LC circuit (Differential Equation)

[rakhil11]

Homework Statement

An inductor with value L and a capacitor with value C are connected in series to a power source. At time t, the voltage of the power source (i.e. the voltage across both the inductor and capacitor) is given by $v(t)=Asin(\frac{2t}{\sqrt{LC}})$. If the voltage across the capacitor at time 0 is 0 and at time $\frac{\pi \sqrt{LC}}{2}$ is B, what is the voltage u(t) across the capacitor?

Homework Equations

$I = \frac{dV}{dt}C$
$V = \frac{dI}{dt}L$

The Attempt at a Solution

Honestly, I'm pretty stuck. I've tried plugging the expression for current from the capacitor into the inductor equation, and then $v(t)=Asin(\frac{2t}{\sqrt{LC}}) = \frac{dI}{dt}L+\frac{1}{C} \int_{t_0}^{t} I dt$, but neither approach got me very far. Any help would be greatly appreciated!

 

[NascentOxygen]

At time t, the voltage of the power source (i.e. the voltage across both the inductor and capacitor) is given by $v(t)=Asin(\frac{2t}{\sqrt{LC}})$. If the voltage across the capacitor at time 0 is 0 and at time $\frac{\pi \sqrt{LC}}{2}$ is B ...

Hi rakhil11.

From exercises similar to this that you have worked on in class, is it concerned with just steady state conditions, or is it including transient conditions, too? That is, are voltages and currents each a pure sinusoid, or is each a mixture of sinusoids?

Does the textbook give the answer?

 

[rude man]

Homework Statement

An inductor with value L and a capacitor with value C are connected in series to a power source. At time t, the voltage of the power source (i.e. the voltage across both the inductor and capacitor) is given by $v(t)=Asin(\frac{2t}{\sqrt{LC}})$. If the voltage across the capacitor at time 0 is 0 and at time $\frac{\pi \sqrt{LC}}{2}$ is B, what is the voltage u(t) across the capacitor?

Homework Equations

$I = \frac{dV}{dt}C$
$V = \frac{dI}{dt}L$

The Attempt at a Solution

Honestly, I'm pretty stuck. I've tried plugging the expression for current from the capacitor into the inductor equation, and then $v(t)=Asin(\frac{2t}{\sqrt{LC}}) = \frac{dI}{dt}L+\frac{1}{C} \int_{t_0}^{t} I dt$, but neither approach got me very far. Any help would be greatly appreciated!

@nascent: there are no transients in this circuit as it comprises a pure inductor and a pure capacitor. No matter when the voltage source was applied, at t = 0 or t → -∞, the current will always comprise two sinusoids, one at ω = 1/√(LC), the other at ω = 2/√(LC).

@rakhil11: you're almost OK with your last equation, but given the problem's initial condition, what is t₀?
Hint: change your integro-differential equation into a second-order ODE and solve conventionally for I(t) and then V_C(t). You will need your initial conditions on I and dI/dt of course.

The problem statement is somewhat misleading: V_C = B cannot be assigned arbitrarily. When you solve the ODE the value of V_C is determined uniquely by setting t = π√(LC)/2.

 

[NascentOxygen]

@nascent: there are no transients in this circuit as it comprises a pure inductor and a pure capacitor.

This being an engineering question, it's possible that the instructor intends the circuit's resistance be considered negligible (i.e., as good as zero) rather than precisely zero. I await clarification from OP.