circuit analysispassive-networks
I am having trouble finding the differential equation of a mixed RLC-circuit, where C is parallel to RL. It is a steady-state sinusoidal AC circuit. I need it to determine the Power Factor explicitly as a function of the components. I know I am supposed to use the KCL or KVL, but I can't seem to derive the correct one.
I am actually a math student, so I apologize if this question is straight forward.
It looks to me like the voltage source at \$t=0\$ is just supplying some charge to the capacitor, and the differential equation would just be the decaying transient response of the RLRC circuit with an initial capacitor voltage of 8V (v_c(0) = 8V).
So yes, at \$t -> ∞\$, \$i(t) = 0\$; there would be no voltage drop across any component, there would be no current flowing through any component.
By the definition of a capacitor:
$$ i_c(t) = C\frac{\mathrm{d}V_c(t)}{\mathrm{d}t}, $$
Since the voltage across C is the same as the voltage across R and RL:
$$ Vc(t) = R{i(t)} + L\frac{\mathrm{d}i(t)}{\mathrm{d}t} = Ri_r(t), $$
And since \$i_r(t) + i(t) = i_c(t) = \frac{v_c(t)}{R} + i(t)\$.
It has been a while since I have solved differential equations, but I think this simplifies to: $$ i_c(t) = C\left[R\frac{\mathrm{d}i(t)}{\mathrm{d}t} + L\frac{\mathrm{d^2}i(t)}{\mathrm{d}t^2}\right]= {i(t)} + \frac{L}{R}\frac{\mathrm{d}i(t)}{\mathrm{d}t} + i(t) $$
$$ i_c(t) = 2\frac{\mathrm{d}i(t)}{\mathrm{d}t} + 2\frac{\mathrm{d^2}i(t)}{\mathrm{d}t^2} = {i(t)} + \frac{\mathrm{d}i(t)}{\mathrm{d}t} + i(t) $$
$$ \frac{\mathrm{d^2}i(t)}{\mathrm{d}t^2} + 0.5\frac{\mathrm{d}i(t)}{\mathrm{d}t} - i(t) = 0 $$
What you have is a series RLC circuit. One of the most common introductory circuit in EE. For example, look under Series RLC Circuit in this wikipedia page:
en.wikipedia.org/wiki/RLC_circuit
These are equations copied from the same page for the under-damped case. Given R, C along with measured \$\omega_d\$, L can be solved from these equations. (You don't even need to measure the attenuation factor \$\alpha\$.)
If the RLC do want to oscillate, the node where D1 cathode is connected will go negative. Therefore, depending on how D1 is driven, it may conduct during the negative half cycle and kill the oscillation.
Best Answer
The voltage, \$ v\$, across \$\small C\$ is equal to the the voltage across the \$\small R, L\$ series combination.
The current in \$\small C\$ is \$i_1=\small C\large \frac{dv}{dt}\$
The current in \$\small RL\$ is related to \$v\$ by; \$ v=\small R\large i_2+\small L\large\frac{di_2}{dt}\$, where \$i_2\$ is the current through \$\small R\$ and \$\small L\$.
Solve both ODEs for \$ i_1\$ and \$ i_2\$, and the total current in the \$\small RLC\$ circuit is then \$i=i_1 + i_2\$.
Let \$\small t\rightarrow \infty\$ in the real parts of the exponentials to remove any transients, and you now have \$ i\$ and \$v\$ in the required steady-state sinusoidal forms.