4.4 KiB
Chapter 8 - Quasi-static Electrodynamics and Alternating Current Circuits
Section 8.1 - The Quasi-static Regime of Electrodynamics
Section 8.1.1 - RC Circuit - Time-dependent Circuits with Resistor and Capacitor in Series Transient Currents
Here, we have a DC voltage source followed by a switch, capacitor with capacitance C, and a resistor with resistance R.
By Kirchoff's Voltage Law, we know that V_{cell} = V_capacitor + V_resistor, so V_{cell} = IR + \frac{Q}{C}. As current is simply charge over time,
V_{cell} = R\frac{dQ}{dt} + \frac{Q}{C}
Assuming the capacitor is uncharged at t = 0, we obtain the initial condition Q(0) = 0. Solving the differential equation, we see that
Q(t) = C V_{cell} (1 - e^{-\frac{t}{RC}})
Then, taking the derivative, we see that
I = \frac{dQ}{dt} = \frac{V_{cell}}{R} e^{-\frac{t}{RC}}
If instead we assume an AC voltage source, we know that V_{cell} = V_0 \cos \omega t. Then, by KVL,
{V_0 \cos \omega t = I(t)R + \frac{Q(t)}{C}}
Now, assume that I(t) = I_0 \cos(\omega t + \phi), as the current may be out-of-phase with voltage. Additionally, we assume that charge cannot accumulate outside of the capacitor. Then, with \frac{d}{dt} Q(t) = I(t), we see that
Q(t) = Q_0 + \frac{I_0}{\omega} \sin(\omega t + \phi)
Substituting into the formula for voltage, we see that
V_0 \cos \omega t = I_0 (R \cos (\omega t + \phi) + \frac{1}{\omega C} \sin(\omega t + \phi)) + \frac{Q_0}{C}
With trigonometric identities, this becomes
[V_0 - I_0 \cos \phi - \frac{I_0}{wC} \sin \phi]\cos(\omega t) + [I_0 R \sin \phi - \frac{I_0}{\omega C} \cos \phi]\sin(\omega t) + \frac{Q_0}{C} = 0
Now, assume let the \frac{Q_0}{C} vanishes, as there is no constant charge on the capacitor (which would imply an additional constant voltage). Then, for this equation to hold true at all times, the coefficients of \sin \omega t and \cos \omega t must equal zero. This will allow us to see that
\cos \phi = \frac{R}{\sqrt{R^2 + X_C^2}}
Definition. Here, the term X_C = (wC)^{-1} is the capacitive reactance of the circuit.
Definition. We also can solve for current to see that I_0 = \frac{V_0}{Z}, where Z = \sqrt{R^2 + X_C^2}, where Z is the impedence of the circuit.
Section 8.1.2 - Quasi-Static Error for a Parallel Plate Capacitor
Consider a parallel-plate capacitor. We know that within the capacitor, the electric flux is \vb{D}(t) = \epsilon_0 \vb{E}(t). With the charge on a plate given by Q(t), we can say that \vb{D}(t) = \frac{Q(t)}{\pi R^2}.
We also know by Ampere's law that \curl \vb{H} = \frac{\partial}{\partial t} \vb{D}. From this, given circular platFrom this, applying Stokes to Ampere's Law, we see that
\vb{H} = \frac{\partial Q}{\partial t} \frac{s}{2 \pi R^2} \vu {\varphi}
This continues on in this manner, however the rest is omitted.
Section 8.1.3 - Inductance
For an inductor, the voltage drop across an inductor is directly proportional to the change in current. That is,
\Delta V_L = L \frac{d^2Q}{dt^2} = L\frac{dI}{dt}
Consider a circular current loop with a voltage source in the x-y plane. Then, applying Faraday's law, \int_{circle} = (\curl \vb{E}) \cdot \vu{z} dS = -\frac{\partial}{\partial t} \int_{circle} \vb{B} \cdot \vu{z} dS.
Apply Stokes' law to the left hand side to see that \int_{circumference} \vb{E} \cdot d\vb{l} = -\frac{\partial \Phi_B}{\partial t}.
With \Phi_B = LI, we see the "back EMF" opposing the increasing current will be \vb{\varepsilon} = -L \frac{\partial I}{\partial t}.
Consider a circuit with a voltage source, a switch, an inductor, and a resistor in series. Then, by KVL, V_{cell} = L \frac{dI}{dt} + IR.
This is a differential equation. Assuming I(0) = 0, we can see that
I(t) = \frac{V_{cell}}{R}(1 - e^{-\frac{R}{L}t})
Now, consider an alternating current voltage source. Then, by KVL, V_0 \cos \omega t = R I(t) + L \frac{dI(t)}{dt}. If we assume that current is also sinusoidal, we can say that I(t) = I_0 \cos(\omega t + \phi). Substituting into the voltage expression,
V_0 \cos \omega t = R I_0 \cos(\omega t + \phi) + L \omega I_0 \sin(\omega t + \phi)
After expanding using trigonometric identities and setting coefficients to zero, we see that
\sin \phi = -\frac{X_L}{Z}; \cos \phi = \frac{R}{Z}; I_0 = \frac{V_0}{Z}
Definition. Here, X_Z = \omega L is the impedence of the circuit.