1
\$\begingroup\$

enter image description here

We are supposed to find \$i_o\$. Looking at the 25-V voltage source and the \$5k\Omega\$ resistor, the current is \$i_x=\frac{25}{5\times10^{-3}}=5mA\$ because the voltage across both inputs of the op-amp are equal (ideal) and zero because the positive terminal is connected to the ground. Using node votlage equations at node \$V_1\$ and \$V_0\$, $$\frac{0-v_1}{50\times10^{3}} +\frac{0-v_1}{10\times10^{3}} = \frac{v_1-v_0}{40\times10^{3}}$$

Knowing \$i_x\$ allows us to determine \$V_1\$ which is \$50\times10^{3} \times 5\times10^{-3}=250V\$ Is it \$-250\$V?

Solving for \$v_0\$ I got \$-550V\$. Using KCL at node \$v_0\$, \$i_0=i_1+i_2=23/600 A\$

But, for some reason, the answer is wrong. Am I missing something? Did I do something wrong somewhere?

\$\endgroup\$
1
  • \$\begingroup\$ Never mind -- for an ideal opamp v1 is about -250 volts. Try this circuit on a breadboard :-\ \$\endgroup\$
    – HL-SDK
    Commented Jan 30, 2014 at 18:18

1 Answer 1

3
\$\begingroup\$

V1 is -250V, so then V0 is -1450V. I0 is 30mA through the 40k, plus 48.3mA through the 30k.

\$\endgroup\$
2
  • \$\begingroup\$ The voltage drop due to resistor takes it down to -250V. My rookie mistake :p \$\endgroup\$
    – user29568
    Commented Jan 30, 2014 at 18:18
  • 1
    \$\begingroup\$ That's correct. 5mA through 50k is 250V, below 0V at the inverting input. \$\endgroup\$
    – user28910
    Commented Jan 30, 2014 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.