Solutions 2.7-Page 176
Problem 1
In the circuit of Fig. 2.7.7, suppose that L = 5 H, R = 25 Ω , and that the source E of emf
is a battery supplying 100 V to the circuit. Suppose also that the switch has been in
position 1 for a long time, so that steady current of 4A is flowing in the circuit. At time
t = 0 , the switch is thrown to position 2, so that I (0) = 4 and E = 0 for t ≥ 0 . Find I (t ) .
It is given that LI ′ + RI = E (t ) describes the circuit. Substituting given information
yields:
5 I ′ + 25 I = 0
Separating variables and integrating yields:
∫
dI
= −5∫ dt
I
ln I = −5t + K
~
I = Ke − 5 t
The initial condition I (0) = 4 is used to solve for the integration constant as follows:
~
I (0) = 4 = K (1)
~
K =4
∴
I = 4e −5t
Problem 3
Suppose that the battery in Problem 2 is replaced with an alternating current generator
that supplies a voltage of E (t ) = 100 cos 60t volts. With everything else the same, now
find I (t ) .
The differential equation describing the circuit is 5 I ′ + 25 I = 100 cos 60t .
A particular and homogeneous solution must be found to find the entire solution.
Particular solution
The assumed form of the particular solution is I p = A cos 60t + B sin 60t . Substituting
this solution into the differential equation yields:
d ( A cos 60t + B sin 60t )
+ 25( A cos 60t + B sin 60t ) = 100 cos 60t
dt
5(60 B cos 60t − 60 A sin 60t ) + 25 A cos 60t + 25 B sin 60t = 100 cos 60t
5
Equating coefficients yields:
300 B + 25 A = 100
25 B − 300 A = 0
∴
A = 4 / 145 and B = 48 / 145
Substituting these coefficients into the particular solution yields:
I p = (4 cos 60t + 48 sin 60t ) / 145
I p = 4(cos 60t + 12 sin 60t ) / 145
Complementary solution
The homogeneous equation is 5 I ′ + 25 I = 0 . From Problem 1 this equation has the
~
solution of the form I = Ke −5t .
The complementary solution is added to the particular solution and the initial condition is
~
now used to find K .
~
I (0) = 0 = 4(cos 60t + 12 sin 60t ) / 145 + Ke −5t
~
0 = (4 / 145) + K
~
K = −4 / 145
I = 4(cos 60t + 12 sin 60t − e −5t ) / 145
Problem 7
(a) Find the charge Q(t) and current I(t) in the RC circuit if E (t ) = E 0 (a constant voltage
supplied by a battery) and the switch is closed at time t = 0, so that Q(0) = 0.
1
Q = E 0 describes the circuit. The particular and homogeneous
C
solutions must be found.
It is given that RQ ′ +
Q p = A is the assumed form of the particular solution. Substituting back into the
differential equation and solving for the constant yields A = E 0 C .
The homogeneous equation is RQ ′ +
1
Q = 0 . Separating variables and integrating
C
yields:
∫
dQ
1
=−
∫ dt
Q
RC
ln Q = −
1
t+K
RC
1
~ − t
Q = Ke RC
The homogeneous solution is added to the particular solution and the initial condition is
~
now used to find K .
~
Q(0) = 0 = E 0 C + K (1)
~
K = − E0 C
∴
Q = E 0 C (1 − e − t / RC )
Recall that I = Q ′ . Therefore,
I=
E 0 −t / RC
e
R
(b) show that
lim Q(t ) = E 0 C and that lim I (t ) = 0 as t → ∞
As t → ∞, e − t / RC → 0 ∴ Q(t ) → E 0 C
As t → ∞, e − t / RC → 0 ∴ I (t ) → 0
Problem 11
The parameters of an RLC circuit with input voltage E (t ) are given. Substitute
I sp (t ) = A cos ω t + B sin ω t
in Eq.(4) using the appropriate value of ω , to find the steady periodic current in the form
I sp = I 0 sin(ω t − δ ) .
R = 30 Ω , L = 10 H , C = 0.02 F , E (t ) = 50 cos 2t
From Eq.10 and Eq.12,
I0 =
E0
=
Z
E0

1 

R 2 +  ω L −
ω C 

2
=
120


1

(30) 2 +  2(10) −
(2)(0.02) 

2
= 1.64399 Ω
From Eq. 15,
LCω 2 − 1 (10)(0.02)(2 2 ) − 1
− 0.2
δ = tan −1
=
= tan −1
= −.165149 rad = 2π − .165149 = 6.118 rad
ω RC
2(30)(.02)
1.2
I sp = 1.64399sin(2t − 6.118)
Problem 17
An RLC circuit with input voltage E(t) is described. Find the current I(t) using the given
initial current (in amperes) and charge on the capacitor (in coulombs).
R = 16 Ω , L = 2 H , C = 0.02 F , E (t ) = 100 V ; I (0) = 0 , Q(0) = 5
First we need an initial condition for I ′ . This can be found from Eq.16 in the text.
1
Q(0) = E (0)
C
2 I ′(0) + 16(0) + 50(5) = 100
I ′(0) = −75
LI ′(0) + RI (0) +
Now Eq.4 can be solved for current.
1
I = E ′(t )
C
2 I ′′ + 16 I ′ + 50 I = 0
LI ′′ + RI ′ +
The solution can be found by solving the characteristic equation.
r 2 + 8r + 25 = 0
− 8 ± 64 − 4(25) − 8 ± 6i
=
= −8 ± 3i
2
2
Therefore
I (t ) = e −8t ( A cos 3t + B sin 3t )
r=
Now the initial conditions are used to find the constants.
I (0) = 0 = A
I ′(0) = −75 = −8e −8t ( B sin 3(0)) + e −8t (3B cos 3(0))
− 75 = 3B
B = −25
∴
I (t ) = −25e −8t sin 3t