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Could someone please tell me what is the value of Ux? How can i calculate it?

is Ux = (U1 - U2) * R2/(R1+R2) correct?

  • 1
    \$\begingroup\$ Your question doesn't really have anything to do with operational amplifiers. Were you aware of this? \$\endgroup\$
    – Andy aka
    Commented Feb 5 at 16:03
  • \$\begingroup\$ You need to show your work so far. This isn't a homework service. \$\endgroup\$
    – Null
    Commented Feb 5 at 16:22
  • \$\begingroup\$ For homework types of questions we expect you to show that you have made a substantial effort to solve the problem yourself. Show us all of your work and then ask a specific question. \$\endgroup\$ Commented Feb 5 at 16:23
  • \$\begingroup\$ to solve this problem i know that i need to use the voltage divider formula, but since there are two voltage sources i am confused how should i apply the formula , i am sorry i am still in school, not a great researcher like yourself. just asked for a simple explanation, got so much backlash. thanks anyways! \$\endgroup\$
    – majid khan
    Commented Feb 5 at 16:30
  • \$\begingroup\$ @Andyaka, you have to know what the ideal properties of the op amp inputs are to solve the problem. \$\endgroup\$ Commented Feb 5 at 16:31

1 Answer 1


The opamp should be seen in such way ideal that its input current can be assumed to be zero. Hopefully you in some phase learn the cases when such assumption can be done without causing too much error nor totally blocking the functionality of the opamp. That's very important in making practical designs.

If the input current of the opamp can be seen = 0 the circuit is double input voltage divider. Solve it by using superposition. Assume at first U2 = 0V i.e. connected to the circuit ground. Calculate how much U1 generates to the opamp input. Assume it's = Ua.

Assume then U1 = 0V i.e. connected to the circuit ground. Calculate how much U2 generates to the opamp input. Assume it's = Ub.

The total input voltage is Ua+Ub. Your own formula is wrong.

The end.

  • \$\begingroup\$ The "double input voltage divider" is actually a "parallel voltage summer" with weighted inputs. \$\endgroup\$ Commented Feb 5 at 21:46

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