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日期:2018-11-26 10:45

GNG1106 – Fundamentals of Engineering Computation

Course Project

Electrical Engineering

Design of an RLC Electrical Circuit

Yuhan Zhang 300057292

Xiaomeng Yang 300028425

Date : Nov 13th, 2018

1. Problem Identification and Statement

This project is aimed to determine the resistance of the resistor R in an RLC

circuit shown in ‘ELG project’ and reveals whether the charge dissipates to the its

initial percentage of charge in the during the given time. As required, users

should input the following values : the inductance of components (L), the

capacitance of components (C ), the battery voltage, the rate of dissipation (td)

and the ideal percentage of the original charge to reach (Pc) in dissipation time

(td). After calculation, the software can give the value of the resistor. Meanwhile,

at most 5 results can be saved into the files. In other words, users can select the

stored data or input a new value. Aiming to reach this function, we need to

determine the exact interval to find roots. The software will provide an upper

bound for resistor and a plotted function for users searching the root of function

by selecting an appropriate interval. ‘g(R) = e

-Rt/(2L)

- pc

’ is utilized to determine

the value of R, and when time varies, q will be calculated and the relationship

between it and time will be found and plotted by graph.

2. Gathering of Information and Input/Output Description

2.1 subsection 1

The study of transient and steady-state behavior of electric circuits is usually based

on Kirchhoff’s laws. In this project, our concentration is on circuits that current and

voltage vary over time until they become a steady state. And that occurs when we close

the switch.

Figure a

For the values users input should be in a certain range.

The values of R (resistance in Ohms): 1 ohm to 1*10

6 ohms ( 1 Mohm)

The values of inductor (L in Henrys H): 1*10

-9 Henrys (1 nH) to 1*10

-2 Henrys (1000μH)

The values of capacitance (C in Farads F ): 1*10

-12 Farads (1 pF) to 1*10

-2 Farads (1000μF)

Then the voltage drop can be determined across every device.

VR = iR Equation 1

Equation 1 is utilized to determine the voltage drop across the resistor.

VL = L (di / dt) Equation 2

Equation 2 is utilized to determine the voltage drop across the inductor.

VC = q / C Equation 3

Equation 3 is utilized to determine the voltage drop across the capacitor.

2.2 subsection 2

L (di / dt) + Ri + q / C = 0 Equation 4

Kirchhoff's second law shows that the sum of voltage drops around the closed circuit is

zero.

i = dq / dt Equation 5

In this equation, i means current and we got another equation to express the

relationship between current and the amount of charges in the circuit.

Thus,

L (d2q/ dt2) + R(dq / dt) + q / C = 0 Equation 6

All the values we calculate are known.

Equation 7

The value of q can be solved by utilizing a second-order linear ordinary differential

equation. When t = 0, q = q0 = V0C, and V0 is equal to the voltage from the charging

battery. And the relationship between t and q0

is graphically shown below. The values

of L and C are selected and in a certain rate, the energy is dissipated by a proper

resistor. The rate of dissipation would be defined as the time, td, it takes for the charge

on the capacitor to reach a desired percentage of its original charge, pc = q/q0

.

Figure b

q(t) = q0e

-Rt/(2L) Equation 8

Due to the fact that we focus on the rate of dissipation, the cosine function is reduced

and we get a new function which matches dashed lines in Figure b.

2.3 subsection 3 (Solving for the Resistor R)

pc=q/q0

Equation 9

q(td

) = pcq0 = q0e

-Rt/(2L) Equation 10

pc=e

-Rt/(2L) Equation 11

We combine Equation 9 and Equation 10 to determine the value of R (the proper

resistor) by the known value of inductance and the percentage of original charge pc

.

g(R) = e

-Rt/(2L)

- pc Equation 12

Equation 12 provides better value of R by using numerical calculation method, though,

we've already got the answer from Equation 11 by analysing.

Using the bisection method to find roots in software

Firstly, setting a range of interval by the desired lower value of xL and upper value of

xU is aimed to changing signs and it also can be checked by f(xL)f(xU)< 0. Secondly, the

estimate of the root xR is determined by xR = (xL + xU)/2. Thirdly, it should follow the

following equations

a). If f(xL)f(xR)< 0, the root lies in the lower subinterval. Therefore, set xU= xR and

return to the previous step.

b). If f(xL)f(xR)> 0, the root lies in the upper subinterval. Therefore, set xL = xR and

return to the previous step.

c).If |f(xL)f(xR)| < some small tolerance value (that is, can be considered equal to 0), the

root equals xR; The computation would be terminated.

2.4 subsection 4 (Input and Output)

● Input ( values are all collected by users)

1. L = the value of the inductor (Henrys),

2. C = the value of the Capacitor (Farads),

3. V0 = the voltage of the battery across the capacitor at time t = 0 (volts),

4. td = the time of dissipation (s).

5. pc = percentage value of the original charge on the capacitor after the

dissipation time td

.

Equation 13

6. Equation 13 is utilized to find the precise value of R after software determine

the range of R by calculating the upper bound by square root term in the

function g(R).

7. The solution of g(R) can plot a graph that user is able to choose an interval by

looking at the graph.

● Output ( values calculated by software)

The value of R and the function graph plotted from t = 0 to time td.


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