Calculate V0 for the circuit shown in figure.

Steps to Solve

1. Identifying the Basic Configuration The circuit shows an operational amplifier (op-amp) with a non-inverting input connected to a voltage divider formed by the 4 kΩ and the 8 kΩ resistors. We need to first calculate the voltage at the non-inverting terminal.

2. Calculating the Voltage at the Non-Inverting Terminal Using the voltage divider formula: $$ V_{+} = \frac{R_2}{R_1 + R_2} \cdot V_{in} $$ where:

• ( R_1 = 4 , k\Omega )
• ( R_2 = 8 , k\Omega )
• ( V_{in} = 3 , V )

Substituting the values: $$ V_{+} = \frac{8 , k\Omega}{4 , k\Omega + 8 , k\Omega} \cdot 3 , V = \frac{8}{12} \cdot 3 , V = 2 , V $$

3. Understanding the Op-Amp Configuration The op-amp is set up to amplify the difference between its inverting (-) and non-inverting (+) inputs. The output ( V_0 ) for an ideal op-amp can be expressed as: $$ V_0 = A \cdot (V_{+} - V_{-}) $$ where ( A \to \infty ) implies ( V_{+} = V_{-} ). Therefore, we need to find ( V_{-} ).

4. Finding the Inverting Voltage ( V_{-} ) Here, ( V_{-} ) can be found using another voltage divider involving the 5 kΩ and 2 kΩ resistors, where ( V_{-} ) can be expressed as: $$ V_{-} = \frac{2 , k\Omega}{2 , k\Omega + 5 , k\Omega} \cdot V_0 $$

Now, we can express ( V_0 ) in equation form by substituting for ( V_{-} ): $$ V_0 = 2 + (V_0 \cdot \frac{2}{7}) $$

5. Solving the Equation for ( V_0 ) Rearranging the equation involves multiplying both sides by 7 to eliminate the fraction: $$ 7V_0 = 14 + 2V_0 $$ Thus, we have: $$ 7V_0 - 2V_0 = 14 $$ $$ 5V_0 = 14 $$ Now, isolating ( V_0 ): $$ V_0 = \frac{14}{5} = 2.8 , V $$