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Old Apr 30th 2018, 06:50 AM   #11
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Originally Posted by Musawwir View Post
yes you just take the equivalent resistance divide by the voltage
You need to be more careful in your working, particularly what you divide by what.

Current = Voltage / resistance.

To move on, bearing in mind you have correctly calculated the current in R4.

So once you have calculated the total current can you see what the current in R3 is?

One of the various stages I labelled 1 to 6 should help,Can you see which one?


Note I am trying to make sure you understand where each part of the calculation comes from so you can do the next one yourself.

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Old Apr 30th 2018, 07:10 AM   #12
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yes no prob
so the current in R3 is the total current - current in R4 using Kirchhoff's current law
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Old Apr 30th 2018, 07:16 AM   #13
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Originally Posted by studiot View Post
You need to be more careful in your working, particularly what you divide by what.

Current = Voltage / resistance.

To move on, bearing in mind you have correctly calculated the current in R4.

So once you have calculated the total current can you see what the current in R3 is?

One of the various stages I labelled 1 to 6 should help,Can you see which one?


Note I am trying to make sure you understand where each part of the calculation comes from so you can do the next one yourself.


ps if you can help me for this one it would be great
Resistance combinations help-opamp.png
do you get 2.91 V
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Old Apr 30th 2018, 07:27 AM   #14
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Have you given up on the much simpler circuit?

So far you have two part correct and two parts incorrect.

As to the diff amp circuit, can you see the value of labelling things?

You should be able to tell V4 by inspection, I would label that amp as A2.
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Old Apr 30th 2018, 07:48 AM   #15
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which parts are incorrect?
for the opamp v4 is equal to V2 but is Vout 2.91?
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Old Apr 30th 2018, 08:05 AM   #16
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current through 18 ohms is just 6/18 a YES
and 12 ohms is just 6/12 NO
and power dissipation is (r3)^3(36)/r1+r3 NO where did this come from?
And equivalent resistance is 7.41 ohm? YES
I am not going to simply check your arithmetic for you.
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Old Apr 30th 2018, 09:19 AM   #17
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current through 12 ohm is (6/12+4.5)
second part
this came from the voltage divider rule since 4.5 and 12 ohms are in series
V(4.5)=(4.5/4.5+12)*6
then power dissipated is V^2/R
which equals to
(V(4.5))^2/4.5
and that power dissipated in 4.5 ohm is also eqaul to the I2R
then current though 4.5 ohm using kirchofff law adds with current of 18 ohms which then equals to total current in circuit thus using this relationship current through 4.5 ohm can be find and then simply use P=I^2R this is my reasoning
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Old Apr 30th 2018, 10:23 AM   #18
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current through 12 ohm is (6/12+4.5)
second part
this came from the voltage divider rule since 4.5 and 12 ohms are in series
This is still not right, because some of the current in R3 does not pass through R1, so they are not technically in series.

If you were to cooperate then we would have finished this question by now and I like to think you would also be understanding what you are doing.

I asked for the total current which you are right in saying is given in my figure 7 as the total voltage divided by the R7, the equivalent resistance, but you did not calculate it.

But you need to be more precise in what you are saying

yes you just take the equivalent resistance divide by the voltage
The voltage?
Which voltage?

Note I said the total voltage (6 volts)

So the total current = 6/7.415 = 0.81 amps.

Now I also asked you

So once you have calculated the total current can you see what the current in R3 is?

One of the various stages I labelled 1 to 6 should help,Can you see which one?
So I labelled on of the junctions or nodes in figs 1 and 2 with an A.

This junction has 3 branches

The total current into the battery
The current into R4
The current into R3

So the current into R3 + Current into R4 must = Total current

Can you now calculate the current in R3?
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Old Apr 30th 2018, 12:30 PM   #19
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Originally Posted by studiot View Post
This is still not right, because some of the current in R3 does not pass through R1, so they are not technically in series.

If you were to cooperate then we would have finished this question by now and I like to think you would also be understanding what you are doing.

I asked for the total current which you are right in saying is given in my figure 7 as the total voltage divided by the R7, the equivalent resistance, but you did not calculate it.

But you need to be more precise in what you are saying



The voltage?
Which voltage?

Note I said the total voltage (6 volts)

So the total current = 6/7.415 = 0.81 amps.

Now I also asked you



So I labelled on of the junctions or nodes in figs 1 and 2 with an A.

This junction has 3 branches

The total current into the battery
The current into R4
The current into R3

So the current into R3 + Current into R4 must = Total current

Can you now calculate the current in R3?
here it is i did it using two different ways one of which you asked me and the other one using the third diagram you provided and now if you tell me its wrong then physics must be broken
Resistance combinations help-dsc_0047.jpg
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Old Apr 30th 2018, 12:50 PM   #20
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Well done.

Not only have you correctly calculated the current in R3 you have found another way I was going to come to.

So now you know the current in R3 and its value, you can calculate two things.

1) You can calculate the voltage dropped across it

2) You can calculate the power dissipated by it.


1) You can use this voltage to calculate the voltage across R5 in my fig 4 which is therefore the voltage across R1 and R2 in my fig 2
(There is also another way, can you spot it ?)

2) Is a value you are asked to calculate.


Having calculated the voltage across R1 you can to find your last answer.




With regards to your compound amp,

Call the top kleft amp A1
Call the middle amp A2
Call the top right amp A3


You were right about V4 = V2.

Can you describe the function of each amp and calcualte the gain of that amp?
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