Simulated Inductor

The currents and voltages at points of interest are indicated in the diagram. The circuit equations have been written assuming that $R_2$ and the source voltage are connected to the ground. The circuit equations are:

$V_1-V_0 = R_1I_1 = \frac{Q}{C}$ $I_2 = \dot{Q}$ $V_S = \frac{Q}{C} + I_2R_2$ $I = I_1 + I_2$

Assuming that the capacitor initially has no energy, and rearranging the equations:

$V_S = \frac{Q}{C} + \dot{Q}R_2$ $\implies 10^{-6}V_S = Q + \dot{Q}$ $\implies 10^{-6}\sin{t} = Q + \dot{Q}$ $\implies 10^{-6} \mathrm{e}^{t} \sin{t} = \mathrm{e}^{t} Q + \mathrm{e}^{t} \dot{Q}$ $\implies \frac{d}{dt}\left(\mathrm{e}^{t} Q \right) = 10^{-6} \mathrm{e}^{t} \sin{t}$ $\implies Q =10^{-6} \mathrm{e}^{-t} \int_{0}^{t} \mathrm{e}^{s} \sin{s} \ ds$

$\implies Q = 10^{-6} \mathrm{e}^{-t}\left(\dfrac{\mathrm{e}^t\sin\left(t\right)-\mathrm{e}^t\cos\left(t\right)+1}{2}\right)$ $Q = 10^{-6}\left(\dfrac{\sin\left(t\right)-\cos\left(t\right)+\mathrm{e}^{-t}}{2}\right)$

Now, from the circuit equations:

$I = I_1 + I_2$ $\implies I = \frac{Q}{R_1C} + \dot{Q}$

Plugging in expressions:

$\implies I = \dfrac{\sin\left(t\right)-\cos\left(t\right)+\mathrm{e}^{-t}}{2} + 10^{-6}\left(\dfrac{\sin\left(t\right)+\cos\left(t\right)-\mathrm{e}^{-t}}{2}\right)$

Neglecting the term multiplied by $10^{-6}$ gives:

$I \approx \dfrac{\sin\left(t\right)-\cos\left(t\right)+\mathrm{e}^{-t}}{2}$

At steady state:

$I \approx \dfrac{\sin\left(t\right)-\cos\left(t\right)}{2}$

At this point, it is easy to conclude that the peak value of current is approximately:

$I_{\mathrm{max}} \approx \frac{1}{\sqrt{2}}$

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As for the bonus question, consider the circuit equations as shown above. Assuming zero initial conditions, the Laplace transform is applied to each of the equations yielding:

$V_S(s) = \left(\frac{1}{C} + sR_2\right)Q(s)$ $sQ(s) = I_2(s)$ $I(s) = I_1(s) + I_2(s)$ $I_1(s) = \frac{Q(s)}{R_1C}$

Solving for $I(s)$ gives:

$I(s) = \left(\frac{1 + sCR_1}{R_1 + sCR_2R_1}\right)V_S(s) \dots \ (1)$

Now, for a conventional RL circuit, the circuit equation in the Laplace domain is:

$I(s) = \frac{V_S(s)}{Ls + R} \dots \ (2)$

For (1) to behave similar to a RL circuit, it must resemble the mathematical structure of (2). This is only possible when the product $R_1C$ is a small number in magnitude. For the given set of parameters, this is indeed satisfied. For the case when $R_1=R_2=C = 1$ , this condition is not satisfied as the circuit behaves as a purely resistive circuit.

So, one can see that the given circuit is equivalent to an RL circiut provided $R_1C$ is a small number in magnitude. The magnitude and phase response in the frequency domain will start to resemble that of an RL circuit when this condition is satisfied.