﻿ First Order Resistor Capacitor Series Circuit

First Order Resistor Capacitor (RC) Series Circuits

Strings (SiPjAjk) = S7P4A41     Base Sequence = 12735     String Sequence = 12735 - 4 - 41

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First Order Resistor Capacitor (RC) Series Circuits
Math

The circuit illustrated in figure 7.25 is a first order circuit. The capacitor has an initial charge Q0 = 800 μC; and capacitance = 4μF. Calculate the current flowing through the capacitor and its charge for t > 0, if the switch S, is closed at t = 0.

The string:
S7P4A41 (Linear Motion - Current Flowing Into Passive Elements). The motion of current flow in an electric circuit can be viewed as piece-wise-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)
The math:
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.7.25 we have:

Time constant, τ = RC = 10 x 4 x 10-6 = 4 x 10-5
Initial voltage across capacitor, v0 = initial charge/capacitance
So, v0 = (800 x 10-6)/(4 x 10-6) = 200
Final voltage across capacitor, v = voltage of voltage source = 100 V.
So, A = 200 - 100 = 100 and B = 100
So, from equation (1), voltage across capacitor at time t, vc(t) is:
vc(t) = 100e-t/(4 x 10-5) + 100 = 100(1 + e-25000t))
So, charge of capacitor for t >: 0, q is as follows:
q = CV = (4 x 10-6) x 100(1 + e-25000t) = (4 x 10-4)(1 + e-25000t) Coulombs
Current in capacitor, i = dq/dt (the rate of change of charge)
So, i = d(4 x 10-4)(1 + e-25000t)/dt = -10e-25000t Amperes

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