# CIRCUIT ANALYSIS: 8 Resistors, 1 IVS, 1 Differential Op-Amp - Find Vo

## Homework Statement

The circuit below is a differential amplifier driven by a bridge, find $V_0$.

http://img476.imageshack.us/img476/8827/chapter5problem48fi2.jpg [Broken]

## Homework Equations

KCL, v = i R, Ideal Op-Amp relationships

## The Attempt at a Solution

http://img259.imageshack.us/img259/8314/chapter5problem48part2fs0.jpg [Broken]

By the Ideal Op-Amp relationships, we know that there is 0 current at both input terminals of the Op-Amp.

KCL at $V_2$)

$$\frac{0.005\,-\,V_2}{40000}\,=\,\frac{V_2}{60000}\,+\,\frac{V_2\,-\,V_4}{20000}$$

$$18\,V_2\,-\,12\,V_4\,=\,0.003$$ < ------ Equation 1

KCL at $V_4$)

$$\frac{V_2\,-\,V_4}{20000}\,=\,\frac{V_4}{80000}$$

$$4\,V_2\,-\,5\,V_4\,=\,0$$ <----- Equation 2

Now, using equations 1 and 2, I get $V_2\,=\,0.003571\,V$ and $V_4\,=\,0.002857\,V$. Does that seem right?

By the relationships of an ideal Op-Amp, $V_3\,=\,V_4$, so $V_3\,=\,0.002857\,V$.

KCL at $V_1$)

$$\frac{0.005\,-\,V_1}{10000}\,=\,\frac{V_1}{30000}\,+\,\frac{V_1\,-\,V_3}{20000}$$

$$11\,V_1\,-\,3\,V_3\,=\,0.03$$ <----- Equation 3

KCL at $V_3$)

$$\frac{V-1\,-\,V_3}{20000}\,=\,\frac{V_3\,-\,V_0}{80000}$$

$$4\,V_1\,-\,3\,V_3\,+\,V_0\,=\,0$$ <----- Equation 4

Using equations 3 and 4, I get $V_0\,=\,0.004363\,V$.

$$V_0\,=\,4.363\,mV$$ <------ Is that right?

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Bump - Bump:)

Looks good. Quick question though, why don't you use linear algebra instead of messy substitutions?