Resonant RLC Circuit Homework - Purcell 8.4

[tjd_maximum]

Homework Statement

Purcell 8.4: In the resonant circuit of the figure(I'll reproduce the image below) the dissipative element is a resistor R' connected in parallel, rather than in series, with the LC combination. Work out the equation analogous to Eq 2. (d^2V/dt^2 + (R/L)(dV/dt) + (1/LC)V = 0, this was for a series circuit), which applies to this circuit. Find also the conditions on the solution analogous to those that hold in the series RLC circuit. If a series RLC and a parallel R'LC circuit have the same L, C, and Q, how must R' be related to R

Crude drawing of the circuit:
_____C___
|________|
|____R'__ |
|________|
|____L___|
(ignore the white lines)

Homework Equations

Well, in deriving Eq 2 the book uses the following equations:
I = -dQ/dt
Q = CV
V(inductor) = L(dI/dt)
V(resistor) = IR

The Attempt at a Solution

I really only think I need help on the first part of the problem (finding the differential equation).
The problem that arises from the fact that the I over R' is different from the I over L and that the proportion changes (I(R') + I(L) = I(total) but all of these change don't they? The capacitor runs out of charge and the Inductor depends on the changing current.
Any push in the right direction would be appreciated, Thanks.

 

[tjd_maximum]

I think I may have made a little progress.

Loop 1:
When V is Q/C
V = I$$_{1}$$R
0 = I$$_{1}$$R - V

Loop 2:
V = LdI$$_{2}$$/dt
0 = LdI$$_{2}$$/dt - V

Combining:
I$$_{1}$$R = LdI$$_{2}$$/dt
(I'm having a problem with latex it seems, those superscripts are supposed to be subscripts)

Now, this looks like it may end up giving me I$$_{1}$$ in terms of I$$_{2}$$, but that's not really what I'm looking for. . . I suppose I may be able to re-plug into one of these equations when I'm done. Does this seem right at all?

 

[Moetasim]

I would approach this problem by first understanding the concept of resonance in RLC circuits. In a series RLC circuit, the inductive reactance of the inductor and the capacitive reactance of the capacitor cancel each other out at the resonant frequency, resulting in a purely resistive circuit. This leads to a peak in the current and a minimum in the impedance of the circuit.

In a parallel R'LC circuit, the resistor R' is connected in parallel with the LC combination. This means that the current will split into two branches, one through the LC combination and one through the resistor R'. At resonance, the current through the LC combination will be maximum, while the current through the resistor R' will be minimum. This is because the inductive reactance of the inductor and the capacitive reactance of the capacitor will cancel each other out in the LC branch, leaving only the resistance of the resistor R' in the other branch.

Now, to derive the differential equation for this circuit, we can use Kirchhoff's laws. The current through the LC branch can be written as I(LC) = V/C, where V is the voltage across the LC combination. The current through the resistor R' can be written as I(R') = V/R'. Using Kirchhoff's current law, we can write the total current as I(total) = I(LC) + I(R'). At resonance, the total current will be maximum, so we can write I(total) = V/R' = I(LC) + I(R'), which gives us V/R' = V/C + V/R'. Rearranging this equation, we get V = (1/R' + 1/C)V. Now, using the equation V = L(dI/dt), we can write L(dI/dt) = (1/R' + 1/C)V. Substituting in the values for V and I(LC), we get L(dI/dt) = (1/R' + 1/C)(I(LC)/C). Simplifying this equation, we get L(dI/dt) = (I(LC)/R'C). Now, using the equation I = -dQ/dt and Q = CV, we can write I(LC) = -d(CV)/dt. Substituting this into our equation, we get L(dI/dt) = (-d(CV)/dt