Download Physics 270, Assignment 4

Physics 270, Assignment 4
31.14
We wish to know what the power dissipated by each resistor in the con…guration is. We know that the
power in a resistor is given by P = I 2 R. We know that both resistors will have the same current because
they are in series. Furthermore, their equivalent resistance Req is given by
Req = R1 + R2 = 12
+ 15
= 27
The current is then obtained by Ohm’s law Kircho¤’s loop rule
9V
IReq = 9 V
I (27
)=0
or
I = 1=3 A
Therefore, the powers dissipated in R1 and R2 are given by
P1
P2
2
= 4=3 J/s
2
= 5=3 J/s
= I 2 R1 = (1=3 A) 12
2
= I R2 = (1=3 A) 15
31.32
We now want the equivalent resistance between the points a and b in the …gure. We can model the system as
three resistors in parallel. The top wire con…guration has three resistors in series and therefore an equivalent
resistance of Rt = 3 (100 ) = 300 . The middle wire has only two resistors in series. Thus, we have an
equivalent resistance of Rm = 2 (100 ) = 200 . The bottom wire has resistance Rb = 100 . We can
think of these three wires as three resistors in parallel. Hence, the total equivalent resistance Req of the
system is given by:
1
1
1
1
=
+
+
Req
Rt
Rm
Rb
or
1
1
600
Req = 1
= 54:54
=
1
1
1 =
1
1
11
+
+
+
+
Rt
Rm
Rb
300
200
100
33.17
Now we would like the energy stored in a 0:03 m diameter, 0:12 m long solenoid with 200 turns and a current
of 0:80 A. We can think of the solenoid as an inductor with an energy given by
UL =
1 2
LI
2
where L is the inductance of a solenoid given by
Lsolenoid =
0N
2
A
l
This gives us
1
UL =
2
0N
2
l
AI 2
1
=
2
0N
2
R2 I 2
l
1 1:26
=
2
10
6 Tm
A
2002
(0:12 m)
1
2
(0:015 m)
2
(0:80 A)
= 1:90
10
4
J
33.66
We are now given that the current rhough inductance L is given by I = I0 sin (!t).
expression for the potential di¤erence VL across the inductor. This is given by
VL =
L
dI
=
dt
We would like an
LI0 ! cos (!t)
Next we are given that the maximum voltage across the inductor is 0:20 V when L = 50 10 6 H and
f = 500 kHz. We now want I0 . We may merely plug these values into the formula above. Note that the
maximum voltage is achieved when cos (!t) = 1.
VL,m ax = LI0 !
Thus,
I0 =
VL,m ax
VL,m ax
=
=
L!
L (2 f )
(50
6
10
0:20 V
H) (2 (500
103 s
1 ))
= 0:0012 A
33.68
Next, we are given an LC circuit with a 10 10 3 H inductor and an 8:0 10 6 F capacitor. The current
attains its maximum value of 0:5 A at t = 0. We want to know how long it will be before the capacitor is
fully charged. We know that when the current is at its maximum, the charge across the capacitor is zero.
Therefore, the charge will be at a maximum after 1/4 of the period of oscillation has passed. That period
T is given by
T =
2
2
=q
!
=2
1
LC
p
LC = 2
p
(10
10
3
H) (8:0
10
6
F) = 0:0017 s
Hence, the capacitor will be charged after T =4 where
T
= 4:44
4
10
4
s
We would also like the voltage across that capacitor at that time. By conservation of energy, we know
that the energy of the inductor at t = 0 will be equal to the energy of the capacitor when it is fully charged
(and therefore no current goes through the inductor.) That energy relation is
1 2
1
LI
= CV 2
2 max
2
or
V =
2
10 10
LImax
=
C
(8:0
3
2
H (0:5 A)
= 312:5 V
10 6 F)
33.72
We want to know the maximum voltage to which we can charge the 1200 F capacitor is. We know that
however we do it, the voltage will be transfered by means of a circuit that can be modelled as an LC circuit.
As such, we can be sure that the maximum voltage on the1200 F capacitor will be attained when the voltage
across the inductor is zero. Hence, we want all the energy of the system to be on the1200 F capacitor,
under the constraint that energy of the total system is conserved. Initially, the energy is
E=
1
C1 V12
2
2
where C1 = 300 F and V1 = 100 V. At a later time, we must have
E=
1
1
C1 V12 = C2 V22
2
2
where C1 = 1200 F and V2 is unknown. This gives us
s
r
C1 2
300 F 2
1
1
V2 =
V1 =
V1 = V1 = 100 V = 25 V
C2
1200 F
4
4
33.74
We are given that the switch in the …gure has been open for a long time, but it is closed at t = 0. We want
the current through the 20 resistor. Initially, when the switch is just closed dI
dt through the inductor is
very big. As such, all the current ‡ows through the path of least resistance: through the 20 resistor as
if the inductor were cut out of the circuit. Hence the current is
I=
30
+ 20
V
=
Req
10
=1A
When the switch is closed for a long time, the voltage di¤erence across the inductor is zero because
= 0. Thus, all current ‡ows through the inductor and nothing at all ‡ows through the 20 resistor.
Lastly, if the switch is immediately reopened, then we have an LR circuit. Hence, there will be an
exponential decay with an initial current (the current we are trying to …nd) of
dI
dt
I0 =
30 V
= 3=2 A
20
33.81
We will now …nd the inductance per meter of coaxial cable. We remember that inductance is de…ned to be
the magnetic ‡ux through a surface per unit current applied. They tell us to use a rectangle with length l
spanning the gap between inner and outer conductors. The magnetic …eld induced by the inner conductor
can be derived from Ampere’s law.
I
~ d~l = B2 r = I
B
0
or B =
0I
2 r.
The magnetic ‡ux through the rectangle will then be the integral.
I
Z r2 Z l
Z r2
Il
I
Il
~ dA
~=
= B
dr
dx 0 =
dr 0 = 0 ln
2 r
2 r
2
r1
0
r1
r2
r1
Thus, the inductance is
L=
I
=
0l
2
r2
r1
ln
and the inductance per unit length is
L
= 0 ln
l
2
r2
r1
For the parameters given in the problem, this is
L
= 0 ln
l
2
r2
r1
=
1:26
10
2
6 Tm
A
3
ln
3 mm
0:5 mm
= 3:59
10
7
Tm
A
31.38
We want the time-constant for the con…guration in the …gure. We may model this as a system with an
equivalent resistance Req and an equivalent capacitance Ceq where
1
+
=
Req
=
Ceq
= C1 + C2 = 2
1
R1
1
R2
1
+
1
1 103
10
6
= 500
1
1 103
F+2
10
6
F=4
10
6
F
Thus, the time constant is
= Req Ceq = (500
) 4
6
10
F = 0:002 s
31.80
We are given an RC circuit driven by a battery. We want the total energy supplied to by the battery as the
capacitor is being charged. The charge on the capacitor is given by equation 31.40
Q = Qmax 1
where
e
t=
= RC and Qmax = CV . Di¤erentiating this expression gives us the current through the system.
I=
CV
dQ
=
e
dt
t=
=
V
e
R
t=
To get the total energy supplied by the battery, we shall need to integrate the power Pbatt = IV from t = 0
to t = 1. This gives us
Z 1
Z 1
V2
V2
V2
Ebatt =
dtIV =
dt e t= =
=
RC = CV 2
R
R
R
0
0
Next, we want the total energy dissipated by the resistor. We may do the same thing: integrate power
Pres = I 2 R.
Eres =
Z
0
1
dtI 2 R =
Z
0
1
dt
V
e
R
2
t=
R=
V2
R
Z
1
dte
2t=
=
0
1
V2
= CV 2
R 2
2
The capacitor’s energy, once fully charged, is well-known:
Ecap =
1
CV 2
2
Lastly, we note that our results show that energy is indeed conserved. The energy supplied by the batter
gets divided evenly between the resistor and the capacitor.
Ebatt = CV 2 =
1
1
CV 2 + CV 2 = Eres + Ecap
2
2
4