Driven RLC circiut in parallel

[silverdiesel]

I need some help setting up the DE for this system. I have a voltage source pumping at V_0cos(wt)

Connected to a capacitor, inductor and resistor all three in parrellel. I am guessing then, that the voltage drop among all the parts are equal.

<br /> V_0 cos(wt) = \frac{Q}{C} = L \ddot{Q} = \dot{Q} R<br />

Then, The total charge on all parts would be equal to the sum of charges, right?

<br /> Q_{Total} = \\Q_C + Q_L +Q_R<br />

But what is the Q total? I have these two equations, but I am having a hard time understanding how to combine them into one single DE that I can solve.

 

[Andrew Mason]

I need some help setting up the DE for this system. I have a voltage source pumping at V_0cos(wt)

Connected to a capacitor, inductor and resistor all three in parrellel. I am guessing then, that the voltage drop among all the parts are equal.

<br /> V_0 cos(wt) = \frac{Q}{C} = L \ddot{Q} = \dot{Q} R<br />

Then, The total charge on all parts would be equal to the sum of charges, right?

<br /> Q_{Total} = \\Q_C + Q_L +Q_R<br />

But what is the Q total? I have these two equations, but I am having a hard time understanding how to combine them into one single DE that I can solve.

If you just give it a one time zap and remove the voltage, you will have:

L\frac{dI}{dt} + \frac{Q}{C} = IR or

L\frac{dI}{dt} + \frac{Q}{C} - IR = 0

This is because the current through the resistor has to come from the emf produced by the capacitor and the inductor.

Now take the same circuit and apply a forcing voltage.

L\frac{dI_L}{dt} + \frac{Q_c}{C} + V_0\cos\omega t = IR or

IR - (L\frac{dI_L}{dt} + \frac{Q_c}{C}) = V_0\cos\omega t

Of course: I_L = \dot{Q} \text{ and } \frac{dI_L}{dt} = \ddot{Q_L}

and I_L + I_C = I

AM

 

[silverdiesel]

okay, I appreciate the help. It seems that this is the same as setting up the circuit in series. (V1+V2+V3=V, Q1=Q2=Q3=Q) Is there no difference when setting up the circuit in parallel? I was thinking V1=V2=V3=V, and Q1+Q2+Q3=Q.

 

[Andrew Mason]

okay, I appreciate the help. It seems that this is the same as setting up the circuit in series. (V1+V2+V3=V, Q1=Q2=Q3=Q) Is there no difference when setting up the circuit in parallel? I was thinking V1=V2=V3=V, and Q1+Q2+Q3=Q.

The difference is that the currents are not the same. I_L, I_C \text{ and } I_R will be different. They are necessarily the same when in series.

AM

 

[SGT]

Since all elements are in parallel they have the same voltage drop.
You have three currents:
i_R(t) = \frac{V_s(t)}{R}
i_C(t) = C\frac{dV_s(t)}{dt}
i_L(t) = i_L(0) + \frac{1}{L}\int_{0}^{t}V_s(\tau)d\tau
Where V_s(t) = V_0 cos(\omega t)