# Driven RLC circiut in parallel

1. Sep 17, 2006

### 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.

$$V_0 cos(wt) = \frac{Q}{C} = L \ddot{Q} = \dot{Q} R$$

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

$$Q_{Total} = \\Q_C + Q_L +Q_R$$

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.

Last edited: Sep 17, 2006
2. Sep 17, 2006

### Andrew Mason

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

Last edited: Sep 17, 2006
3. Sep 17, 2006

### silverdiesel

okay, I appriciate 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.

4. Sep 18, 2006

### Andrew Mason

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

5. Sep 20, 2006

### 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 $$V_s(t) = V_0 cos(\omega t)$$