I want to understand how do you approach this question after forming the KVL equation and differentiating. We were given the values for R = 1.5 Ohms, L = 0.9 H, C = 9 mF and V = 69V. I just want to know how do we go about it?

 

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I want to understand how do you approach this question after forming the KVL equation and differentiating. We were given the values for R = 1.5 Ohms, L = 0.9 H, C = 9 mF and V = 69V. I just want to know how do we go about it?

The input from the DC source V, is a unit step function scaled by the magnitude of V. The forcing function for the differential equation is:

\( V(t)\;=\;V_{dc}u(t) \)

where u(t) is 0 for t<0 and 1 for t≥1

 

From Kirchhoff`s law we know that the sum of all voltages across the 3 componenets is identical to V.
Therefore:
i*R + L*d(i)/dt + (1/C)*Int(i*dt)=V.
After multiplying with C and differentiating again (d/dt), you arrive at the classical 2nd-order differential equation.
Such an equation can be solved with i(t)=Io*exp(st).
As the last step, solve for d(i)/dt for t=0.

 

The input from the DC source V, is a unit step function scaled by the magnitude of V. The forcing function for the differential equation is:

\( V(t)\;=\;V_{dc}u(t) \)

where u(t) is 0 for t<0 and 1 for t≥1

Thank You

 

Depending on the relative values of R, L, & C you would expect to see:

1. An underdamped sinusoidal waveform with a decaying exponential waveform
2. A critically damped exponential waveform where the output approaches asymptotically the limiting value of Vdc without ever actually reaching it.
3. An overdamped waveform where the output approaches the limiting value of Vdc, very slowly without ever actually reaching it.

 

asymptotically the limiting value of Vdc without ever actually reaching it.

That's a pathological case.
In reality, when the voltage reaches to within the thermal voltage noise value of the circuit it has practically reached the input voltage value.

 

From Kirchhoff`s law we know that the sum of all voltages across the 3 componenets is identical to V.
Therefore:
i*R + L*d(i)/dt + (1/C)*Int(i*dt)=V.
After multiplying with C and differentiating again (d/dt), you arrive at the classical 2nd-order differential equation.
Such an equation can be solved with i(t)=Io*exp(st).
As the last step, solve for d(i)/dt for t=0.

Thank You

 

Whoever wrote the question missed the fact that, in general, when the switch is opened and everything reaches steady state there can still be a residual charge on the capacitor that will sit there indefinitely.

 

That's a pathological case.
In reality, when the voltage reaches to within the thermal voltage noise value of the circuit it has practically reached the input voltage value.

It is true that there is mathematics and there is engineering. Both viewpoints are worth understanding.