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Op-amp task

Hello I was wondering how the approach is to solve this exercise?

I wonder how to find u+ = 2.5 V?

Any suggestion ?
 

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R5 and R6 are connected in series with a positive voltage at on end and a negative voltage at the other end.
You can find the current from the total voltage and the total resistance.
Then calculate the voltage across each resistor and find U+ relative to one end and then shift the reference to the common.

The values are chosen so the sums can be done in your head.
 

Harald Kapp

Moderator
Moderator
You can find the current from the total voltage and the total resistance.
Then calculate the voltage across each resistor and find U+ relative to one end and then shift the reference to the common.
Or use the common equation for a voltage divider.

This is so obvious:rolleyes:
 
When I used the voltage divider law:

(7,5+12,5)*3/20+10 = 2 V

I know this is wrong, could someone explain too me why?

Break it down into simple steps.
Total resistance of R5 and R6.
Total voltage across R5 and R6.
Current through R5 and R6.
Voltage drop across R5. subtract that from +10 V.
 
Break it down into simple steps.
Total resistance of R5 and R6.
Total voltage across R5 and R6.
Current through R5 and R6.
Voltage drop across R5. subtract that from +10 V.
Thank you for a good answer. Maybe this question is very easy anyhow, how do you find "Total voltage across R5 and R6". Why do you need solve only the voltage drop across R5, why not R5+R6?



Rotal = 7,5+12,5 =20 ohm
 
Thank you for a good answer. Maybe this question is very easy anyhow, how do you find "Total voltage across R5 and R6". Why do you need solve only the voltage drop across R5, why not R5+R6?



Rotal = 7,5+12,5 =20 ohm

See where R5 and R6 go. One side goes to +10 V the other side goes to -10 V, so a total of 20 V across a total of 20,000 ohms. Easy calculation, 1 mA. Your point of interest is 7,500 ohms down from +10 V, at 1 mA that puts it at+2.5V. It works out the same if you use the other resistor.
 

Harald Kapp

Moderator
Moderator
Why? It doesn't tell you what the op amp is so there is no way to know. Given an ideal op amp, the impedance of the op amp is of no concern.
The input ompedance of an ideal opamp is infinity, that's why it is of no concern and that was where I was hinting at.
 
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