Consider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go- R 9(1) i(t)↓ After the switches closes, current flows through the circuit and the capacitor begins to discharge. The equation that describes the total voltage in the loop comes from Kirchoff's voltage law: L di(t) + Ri(t)+(0) = 0, (1) where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance, R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of change of the capacitor charge, and dividing by L, we can write the following homogeneous (no input source) differential equation for the charge on the capacitor: 4(1) +29(1)+w79(1)=0, ཀྱི where a= R 2L and The solution to this second order linear differential equation can be written as: 9(1) =Aent - Beat, where (3) (4) (5) A= (81+20)90 +90 (82+20)90 +90 and B= (6) 81-83 81-83 5. (10 marks) Using values of L = 1mH, C = 10μF, choose input values such that a = wo and Vc(0)=1V, and use your modified function to plot the voltage drop across the resistor, capacitor, and inductor, and the sum of all three, and plot the energy stored in the inductor and capacitor, and the total energy stored, as a function of time. Label the R, L, C, and go values on each graph.
Consider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go- R 9(1) i(t)↓ After the switches closes, current flows through the circuit and the capacitor begins to discharge. The equation that describes the total voltage in the loop comes from Kirchoff's voltage law: L di(t) + Ri(t)+(0) = 0, (1) where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance, R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of change of the capacitor charge, and dividing by L, we can write the following homogeneous (no input source) differential equation for the charge on the capacitor: 4(1) +29(1)+w79(1)=0, ཀྱི where a= R 2L and The solution to this second order linear differential equation can be written as: 9(1) =Aent - Beat, where (3) (4) (5) A= (81+20)90 +90 (82+20)90 +90 and B= (6) 81-83 81-83 5. (10 marks) Using values of L = 1mH, C = 10μF, choose input values such that a = wo and Vc(0)=1V, and use your modified function to plot the voltage drop across the resistor, capacitor, and inductor, and the sum of all three, and plot the energy stored in the inductor and capacitor, and the total energy stored, as a function of time. Label the R, L, C, and go values on each graph.
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter19: Capacitors
Section: Chapter Questions
Problem 2PA: You are an electrician working in an industrial plant. You discover that the problem with a certain...
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Transcribed Image Text:Consider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before
the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go-
R
9(1)
i(t)↓
After the switches closes, current flows through the circuit and the capacitor begins to discharge.
The equation that describes the total voltage in the loop comes from Kirchoff's voltage law:
L
di(t)
+ Ri(t)+(0) = 0,
(1)
where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance,
R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of
change of the capacitor charge, and dividing by L, we can write the following homogeneous (no
input source) differential equation for the charge on the capacitor:
4(1) +29(1)+w79(1)=0,
ཀྱི
where
a=
R
2L
and
The solution to this second order linear differential equation can be written as:
9(1) =Aent - Beat,
where
(3)
(4)
(5)
A=
(81+20)90 +90
(82+20)90 +90
and B=
(6)
81-83
81-83
Transcribed Image Text:5. (10 marks) Using values of L = 1mH, C = 10μF, choose input values such that a = wo
and Vc(0)=1V, and use your modified function to plot the voltage drop across the resistor,
capacitor, and inductor, and the sum of all three, and plot the energy stored in the inductor
and capacitor, and the total energy stored, as a function of time. Label the R, L, C, and go
values on each graph.
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