Fall 2003 Math 308/501–502
3 Modeling and Applications
3.5 Electrical Circuits
c
Mon, 22/Sep
2003,
Art Belmonte
Solution
The SLF I 0 +
R
L
I =
E
L
R
I
L
=
(e Rt/L I )0
=
e Rt/L I
=
e Rt/L I 0 + e Rt/L
Summary
reveals integrating factor µ = e Rt/L . Thus
E Rt/L
e
L
E Rt/L
e
L
E Rt/L
e
+K
R
E
+ K e−Rt/L
R
E
+K
R
E
I0 −
R
E
E
+ I0 −
e−Rt/L
R
R
Electric charge flows through a circuit usually containing a
I =
voltage source, resistor, capacitor, and inductor. (Some of these
may be absent.) A battery is a constant voltage source; a generator
I0 = I (0) =
a variable one (typically sinusoidal). An arbitrarily chosen
direction around the circuit (say clockwise or counterclockwise) is
K =
taken as the positive direction. The opposite direction is regarded
as the negative direction. Here is a table of electrical concepts.
I =
Using these concepts and the laws that follow permit us to set up a
model involving a differential equation that governs the circuit.
CONCEPT
UNITS
MATLAB Examples
C, capacitance
F, farads
E, voltage
V, volts
Example B
I, current
A, amps
L, inductance
H, henrys
A resistor (20 ) and capacitor (0.1 F) are joined in series with an
q, charge
C, coulombs
emf E = E(t) = 100 cos 3t V. (See figure below.) If there is no
R, resistance
, ohms
charge on the capacitor at time t = 0, find the ensuing charge on
The rate at which charge flows is called the current; hence
the capacitor at time t.
I = dq/dt. In the following, E k represents the voltage drop due
Resistance
to component k.
R
I
Component Laws
1. Ohm’s law: E R = R I .
E
2. Faraday’s law: E L = L
dI
.
dt
3. Capacitance law: E C =
1
q.
C
Voltage
source
Solution
P
2. Current law:
Capacitance
I
Kirchhoff’s Laws
1. Voltage law:
C
Kirchhoff’s voltage law gives E R + E C − E = 0. Applying
1
Ohm’s law and the Capacitance law, we have R I + q = E.
C
Finally, recalling that I = dq/dt, we obtain the differential
1
equation Rq 0 + q = E or 20q 0 + 10q = 100 cos 3t.
C
E k = 0 around any closed loop in a circuit.
P
In = 0 into any junction.
Thus q 0 + 12 q = 5 cos 3t with q(0) = 0. Via dsolve, we have
60
10 −t/2
q(t) = 10
. (See top of next page.)
37 cos 3t + 37 sin 3t − 37 e
Hand Examples
Example A
%
% T-156/04
%
sol = dsolve(’Dq + Q/2 = 5*cos(3*t)’, ...
’q(0) = 0’, ’t’);
pretty(sol)
10
60
10
-- cos(3 t) + -- sin(3 t) - -- exp(- 1/2 t)
37
37
37
Solve the prototypical IVP that models the L R circuit
L
dI
+ R I = E,
dt
I (0) = I0
where E is a constant source of electromotive force (emf).
1
Example C
Solution
A inductor (1 H) and resisitor (0.1 ) are joined in series with an
emf E = E(t) = e−0.1t V. (See figure below.) If there is no
current in the circuit at time t = 0, find the ensuing current in the
circuit at time t.
Kirchhoff’s voltage law gives E R + E C − E = 0. Applying
1
Ohm’s law and the Capacitance law, we have R I + q = E.
C
Finally, recalling that I = dq/dt, we obtain the differential
1
equation Rq 0 + q = E or 10q 0 + 5q = 10 sin(2πt).
C
Resistance
R
Thus q 0 + 12 q = sin(2πt) with q(0) = q0 for the various initial
charges. Via dfield7, we have the following graph. Irrespective of
the initial condition, the graphs eventually settle in to the so-called
steady-state response. Beauty, eh?
I
E
L
q ’ = sin(2πt) − q/2
1
Inductance
Voltage
source
q
0.5
I
0
−0.5
Solution
Kirchhoff’s voltage law gives E L + E R − E = 0. Applying
dI
+ R I = E or
Faraday’s Law and Ohm’s law, we have L
dt
1
0
−t/10
with I (0) = 0. Via dsolve, we have
I + 10 I = e
−t/10
I (t) = te
.
−1
0
%
% T-156/08
%
sol = dsolve(’DI + I/10 = exp(-t/10)’, ...
’I(0) = 0’, ’t’);
pretty(sol)
exp(- 1/10 t) t
Example D
An RC circuit (see figure below) contains a resistor (10 ) and a
capacitor (0.2 F) linked in series with an emf
E = E(t) = 10 sin(2πt). Use a numerical solver to sketch the
graph of the charge on the capacitor versus time for the ICs
q(0) = q0 = −1, −0.75, −0.5, . . . , 1. Use a time interval of
0 ≤ t ≤ 15 and superimpose all solutions on the same graph.
Resistance
R
E
I
C
Voltage
source
Capacitance
I
2
5
t
10
15