# Simple Circuit with a Op Amp

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?

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