16, 2014
15
I=V/R= (9+3)/(100+300) = 3/100 amps
V100 ohm= (3/100) * 100 = 3 volts so VB = 3V

V ab= Va - Vb= 3-3 =0volt
so I should put 0 volts instead of 3 v

but how about part B?
 

WBahn

Joined Mar 31, 2012
29,976
I=V/R= (9+3)/(100+300) = 3/100 amps
V100 ohm= (3/100) * 100 = 3 volts so VB = 3V

V ab= Va - Vb= 3-3 =0volt
so I should put 0 volts instead of 3 v
You are taking mostly the right approach, but do need to clean things up a little bit.

Starting with something like Vab = Va - Vb is a good, safe way to work from, but you need to recognize that while the voltage between two points (e.g., Vab) is well-defined, that the voltage AT a point (e.g., Va or Vb) is not well-defined. It requires an reference point that is declared to be 0V, and you have not done this. So when you say Vb = 3V, you really haven't said anything because the question that is unanswered is, 'Vb is 3V relative to what?". In fact, I think you got lucky because you just took a current and multiplied it by a resistance and got a voltage and called that Vb. What you actually found is Vbc, the voltage across the 100Ω resistor, where Point B is the side of the resistor that the current is going into and Point C is the side of the resistor that the current is leaving from. Hence, probably without realizing it, you are saying that Point C is your ground reference and this just happens to be the negative terminal of the 3V supply and using that voltage for Va just happens to result in you using the same reference point for both. A lot of "just happens" at play here, so make sure that you understand this.

but how about part B?
Label the resistors R1, R2, and R3 and then write Rth in terms of those resistances. Do you know what value you want Rth to be? Then set that equal to the value that you want it to be and see which, if any, of the values you can change to satisfy that equation. Since you are allowed to change only one value, you can treat it as three different problems of the form: If I only change R1, what would R1 have to be in order to make Rth the value I want?
 

Thread Starter

serkefti

Joined Dec 16, 2014
15
You are taking mostly the right approach, but do need to clean things up a little bit.

Starting with something like Vab = Va - Vb is a good, safe way to work from, but you need to recognize that while the voltage between two points (e.g., Vab) is well-defined, that the voltage AT a point (e.g., Va or Vb) is not well-defined. It requires an reference point that is declared to be 0V, and you have not done this. So when you say Vb = 3V, you really haven't said anything because the question that is unanswered is, 'Vb is 3V relative to what?". In fact, I think you got lucky because you just took a current and multiplied it by a resistance and got a voltage and called that Vb. What you actually found is Vbc, the voltage across the 100Ω resistor, where Point B is the side of the resistor that the current is going into and Point C is the side of the resistor that the current is leaving from. Hence, probably without realizing it, you are saying that Point C is your ground reference and this just happens to be the negative terminal of the 3V supply and using that voltage for Va just happens to result in you using the same reference point for both. A lot of "just happens" at play here, so make sure that you understand this.



Label the resistors R1, R2, and R3 and then write Rth in terms of those resistances. Do you know what value you want Rth to be? Then set that equal to the value that you want it to be and see which, if any, of the values you can change to satisfy that equation. Since you are allowed to change only one value, you can treat it as three different problems of the form: If I only change R1, what would R1 have to be in order to make Rth the value I want?
thanks very much. I got the answer of part B as well. It happened to be 60 ohms. I've changed the value of 100 ohms
 

WBahn

Joined Mar 31, 2012
29,976
thanks very much. I got the answer of part B as well. It happened to be 60 ohms. I've changed the value of 100 ohms
Good. So why did you change the 100Ω resistor? That was part of the question, too. Also, what is the Vth for the new circuit?
 
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