I got a question regarding the following system.

enter image description here

The only given value is the input voltage $$U_e = 8V.$$ Now the task is to find the output voltage $$U_a.$$ My question now is how do I find this value, if I only know the input voltage and that the computing amplifier is an ideal computing amplifier? I tried to find a formula for the value that is being searched and came up with the following solution: $$U_a = \frac{R_2}{R_1 + R_2} * U_e.$$ $$R_1$$ is the first resistor viewed from above and $$R_2$$ is the second resistor viewed from above. How does this formula help me to find $$U_a?$$

(Bear in mind that the rectangle is a resistor as we use this notation in Germany.)

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    \$\begingroup\$ So, where is the problem because you already found the equation for Ua? \$\endgroup\$ – G36 Jan 16 '19 at 15:30
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    \$\begingroup\$ You have already given the answer. If you assume that both resistors are value "R" then you can simplify it further, but if you don't, it remains the equation you have. \$\endgroup\$ – Phil G Jan 16 '19 at 15:36
  • \$\begingroup\$ So can I assume that the values for both resistors are 0 Ohm and that means that Ua is also 0? \$\endgroup\$ – Nick Jan 16 '19 at 16:04

In the frequency domain, components being represented as rectangular blocks is not uncommon (resistors included!)

You already have U_a. Perhaps a small discussion concerning how the equation is produced will help.

Suppose you have a voltage divider. Would you know how to compute the output voltage (the voltage at the node between the two resistors relative to ground)? The approach is exactly the same here.

The input impedance of the op-amp is huge, so it draws negligible amounts of current. Thus, using the analysis you used for the voltage divider, you would get the same formula for the node between the two resistors in the diagram above, and since that node is connected to one of the two input terminals of the op-amp, the other input-terminal (the - terminal) has the same voltage (remember the three golden rules of op-amps!). Since the other input terminal (the - terminal) is connected to the output, we can therefore conclude that output is equal to the voltage at the node between the resistors.

Of course, the circuit above isn't the same as a voltage divider. If the input of the voltage divider is 100 V and you have two resistors of equal resistance, you get 50 V at the output. For the case of your circuit, the output voltage will be limited by the voltage you supply your op-amps with (with a 100 V input, supplying the op-amp with a measly +- 3 V will not allow the output to be 50 V!).

Concerning your comment about the resistors being 0 Ohms. The equation only holds for when at least one of your resistors isn't 0 Ohms. If you have both of the resistors at 0 ohms, then you'll short your input source! Not good.

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  • \$\begingroup\$ Thanks for answering my question. One last question still remains. When the input impedance of the computing amplifier is huge, does that mean that the input current of the op-amp is equal to 0A? And would that mean that the voltage between the node between the two resistors and ground would be 0V and so U_a is also 0V or is there still something I don't understand? \$\endgroup\$ – Nick Jan 19 '19 at 7:21
  • \$\begingroup\$ Yes, that means the input current into the op-amp is approximately 0 A. This does not mean the current through the resistor is 0 A; it just means you can ignore any possible loading from the op-amp, so you can evaluate your circuit as though the op-amp is not there! \$\endgroup\$ – Metric Jan 19 '19 at 20:23
  • \$\begingroup\$ And if you ignore the op-amp, then you just have a voltage divider left! \$\endgroup\$ – Metric Jan 19 '19 at 20:24

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