# Find output voltage as a function of time

**1. The problem statement, all variables and given/known data**

The op amp in the circuit seen in the figure is ideal. Find the steady-state expression for v_{o}(t) = v_{g} = 1cos(10^{6}t) V. Express your answer in terms of t, where t is in microseconds. Enter the phase angle in radians.

Hints: (1) Let the node above the capacitor be Node a, and the node at the inverting (i.e., minus) input be Node b. (2) Write KCL equations at Node a and at Node b. Hints: This gives equations in terms of Va and Vo. Also note that the 40 K-Ohms load resistor is not part of the equations

**2. Relevant equations**

Current = 0 into input terminals of op amp

Z_{C} = 1/(jωC), where j=[itex]\sqrt{-1}[/itex]

KCL

V = IZ

**3. The attempt at a solution**

V_{g} is given in the format: v(t) = V_{max}cos(ωt+ø), where ω is the frequency and ø is the phase angle.

V_{b} = 5V

For KCL, ∑i_{entering} = ∑i_{leaving}

Node a:

i_{100kΩ} + i_{20kΩ} = i_{5kΩ} + i_{100pF}

Using V=IZ,

[itex]\frac{V_o – V_a}{100k}[/itex]+[itex]\frac{V_b – V_a}{20k}[/itex] = [itex]\frac{V_a – V_g}{5k}[/itex] + V_{a}*(jωC)

[itex]\frac{V_o – V_a}{100k}[/itex] + [itex]\frac{5 – V_a}{20k}[/itex] = [itex]\frac{V_a-1}{5k}[/itex] + V_{a}*j*10^6*100*10^(-12)

Node b:

i_{10pF} = i_{20k}

V_{b}*(jωC) = [itex]\frac{V_b-V_a}{20k}[/itex]

V_{b}*(jωC) = [itex]\frac{5-V_a}{20k}[/itex]

Solving, I get V_{a} = 5-j and V_{o} = 95+j24

Converting V_{o} to polar, V_{o} = 98∠14.2° V

Converting to radians, V_{o} = 98∠0.25 V

Putting this in terms of t,

V_{o}(t) = 98cos(10^6t+0.25) V

Converting to microseconds,

V_{o}(t) = 98cos(10t+0.25) V which is incorrect. Not exactly sure what I’ve done wrong here.

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