First Order Resistor Capacitor (RC) Parallel Circuits
Strings (SiPjAjk) = S7P4A41 Base Sequence = 12735 String Sequence = 12735 - 4 - 41
Figure 5.41 illustrates a first order RC parallel circuit comprising a voltage source = 100 V; two capacitors in parallel each of capacitance 4 F and four resistors.
The switch,s1 is always open. The switch s2 is closed at t = 0.
(a) The capacitor's voltage vc(t)
(b) The current through vc(t).
(b) S7P4A41 (Linear Motion - Current Flowing Into Passive Elements).
The motion of current flow in an electric circuit can be viewed as piecewise-linear (current flow into a device) or looped (current blow in a circuit branch or the entire circuit). In a series connection, the piece-wise-linear current and looped current are the same. In a parallel connection, they are not. The looped current is stringed as S7P4A42 (Curved Motion)
(a) Pj Problem of Interest is of type force. Voltage problems are force problems.
(b) Pj Problem of Interest is of type motion. It could also be viewed as of type change because current is the rate of change of charge flow with respect to time. However, its representation as the flow of charge is more appropriate perspectively.
Any system that is able to store potential or kinetic energy, and to dissipate this energy, is a First Order System. In an electric circuit, a first order circuit is any circuit that consists of a single energy storage element such as a capacitor or an inductor; voltage or current sources and resistors.
Whenever a circuit is switched from one state to another, either by a change in the active elements or a change in the passive elements, there exists a transitional period during which the branch currents and element voltages change from their pre-switched values to their post-switched values. This transitional period is called the transient.
The circuit is said to be in steady state at the end of the transient.
First Order Circuits are described by First Order Ordinary differential equations (equations of first degree non-partial derivatives).
The solution to this first order ordinary differential equation consists of a homogeneous solution (or natural response) which is the solution of the transient; and the >particular solution (or forced response) which is the solution of the steady state. In other words, the complete solution (or complete response) is the sum of the homogeneous solution and the particular solution.
The complete response of a first order circuit is represented as follows:
f(t) = Ae-t/τ + B ------(1).
where A = initial value - final value; B = final value and τ = time constant .
τ = RC for a RC circuit. τ = L/R for a RL circuit.
The time constant is the time at which the function is 36.8% of its initial value. In other words, it is the time at which the function has undergone 63.2% of the change from f(0+) to f(∞)
Now for the problem of fig.5.23 we have:
Replace parallel capacitors with their equivalent:
Equivalent capacitance, C, of the two parallel capacitors = C1 + C2 = (4 + 4) F = 8 F.
Determine the Thevenin equivalent resistance. To calculate Thevenin equivalent resistance (fig.5.42):
(1) Remove the load.
(2) Zero all independent voltage and current sources.
(3) Calculate the total resistance, beginning from the resistances farthest from the load's original position.
In figure 5.42, the load (the capacitors in this case) have been removed. The voltage source is set to zero (a short circuit is in its place) and the current source is set to zero (an open circuit is in its place).
So, Thevenin equivalent resistance = [(3)(6)/(3 + 6)] + 4 = 6 Ω
The Thevenin voltage is calculated as follows (fig.5.43):
(1) Remove the load, and open-circuit its terminals.
(2) Establish the open-circuit voltage vOC accros the open load terminals.
(3) Solve for vOC
(4) The Thevenin voltage, vT = vOC.
In figure 5.43, load is removed and the open-circuit voltage, vOC is established across open-circuited terminals of the load.
So, voltage across the 3 Ω resistor = vOC
Also, voltage across the 6 Ω resistor = vOC
Only the current source is in play. The voltage source is not in play since switch s1 is always open.
So, by the current division rule, current through the 3 Ω resistor = (6/9)4 = (2/3)4
A lSo, voltage across the 3 Ω resistor = (2/3)4 x 3 = 8 V.
So, vOC = 8 V
So, Thevenin voltage, vT = 8 V
Time constant = RC = 6 x 8 = 48 secs; and A = 0 - 8 = -8; B = 8;
So, capacitor voltage vc(t) = 8(1 - e-t/48) for t ≥ 0
Capacitor charge, q = CV = 4 x 8(1 - e-t/48);
So, current in capacitor = dq/dt = (2/3)e-t/48.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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Ordinary Differential Equations (ODEs)
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