http://iml.umkc.edu/physics/wrobel/phy250/homework.htm
Homework 6
chapter 32: 11, 18
chapter 33: 5, 50
Problem 32.11
A 12 V battery is connected into a series circuit containing a 10 Ω resistor and a 2 H
inductor. In What time interval will the current reach (a) 50% and (b) 90 % of its final
value?
S
This circuit has only one loop.
From Kirchhoff’s rule we get
only one (differential) equation
dI
ε − L − IR = 0
dt
ε
1
I
L
I
We can separate the variables
dI
dt
=
ε − IR L
R
and integrate both sides of the equation (within appropriate
limits). With the reference for time at the instant of closing the
circuit
I(t )
t dt
dI
=∫
∫
IR
ε
−
0
0L
(Note that I used one symbol t for two different quantities!)
From the fundamental theorem of calculus
ln
ε − I( t ) ⋅ R
R
= − ⋅t
ε
L
The solution is a time dependent function
R
− ⋅t ⎞
ε ⎛⎜
I( t ) =
1− e L ⎟
⎟
R ⎜⎝
⎠
a) Current reaches its final value after an infinite amount of
time.
− ⋅t ⎞
ε⎛
ε
I(∞ ) = lim ⎜1 − e L ⎟ =
⎟ R
t →∞ R ⎜
⎝
⎠
R
At instant t, the current is a fraction of its final value
R
⎛
− ⋅t ⎞
I(t1 ) = I(∞ ) ⋅ ⎜1 − e L ⎟
⎜
⎟
⎝
⎠
From which
R
− ⋅t ⎞
I(t ) ⎛⎜
= 1− e L ⎟
⎟
I(∞ ) ⎜⎝
⎠
Solving this equation for t
t=−
L ⎛
I( t ) ⎞
ln⎜1 −
⎟
R ⎝ I(∞ ) ⎠
Hence
a)
t 50% = −
2H
ln(1 − 0.5) = 0.14s
10Ω
b)
t 90% = −
2H
ln (1 − 0.9 ) = 0.46s
10Ω
Problem 32.18
The switch in Fig. P32.18 is open for t < 0 then closed at time t = 0s. Find the
current in the inductor and the current in the switch as a function of time thereafter.
4Ω
10V
1
8Ω
2
4Ω
I1
1H
I2
I3
S
Using Kirchhoff’s rules we can write three independent
equations. Consistent with the assumed direction of the three
unknown currents. (For the sake of simplicity I will not write the
units but I will make sure all quantities are in the SI system.)
I1
− I2
− 4Ω ⋅ I1 − 4Ω ⋅ I 2
4Ω ⋅ I 2
=
I3
= − 10V
− 8Ω ⋅ I3
− 1H ⋅
dI3
dt
=
0
The rest is math. From the first two equations we can find the
substitution for I2 in the last equation
1
I3
− 4Ω − 10V
2 Ω 5 V 5 V − 2Ω ⋅ I 3
=
=
I2 =
1
1
1
1
4Ω
− 4Ω ⋅
−1 −1
− 4Ω − 4 Ω
1
I3
− 2⋅
With this substitution only one variable (function) is left in the
third equation
(5V − 2Ω ⋅ I3 ) − 8Ω ⋅ I3 − 1H ⋅ dI3 = 0
dt
We can solve this differential equation separating the variables
dI3 − 10Ω ⋅ I3 + 5V
V
1
A
Ω
=
= −10 ⋅ I3 + 5 = −10 ⋅ I3 + 5
dt
1H
H
H
s
s
dI3
1
A
− 10 ⋅ I3 + 5
s
s
= dt
and integrating both sides (within appropriate limits)
I3
∫
dI3
0 A − 10 1 ⋅ I + 5 A
3
s
s
t
= ∫ dt
0s
(Do not confuse the symbol t in the limit with the integration
variable t! I used one symbol for two different quantities.)
Using the fundamental theorem of calculus
1
A
− 10 ⋅ I3 + 5
1
s
s =t
− s ⋅ ln
A
10
5
s
Hence (including the appropriate SI units)
t
⎛
⎞
−
0
.
1
s
⎟
⎜
I 3 = 0 .5 A 1 − e
⎟
⎜
⎝
⎠
In order to find the current I1 in the switch we have to return to
the first two equations
I
−1
−1
− 2⋅ 3
t
t
−
−
⎞
⎛
5V 2Ω 2Ω ⋅ I3 + 5V
− 10V − 4Ω
=
= 1.25A + 0.25A⎜1 − e 0.1s ⎟ = 1.5A − 0.25Ae 0.1s
I1 =
=
⎟
⎜
1 −1
1
−1
4Ω
⎠
⎝
− 4Ω ⋅
1 1
− 4Ω − 4Ω
I3
Problem 33.5
The current in the circuit shown in Figure 33.2 equals 60% of the peak value of the
peak current at t = 7 ms. What is the lowest source frequency that gives this
current?
The problem is phrased ambiguously. Since there is only
one oscillating quantity, we can assume that the initial phase of
the current is zero and the (instantaneous value of) current is a
sinusoidal function of time
I(t ) = I m sin (2πf ⋅ t )
At the indicated instant t = 7 ms, the current assumes p=60% of
its peak value. It requires that
sin (2πf ⋅ t ) = p
This equation has an infinite number of solutions. The smallest
frequency corresponds to the smallest argument of the sin
function satisfying the equation
2πf ⋅ t = arcsin p
From which, the smallest frequency correspond is
f min =
arcsin p
arcsin 0.6
=
= 14.6Hz
2π ⋅ t
2π ⋅ 7 ⋅ 10 − 3 s
Problem 33.50
Show that the rms value for the sawtooth voltage shown in Figure P33.56 is
Vm 3 .
V(t)
Vm
t
-Vm
T
2T
3T
The root-mean-square value of a function in a time interval
(t1,t2) is defined as
t2
Vrms (t 1 , t 2 ) =
2
V
∫ (t )dt
t1
t 2 − t1
We can solve the problem directly from this definition.
We have to find the explicit form of the sawtooth function.
Consistent with the symbols in the figure,
2V
V(t ) = m ⋅ t − (2 N − 1)Vm
for NT < t < (N+1)T
T
(In each time interval [NT, (N+1)T], the voltage is a linear
function of time and assumes the values -Vm and Vm at the
beginning and the end of the interval, respectively.)
Let's find the root-mean-square value of this function in the
time interval from 0 to t’>>T, so that the time interval is much
longer than one period of the function.
2
⎛ 2Vm
⎞
⋅ t − (2 N − 1)Vm ⎟ dt
∫⎜
T
⎠
=
= 0⎝
t '−0
t'
Vrms
2
2
NT + Δt ' 2V
⎞
⎛ m
⎛ 2V
⎞
N ∫ ⎜ m ⋅ t − Vm ⎟ dt
⋅ t − (2 N − 1)Vm ⎟ dt
∫ ⎜
T
⎠
⎠
0⎝ T
+ NT ⎝
=
t'
t'
T
where N is the number of "full" periods in the considered time
interval, and Δt' is the time remaining after the last "full"
period. The last integral is over a time interval shorter than one
period of the function ( Δt’<T ).
Let's discuss each integral separately beginning with the
second one. As time t' is much longer than one period the
second term is approximately zero.
NT + Δt '
0 ≤ lim
t '→ ∞
∫
NT
2
⎛ 2Vm
⎞
⋅ t − (2 N − 1)Vm ⎟ dt
Vm 2 dt
⎜
∫
⎝ T
⎠
< lim 0
=0
t '→ ∞
t'
t'
T
The first term yields
2
T Vm 2
⎛ 2Vm
⎞
N⋅
N∫ ⎜
⋅ t − Vm ⎟ dt
∫ x dx
2
V
m − Vm
⎝ T
⎠
lim 0
= lim
=
t '→ ∞
t '→ ∞
t'
t'
T
NT
1 x3
= lim
⋅
⋅
t '→∞ NT + Δt ' 2V
m 3
Vm
− Vm
Vm2
=
3
Using our findings in the initial expression for the root-meansquare value of the voltage, we can relate it to the peak value of
the voltage.
Vrms
Vm2
V
≈
+0 = m
3
3
Note that similar to the relationship for the sinusoidal voltage,
we can assume that the root-mean-square value of the voltage
does not depend on time for time intervals much longer than the
period of the voltage.