Find Vout for Op-Amp Circuit with Attached Image

[Metamorphose]

1. Please refer to attached image.



2. The attempt at a solution.

I noted that V_b was grounded and would therefore be equal to 0 and because one of the two op-amp rules states that V_a = V_b, both will be equal to 0.

I wrote an equation at node V_a:

[V_a - V_in]/R₁ + [V_a - V_out]/R₂ = 0

Because V_a is 0, the equation simply became:

-V_in/R₁ - V_out/R₂ = 0

And plugging in the given values for part (a), I acquired - 5V. I'm not sure if this is correct or not as the book does not have a solution manual available yet.

 

[lewando]

-5V is right. In this configuration, the circuit is an "inverter".

 

[Metamorphose]

Thanks! I was just wondering :)

 

[CWatters]

Just for completeness...

Take your equation...

-Vin/R1 - Vout/R2 = 0

then rearrange it to give..

Vout/Vin = -R2/R1

where Vout/Vin is known as the Gain.

a) R2 = R1 so Gain = -1
b) etc

 

[rezeew]

Your approach to solving this problem is correct. Based on the given circuit, Va and Vb are both grounded and therefore equal to 0. Using Kirchhoff's Current Law (KCL) at the inverting input of the op-amp, we can write the following equation:

(Va - Vin)/R1 + (Va - Vout)/R2 = 0

Since Va is equal to 0, the equation simplifies to:

-Vin/R1 - Vout/R2 = 0

Solving for Vout, we get:

Vout = -R2/R1 * Vin

Plugging in the given values for R1 and R2, we get Vout = -5V. This means that the output voltage will be 5V lower than the input voltage.

It is always a good practice to double check your calculations and make sure they make sense. In this case, we can see that the output voltage is indeed lower than the input voltage, which is what we would expect from an inverting op-amp circuit. Keep up the good work!