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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?

  • \$\begingroup\$ Never mind -- for an ideal opamp v1 is about -250 volts. Try this circuit on a breadboard :-\ \$\endgroup\$ – HL-SDK Jan 30 '14 at 18:18

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

| improve this answer | |
  • \$\begingroup\$ The voltage drop due to resistor takes it down to -250V. My rookie mistake :p \$\endgroup\$ – user29568 Jan 30 '14 at 18:18
  • 1
    \$\begingroup\$ That's correct. 5mA through 50k is 250V, below 0V at the inverting input. \$\endgroup\$ – user28910 Jan 30 '14 at 18:24

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