```MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2013
Problem Set 8 Solutions
Problem 1: Inductor
An inductor consists of two very thin conducting cylindrical
shells, one of radius a and one of radius b, both of length h.
Assume that the inner shell carries current I out of the page,
and that the outer shell carries current I into the page,
distributed uniformly around the circumference in both
cases. The z-axis is out of the page along the common axis
of the cylinders and the r-axis is the radial cylindrical axis
perpendicular to the z-axis.
a) Use Ampere’s Law to find the magnetic field
between the cylindrical shells. Indicate the direction
of the magnetic field on the sketch. What is the
magnetic energy density as a function of r for a < r < b?
The enclosed current I enc in the Ampere’s loop with radius r is given by
I enc
Applying Ampere’s law,
ì0, r < a
ï
= íI , a < r < b
ï0, r > b
î
ò B × d s = B(2p r) = m I
0 enc
, we obtain
The magnetic energy density for a < r < b is
2
m0 I 2
B2
1 æ m0 I ö
uB =
=
=
2m0 2m0 çè 2p r ÷ø
8p 2r 2
It is zero elsewhere.
b) Calculate the inductance of this long inductor recalling that U B =
1 2
LI and using
2
your results for the magnetic energy density in (a).
The volume element in this case is 2p rhdr . The magnetic energy is:
b
æ m I2 ö
m I 2h æ b ö
U B = ò uB dVol = ò ç 02 2 ÷ 2p h rdr = 0
ln ç ÷
4
p
è aø
8
p
r
è
ø
V
a
Since
UB =
m0 I 2l æ b ö 1 2
ln ç ÷ = LI ,
4p
è aø 2
the inductance is
L=
m0 h æ b ö
ln
2p çè a ÷ø
.
c) Calculate the inductance of this long inductor by using the formula
F = LI = ò B ×dA and your results for the magnetic field in (a). To do this
open surface
you must choose an appropriate open surface over which to evaluate the magnetic
flux. Does your result calculated in this way agree with your result in (b)?
The magnetic field is perpendicular to a rectangular surface shown in the figure. The
magnetic flux through a thin strip of area dA = ldr is
æm Iö
m Ih
dF B = BdA = ç 0 ÷ ( h dr ) = 0 dr
2p r
è 2p r ø
Thus, the total magnetic flux is
B   dB  
b
a
0 Ih
 Ih b dr 0 Ih  b 
dr  0 

ln  
2 r
2 a r
2
a
Thus, the inductance is
L=
which agrees with that obtained in (b).
m0 h æ b ö
ln
2p çè a ÷ø ,
Problem 2: Self Inductance, Energy, Induced Electric Fields, and Faraday’s Law
A wire is wrapped N = 100 times around a cylinder of non-magnetized material of radius
a = 3.0cm and length l = 2.5´101 cm . At t  0 , the current through the wire is increased
according to
I(t) = bt,
0 < t < 2s
where b = 2.0 ´10-1 A ×s-1 . Assume the direction of the current through each turn is
clockwise when seen from above. After t ³ 2s the current remains constant. A small
wire loop of radius r = 1.0cm is placed on the solenoid’s axis at the mid point of the
solenoid with the normal to the plane of the loop pointing along the solenoid’s axis. The
loop has resistance R = 5.0 W .
a) After t ³ 2s , use Ampere’s Law to find the direction and magnitude of the
magnetic field within the solenoid. (Ignore edge effects).
Answer: The magnetic field inside a solenoid can be found using Ampere’s Law,
7
-1
1
-1
0 NI 0 Nbt  410 T  m  A  100   2.0 10 A  s   2s 
B


 2.0 104 T .
1
l
l
2.5 10 m
b) Sketch the magnetic field lines everywhere including edge effects.
c) What is the self-inductance of the solenoid?
Answer: The magnitude of the magnetic flux through the solenoid is
ò
solenoid
B×dA = N
ò
oneturn
B × d A = NBp a 2 = N ( m0
N
I )p a 2
l
The magnetic flux through the solenoid is proportional to the current,
ò
N
B × d A = LI .
oneturn
The constant of proportionality is called the self-inductance,
ò
L= N
B×dA I .
oneturn
So the self-inductance of the solenoid is
2
Lsolenoid
(
) (
-7
-1
-2
N 2 m0p a 2 (100 ) 4p ´10 T × m × A (p ) 3.0 ´10 m
=
=
l
2.5 ´10-1 m
)
2
= 1.4 ´10-4 H
d) The rate of doing work against the back emf is dW / dt = -e I , where the emf is
given by e = -L(dI / dt) . How much work is done against the back emf in order
to reach a steady current after t  2s in the solenoid?
Answer: The voltage source must do work against this back-emf. The rate of doing work
(power) is
dI
P  dW dt   back I  LI
.
dt
So the total work is the integral
t
t
I
dI
1
W   Pdt   LI dt   LIdI  LI 2
dt
2
t0
t0
0
e) The stored energy in a magnetic field is equal to U magnetic 
1
2 0

B 2 dV . How
all space
much energy is stored in the magnetic field of the solenoid? Do you expect your
result to agree or disagree with your answer in part e)? Briefly explain your
reasoning.
Answer: Since the magnetic field is uniform, the integral is just
U mag 
1
20

all space
B 2 dV 
1 2
B Vvol
20
.
Then we use our result for the magnetic field of a solenoid and self-inductance to find
that
1
1 2
1 2 2 N 2 0 a 2 I 2 1 2
2
U mag 
 B dV  20 B Vvol  20 B  a l  2l  2 LI
20 all space
This agrees with the energy we put into the solenoid to overcome the back-emf.
f) The changing magnetic flux inside the solenoid produces an induced tangential
electric field. During the interval, 0  t  2s , find an expression for the induced
electric field as a function of distance from the central axis of the solenoid.
òE
ind
×ds = -
d
B×dA
dt òò
For 0 < t < 2s , the change in magnetic flux through an open disk of radius r due to the
changing current I(t) = bt in the solenoid is given by

 N dI 2
N
d
d
Bkˆ  dAkˆ    BdA   0
 r   0 b r 2 .

dt
dt
l dt
l
The line integral around the circle of radius r bounded the pen disk is
òE
f =2p
ind
×ds =
ò
Eind r df = Eind 2p r
f =0
where we have assumed that the cylindrically symmetric changing magnetic flux through
will produce an induced tangential electric field,
becomes
mN
Eind 2p r = - 0 bp r 2 Þ
l
N
Eind   0 br .
2l
The negative sign means that the electric field points in the clockwise direction when
seen from above in the figure above (part b).
Problem 3 Mutual Inductance of Two Rings
Consider the circuits shown in the figure below, two coplanar, concentric rings, a small
ring C2 of radius R2 and a much larger ring C1 of radius R1 with current I1 . What is the
mutual inductance? You may assume that R2 << R1 , so that you may neglect the variation
of the magnetic field B1 over the interior of the small ring.
Solution: We begin by calculating the magnetic field B1 at the center of ring C2 using
the Biot-Savart Law
m Id s ´ r̂
dB = 0
.
4p r 2
Choose cylindrical coordinates with unit vectors
. Note is the unit vector that
points radially outward in the plane. We have for the current element
, the
distance from the current element to the center of ring C2 is r = R1 , and the unit vector
from the current element to the field point at the center of the rings is
. Therefore
the Biot-Savart Law becomes
Integrating around the circle we get that
m0 I1 2p
mI
B1 =
dqk̂ = 0 1 k̂ .
ò
4p R1 0
2R1
Because we are assuming that R2 << R1 , the magnetic field B1 is approximately uniform
over the small ring C2 so the magnetic flux F 21 through the ring C2 due to the magnetic
field of ring C1
F21 = B1p R22 =
m0 I1p R22
2R1
Therefore the mutual inductance is given by
F21 m0p R22
.
M 21 =
=
I1
2R1
.
Problem 4 Straight Wire and Rectangular Loop of Wire
At t = 0, there is a current I 0 through a very long wire. (You may assume this is an
infinite wire). No current is flowing in a rectangular loop of wire with dimensions length
l and width w, and has a resistance R . The side of the square loop closest to the wire is
a distance s away from the wire. At t = 0 , the current in the wire is decreased according
to
ì(1- a t)I0 , 0 < t < 1/ a
I(t) = í
ïî0, t > 1/ a
a) In what direction does the induced current in the rectangular loop flow?
b) What total charge passes a given point in the loop during the time this current
flows?
Solution:
a) In what direction does the induced current in the rectangular loop flow?
Answer: For the time interval 0 < t < 1/ a , the current is decreasing, therefore the
magnetic field generated by this current is also decreasing, and hence the magnetic flux
in the rectangular loop is also decreasing into the plane of the figure. By Lenz’z Law,
there is an induced current in the rectangular loop in the clockwise direction to oppose
this change in magnetic flux. We can use Ampere’s Law to calculate the magnetic field
due to the wire, and then Faraday’s Law in the form
I ind R = -
d
òò B × n̂da
dt loop
to calculate the induce current.
In the figure below we show our choice of a circle for the Amperian loop.
Then Ampere’s Law
ò B × ds = m I
0 enc
becomes B2p r = m0 I and so the magnetic field
C
points into the plane of the figure and the magnitude varies with the distance r from the
wire according to
mI
B= 0 .
2p r
In order to apply Faraday’s Law, we choose n̂ = k̂ into the plane of the figure below, we
choose for an integration area element da = ldr , and the magnetic field in that element is
mI
given by B = 0 k̂ .
2p r
Therefore the left hand side of Faraday’s Law becomes
-
d
d
B × n̂da = òò
dt loop
dt
=-
r = s+ w
ò
r =s
m0 I
m l dI r = s+ w dr
k̂ × k̂ldr = - 0
2p r
2p dt r ò= s r
m0 l dI æ s + w ö
ln
2p dt çè s ÷ø
From our function that describes the decreasing current we have that
dI ïì-a I0 , 0 < t < 1 / a
=í
dt ïî0, t > 1 / a
Therefore for the interval 0 < t < 1/ a ,
-
m la I æ s + w ö
d
.
B × n̂da = 0 0 ln ç
òò
dt loop
2p
è s ÷ø
I ind R =
m0 la I 0 æ s + w ö
.
ln ç
2p
è s ÷ø
We can now solve for the induced current
I ind =
m0la I0 æ s + w ö
.
ln
2p R çè s ÷ø
Note that I ind > 0 , and because we choose n̂ = k̂ into the plane of the figure, a positive
I ind means that the current is in the clockwise direction consistent with our argument
using Lenz’s Law.
b) What total charge passes a given point in the loop during the time this current flows?
During the time interval 0 < t < 1/ a , the total charge passing a given point in the loop is
given by
Q=
t =1/ a
ò
t =0
I ind dt =
t =1/ a
ò
t =0
m0 la I0 æ s + w ö
m0lI0 æ s + w ö
.
ln ç
dt
=
ln
2p R è s ÷ø
2p R çè s ÷ø
Problem 5: Alternating-Current Generator
An N-turn rectangular loop of length a and width b is rotated at a frequency f in a
uniform magnetic field B which points into the page, as shown in the figure below. At
time t = 0, the loop is vertical as shown in the sketch, and it rotates counterclockwise
when viewed along the axis of rotation from the left.
a) Make a sketch depicting this “generator” as viewed from the left along the axis of
rotation at a time t shortly after t = 0, when it has rotated an angle  from the
vertical. Show clearly the vector B , the plane of the loop, and the direction of the
induced current.
b) Write an expression for the magnetic flux  B passing through the loop as a
function of time for the given parameters.
Answer: The dot product between the magnetic field and the unit normal is
B × n̂ = Bcosq (t) .
(1)
The angle  (t )  t  0 where  is the angular frequency is related to the frequency f
by   2 f , and  0  0 is the angle between n̂ and B and t  0 . The magnetic flux
through the loop is
Fmagnetic = N
òò
oneturn
B×dA = N
òò
oneturn
Bcos ( 2p ft ) dA = NBabcos ( 2p ft ) .
(2)
c) Show that an induced emf
 appears in the loop, given by
e = 2p fNBabsin(2p ft) = e 0 sin(2p ft) .
Answer: The time derivative of the magnetic flux is
d  magnetic
dt
 NBab
d cos  2 ft 
 2 fNBab sin  2 ft 
dt
(3)
So there is a non-zero electromotive force in the wire loop.
e = - d òò
dt
open
surface
where
B × d A = 2p fNBabsin ( 2p ft ) = e 0 sin ( 2p ft )
(4)
 0  2 fNBab .
d) Design a loop that will produce an emf with  0  120 V when rotated at 60
revolutions/sec in a magnetic field of 0.40 T. (Describe the dimensions of the
loop and the number of wrappings.)
Answer: At first we may want to minimize the length of the wire that we need.
Suppose we have a square loop of side a . The total length of the wire with N turns
is lN  4 Na . If we choose a single square loop of side a1 , then the length of a single
lN 2
l12

. This implies
4N 2 4
that the length of the loop with N turns is lN  Nl1 . The length increases as a
loop is l1  4a1 . Since we want Na 2  a12 . We have that N
function of the N . So to minimize costs we choose N  1. Since the electromotive
force 0  2 fBNa 2 , the length of a side is
 
a 0
 B 2
12


f 


1.2×102 V 



  4.0×10-1 T   2π  60 Hz  


.
This is a pretty big loop, a  89 cm .
12
 8.9×10-1m
(5)
Problem 6
A pair of conducting rails separated by a distance w is inclined at an angle q to the
horizontal plane. The lower ends of the rails are connected by a resistor with resistance
R . A conducting bar of mass m slides along the rails under the influence of gravity.
There is a constant external magnetic field of magnitude B 0 that points vertically
upwards. Assume that there is no friction between the bar and the rails. Use g for the
gravitational constant. The resistance of the bar, rails, and the contact between the bar
and rails is negligibly small.
a) If the bar moves down the rails at a speed v , determine the induced current that
flows through the resistor. Indicate the direction of the current. You may assume
the resistance R is large enough to ignore self-inductance.
Answer: Choose a coordinate as shown in the sketch below, where k̂ is perpendicular to
the plane of the loop pointing in the direction shown in the figure below. Let x(t) be the
distance from the bottom of the rail to the bar. Then the area of the loop is A = wx(t) .
Choose a counterclockwise circulation for the conducting loop formed by the bar and the
rails. Then the area vector for the flat rectangular surface with the loop as the boundary is
A = An̂ = wx(t) k̂ . The magnetic field is B = Bcosq k̂ + Bsinq î . The flux in the flat
rectangular surface is given by
F B = B× A = B(cosq k̂ + sinq î)× wx(t) k̂ = Bcosq wx(t)
Then,
e =-
dF B
dx(t)
= -Bcosq w
= -Bcosq wvx (t) .
dt
dt
.
Note that vx (t) = -v . Therefore
.
e = Bcosq wv
Because the resistance of the circuit is R , the magnitude of the induced current is
I=
e Bcosq wv
Bcosq wvx
=
=R
R
R
(1)
The induced emf is positive therefore the induced current has the same direction as the
circulation direction which is counterclockwise. This agrees with the fact that the upward
flux is decreasing and therefore we need a counterclockwise current to produce more
upward flux to oppose the change.
b) If the bar moves down the rails at a speed v , what is the magnitude and direction
of the induced force on the bar?
Answer: The magnetic force on the bar is given by
Find = Iw ´ B = Iwĵ ´ (Bcosq k̂ + Bsin q î)
Bcosq w2vx
= Iw(Bcosq î - Bsinq k̂) = (Bcosq î - Bsinq k̂)
R
c) Using Newton’s Second Law, derive a differential equation for the velocity v(t)
of the bar as a function of time. Think about choosing the easiest coordinate
system for describing the motion down the inclined plane.
Answer: The total force on the bar is the vector sum of the normal force N = Nk̂ , the
gravitational force mg = mg(cosq (-k̂) + sinq (- î)) , and the induced force, Find .
The vector sum is
N + mg + Find
= (N - mg cosq -
æ B 2 cos 2 q w2v
ö
B 2 sin q cosq w2v
)k̂ + mg sin q (- î)) + ç
- mg sinq ÷ î
R
R
è
ø
The bar is only accelerating in the x -direction, so Newton’s Second Law in that direction
is
æ B2 cos2 q w2vx
ö
dv
- mg sinq ÷ î = m x î .
çR
dt
è
ø
d) Solve the equation in part c) for the velocity as a function of time.
Answer: The equation of motion is
æ B2 cos2 q w2vx
ö dv
- g sinq ÷ = x
çmR
è
ø dt
Let a = B2 cos2 q w2 / mR then
-a vx - g sinq =
dvx
.
dt
Separate variables and rewrite the equation as
-a dt =
dvx
.
vx + g sinq / a
Integrate both sides
t
-a ò dt =
0
Integration yields
vx (t )
dvx
ò v + g sinq / a
v =0 x
x
(2)
æ v (t) + g sinq / a ö
-a t = ln ç x
÷ø .
g sinq / a
è
Exponentiate both sides
æ v (t) + g sinq / a ö
e- a t = ç x
÷ø .
g sinq / a
è
Solve for the x -component of the velocity as a function of time
vx (t) = -(g sinq / a )(1- e-at )
(3)
e) What is the terminal velocity of the bar?
Answer: When t ® ¥ ,
vx,term = lim(vx (t)) = -(g sinq / a ) = t®¥
g sinq mR
B2 cos 2 q w2
(4)
Note that at terminal velocity, dvx / dt = 0 . Therefore at terminal velocity, Newton’s
Second Law in the x -direction (Equation (2)) becomes
-a vx,term - g sinq = 0 .
The terminal velocity is then
vx,term = -g sinq / a ,
agreeing with our result above.
f) What is the rate at which electrical energy is being dissipated through the resistor?
Answer: Using Equation (1), electrical energy is dissipated in the resistor at the rate
P = I 2R =
B2 cos 2 q w2 vx 2
.
R
g) What is the rate of work done by gravity on the bar? The rate at which work is
done is F × v . How does this compare to your answer in (f)? Why?
Answer: Work is done on the bar by the gravitational force according to
dW = mg × ds = (mg(cosq (-k̂) + sinq (- î)))× dxî = -mg sinq dx .
The rate of doing work is
dW
dx
= -mg sinq
= -mg sinq vx .
dt
dt
The rate that the kinetic energy is increasing is
dK m d(vx 2 )
=
= mvx dvx .
dt
2 dt
The rate of doing gravitational work minus the rate that energy is dissipated is equal
to the rate that the kinetic energy is increasing
-mg sinq vx -
B2 cos2 q w2vx 2
dv
= mvx x .
R
dt
We can rewrite this equation as
-g sinq -
B 2 cos 2 q w2 dvx
.
=
mR
dt
This is the identical to Equation (2), which we found by using Newton’s Second Law
in the x -direction.
Problem 7: Pressure Transducers
A common way of measuring small pressure changes (for example, in car transmission
systems, airplane cabin pressure monitors, and Parenchymal catheter tips, used, for
example, to measure intercranial pressure) is to use piezoresistive materials, that is,
materials that change resistance when a pressure is applied to them. Often, four identical
piezoresistors are connected as shown below, in such a way that when a pressure is
applied Ra and Rd increase by some fraction and Rb and Rc decrease by about the same
fraction. The potential difference Vin = V2 -V1 and Vout = V3 -V1
a) Determine an expression for Vout / Vin in terms of Ra , Rb , Rc , and Rd .
Answer: Introduce currents in each branch according to the figure below.
Then
Vin
Ra + Rc
Vin
.
Vin = I 2 (Rb + Rd ) Þ I 2 =
Rb + Rd
Vin = I1 (Ra + Rc ) Þ I1 =
We also know that
V4 -V1 = I1 Rc
V3 -V1 = I2 Rd
Then
Vout = (V3 - V4 ) = (V3 - V1 ) - (V4 - V1 )
= I 2 Rd - I1 Rb
=
.
Vin Rd
V R
- in b
Rb + Rd Ra + Rc
Therefore
Vout
Rd
Rb
=
Vin Rb + Rd Ra + Rc
(1)
Note that this is like the Wheatstone bridge, where an ammeter was connected across
the potential difference Vout = (V3 -V4 ) .
b) If a 1 V battery is connected across points 1 and 2 such that Vin = V2 -V1, what
voltage will be measured at Vout = V3 -V4 if the system is unstrained and all four
resistors have the same resistance R ?
Answer: If they all have the same resistance then by symmetry the output voltage will be
zero.
Vout =
Vin R
V R
- in = 0
R+ R R+ R
c) An unfortunate problem with piezoresistors is that they are also temperature
dependent. If a temperature change causes the resistance of all four resistors to
increase by 5%, what is the new output voltage at Vout = V3 -V4 ?
Answer: They all still have the same resistance so the output voltage will still be zero.
d) Starting from the condition where all four resistors have resistance R , a pressure
is applied that causes Ra and Rd to increase by 1% and Rb and Rc to decrease
by 1%. What now is measured at Vout = V3 -V1?
Answer: We can substitute the values Rd = Ra = 1.01R and Rb = Rc = 0.99R into Eq. (1)
yielding
æ 1.01R 0.99R ö
Vout = (1.0 V) ç
= 0.01 V .
2R ÷ø
è 2R
Problem 8 Design an Ohmmeter.
You own a micro-ammeter that reads 50 m A at full scale deflection, and the coil in the
meter movement has a resistance of 20 W . By adding two resistors, R1 and R2 , and a
1.5 V battery you can convert this into an ohmmeter. When the two outgoing leads of
this ohmmeter are connected together, the meter should register 0 ohms by giving fullscale deflection. When the leads are connected across an unknown resistance Ru , the
deflection will indicate the resistance value if the scale is appropriately marked. In
particular, we want half-scale deflection to indicate 15 W . What values of R1 and R2 , are
required, how should the connections be made, and where on the ohm scale will the
marks be (with reference to the old micro-ammeter calibration) for 5 W and 50 W ? Note
that there is more than one way to choose these resistors. Depending on your values for
the resistors, do you think your design is a feasible design? HINT: Can you make any
approximations that make your calculations far easier?
Solution:
If we have a e = 1.5 V battery and two resistors, we would like to form a circuit so that
we can convert current readings on the ammeter into resistance measurements across
some resistor of unknown resistance Ru .
The ammeter has an internal resistance RA = 20 W. We want I A = 50 mA through the
ammeter at full-scale deflection
We will calibrate our new ‘ohmmeter’ by requiring full-scale deflection when Ru = 0 ,
and half scale deflection when Ru = 15 W . Half scale deflection means that the ammeter
measures 25 m A. We can achieve half-scale deflection when Ru = 15 W by connecting
the resistors and ammeter in some combination such that the equivalence resistance is
Req = 15 W . This will reduce the current everywhere in the circuit by a factor of two,
I new = (1 / 2)I old .
Note that when the unknown resistor is replaced by a wire the current in the circuit is
I old =
1.5 V
= 0.1 A.
15 W
The current I old = 0.1 A is much larger than the I A = 50 mA through the ammeter so we
need a current divider. We can make a current divider by forming two parallel branches.
To reduce the current in the branch containing the ammeter, we connect resistor 2 in
series with the ammeter with R2 >> RA such that the equivalent resistance is then
Req¢ = R2 + RA R2 .
In the other branch where we shunt most of the current, we have resistor 1 with a small
resistance compared to the other branch R1 << Req
¢.
Because we know that the combination of resistors 1, 2, and the ammeter, must have an
equivalent resistance of 15 W , we can solve for the values of R1 and R2 .
We begin by calculating the equivalent resistance of the resistors 1, 2, and the ammeter
and set it equal to 15 W . We require that R1 << Req
¢ so
Req =
R1 Req¢
R1 Req¢
R1 + Req¢
Req¢
= R1 = 15 W .
In order to calculate the resistance of resistor 2, we note that for full scale deflection
Ru = 0 and I A = 50 mA , and so we can solve for the value of the required large
resistance
e
IA
= R2 + RA
R2 .
and thus
R2 =
1.5 V
= 3 ´ 10 4 W
50 m A
This is a reasonable design because we can easily find resistors with values that are
needed. We made some approximations but most resistors are manufactured with 1% 5% deviation from the stated values so our approximations are well within the
manufacturing range of tolerance for the resistors.
We note that a full-scale deflection the electric potential difference across the ammeter is
DVA = I A RA = (20 W)(50 mA) = 1 mV .
Therefore we cannot connect the ammeter directly across the battery which maintains a
1.5 V electric potential difference. This is further justification that we need to add a
resitor with a large amount of resistance in series with the ammeter.
There is another solution which is to shunt the current away from the ammeter with a
very small resistor R2 .
Then the equivalent resistance of the parallel elements is
Req¢¢ =
R2 RA
R2 + RA
R2 RA
= R2 .
RA
Then we need to add an additional resistor with resistance R1 = 15 W to give an overall
equivalent resistance
Req = R1 + Req¢¢
R1 + R2
15 W .
Because we apply an electric potential difference of
DVA = I A RA = (20 W)(50 mA) = mV
across the ammeter, and I old =
e 1.5 V
=
= 0.1 A , the resistor in parallel with the
Req 15 W
ammeter must have a resistance
R2
DVA
mV
=
= 0.01 W .
I old 0.1 A
This is a very small resistance and hard to manufacture so we would prefer our first
solution.
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