Homework 1
P1.1.2
In a one-dimensional flow of current through a semiconductor, positively-charged
holes move in the positive x-direction at a steady rate of 5  1018 holes/min , and
electrons move in the negative x-direction at a steady rate of
2.5  1018 electrons/ min. Determine the total current in mA in: (a) the positive x-
direction, (b) the negative x-direction. (c) What would be the current if the holes
and electrons move in the same direction? Note that the positive charge of a hole
has the same magnitude as the charge on an electron.
Solution: (a) The current due to positive charges moving in the positive x-direction adds to
the current due to negative charges moving in the negative x-direction. The
total current is:
5  1018  1.6  10 19  2.5  1018  ( 1.6  10 19 )
+
 0.02 A  20
60
60
mA..
(b) The sign of the current is reversed to –20 mA.
(c) The currents due to holes and electrons are in opposition. The total current is:
5  10 18  1.6  10 19 2.5  10 18  ( 1.6  10 19 ) 0.4
20

A
mA in the
+
60
60
3
60
direction of movement.
1-1
P1.1.4
In electroplating an object with silver, the object is made the cathode and a silver
plate the anode. Silver ions in solution are attracted to the cathode, gain
electrons from the cathode, and are deposited on the object as silver ions (Figure
P1.1.4). (a) How many silver ions are deposited by a current of 10 A flowing for 1
hr? Note that a silver ion
carries a single electronic
Power
supply
unit of positive charge
(+1.610-19 C); (b) If a
gram-molecular weight of
–
+
Electrons
I
Electrolytic
solution
silver is 0.1079 kg and
contains 6.0251023
atoms (Avogadros’s
number), what is the
Silver ions
Object
(cathode)
Figure P1.4.4
mass of silver deposited?
Solution: (a) The number of deposited silver atoms is
(b) The mass of silver deposited is 0.107 
Silver plate
(anode)
10  3600
 22.5  10 22 .
1.6  1019
22.5  10 22
 0.040 kg  40 g.
6.025  10 23
2-1
P1.2.2
Given an electric field  V/m that is constant in the x-

direction and a charge +q C located at the origin and
x
O
free to move in the x-direction (Figure 1.2.2). (a)
d
Figure P1.2.2
What is the magnitude and direction of the force F
acting on q? (b) If q moves under the influence of F a distance d m, how much
work is done by F? (c) Assuming the voltage at the origin to be zero, what is the
voltage Vd at x = d m, bearing in mind that  = -dv/dx? (d) How is the loss in
electric potential energy related to the work done by F? (e) Assuming the charge
has a mass m kg and zero velocity at the origin, show that the KE of the charge
at x = d is equal to the loss in electric potential energy.
Solution: (a) From basic electrostatics, F = q, in the direction of the electric field, where in
SI units, q is in coulombs,  is in volts/meter, and F is in newtons.
(b) Since F is independent of x, work done by F is W = Fd.
d
(c) Vd    dx  d V, the minus sign indicating that v decreases in the x0
direction.
(d) The loss in electric PE is qVd = qd, which is equal to Fd.
(e) The acceleration is a = F/m = q/m m/s2, which is constant. Hence, velocity is
t
t


u  adt  at. The distance travelled during the interval t is d  udt 
0
t

atdt 
0
0
1 2 1 u2 1 u2
. This gives: u 2  2ad . The KE is
at  a 2 
2
2 a
2 a
1
q
mu2  mad  m
d  qd , which is the loss in electric PE.
2
m
P1.2.3
Consider the system of Figure P1.2.3,
+q
where the voltage VAB V between the two
metal plates is maintained constant. Let
+
there be a mechanism for moving positive
–
charge from the lower plate to the upper
+
VAB
–
A
B
+q
plate. (a) How much work is done in
moving an amount of charge +q C? What
Figure P1.2.3
is the graph of W vs. q?
Solution: (a) In moving a charge q through a voltage rise VAB, the increase in electric PE is
qVAB, which is equal to the work done in moving the charge.
(b) Since W = qVAB, with VAB constant, the graph of W vs. q is straight line of
slope VAB passing through the origin.
3-1
P1.2.4
Suppose that the voltage supply in Figure P1.2.3 is removed (Figure P1.2.4), but
that positive charge is still moved from the lower
plate to the upper plate. Assume that when the
+q
charge on the upper plate is +q C, with an equal and
opposite charge -q C on the lower plate, the voltage
between the plates is v = Kq V, where K is a
+
VAB
–
A
B
constant. It is required to determine the work done in
+q
moving the amount of charge qf that establishes the
Figure P1.2.4
voltage VAB = Kqf. To do this, consider an
intermediate state when the charge is q C and the voltage is v V. Let an
infinitesimal charge dq be moved, where dq is small enough to keep v constant
while dq is moved. (a) What is the work dW that is done in moving the charge dq?
(b) Substitute Kq for v and determine W by integrating between 0 and qf. (c)
Express W as a function of VAB. (d) What is the graph of W vs. q in this case?
P1.2.4
(a) The work done in moving dq through a voltage rise v is dW = vdq.
qf
(b) Substituting v = Kq, dW = Kqdq, and W   Kqdq 
0
1
Kqf2 J.
2
2
2
1 V 
1 VAB
dq
(c) Substituting qf = VAB/K, i 
.
 qm cos t W  K  AB  
2  K 
2 K
dt
(d) The graph of W vs. q is parabola centered at the origin.
4-1
P1.3.2
The voltage v across an
v, V
element A is a rectangular
5
waveform of 5 V amplitude
and 4 s duration (Figure
+
0
v
P1.3.2).The current i through
A in the direction of the
voltage drop v is a biphasic
pulse that
A
i
2
1
3
4
t, s
i, A
2
–
0
Figure P1.3.2
has an amplitude of +2 A, 0 <
1
2
3
t, s
4
-2
t < 2 s, and an amplitude of -2 A, 2 < t < 4 s. Determine the power delivered or
absorbed by A during each 2 s interval. What is the total charge that has passed
through A at t = 4 s?
Solution: During the interval 0 < t < 2 s, i
v, V
is in the direction of a voltage
drop v. Hence, A absorbs
5
+
0
power equal to (5 V)(2 A) =
10 W. During the interval 2 < t
v
A
i
< 4 s, i reverses sign and
becomes a current in the
direction of a voltage rise v.
Hence, A delivers power equal
2
1
3
4
t, s
i, A
2
–
0
Figure P1.3.2
1
2
3
4
t, s
-2
to (5 V)(2 A) = 10 W. The charge is the area under the i vs. t graph. This charge
is 4 C in one direction during the first 2 s and 4 C in the opposite direction during
the next 2 s. The total charge though A is therefore zero at t = 4 s.
5-1
P1.3.5
v, V
The voltage drop across a certain
device, and the current through it,
in the direction of voltage drop, are
shown in Figure P1.3.5. Determine:
6
3
(a) the charge q through the device
at the end of each 1s interval from
t , ms
0
t  0 to t  6 s; (b) the
3
6
0
3
Figure P1.3.5
6
i , mA
instantaneous power p during the
aforementioned intervals; and (c)
3
the total energy consumed by the
device.
1
Solution: (a) 0 < t < 1: q =
t = 2: q =

1
1
0
0
 idt   0dt  0 ;
2
2
 (1  t )dt 
idt 
t , ms
1
1
2
 t2 
t   = 2.5 mC;
2 1


3
t = 3: q = 2.5 + idt  2.5
2

3
+ 3dt  5.5 mC;
2

4
t = 4: q = 5.5 + idt  5.5
3

4
+ 3dt  8.5 mC;
3
5
5
5
 t2

t = 5: q = 8.5 + idt  8.5 + (t  7)dt  8.5 +   7t  = 11 mC;
4
4
 2
4


6
6
5
5
 idt   0dt  11 mC.
t = 6: q = 11 +
(b) 0  t  1: p  0 ;
1  t  2 : p  2t (t  1)  2t 2  2t mW;
2  t  3 : p  2t  3  6t mW;
3  t  4 : p  (2t  12)  3  6t  36 mW;
4  t  5 : p  ( 2t  12)(t  7)  2t 2  26t  84 mW;
5  t  6 : p  0.
6

t

(c) w (t )  pdt  0dt 
0
0
2
 (2t
1
2
3


4
 2t )dt  6tdt  (6t  36)dt +
2
3
6-1
2
    3t
 2t 3

 t 2   3t 2
(2t  26t  84)dt + 0dt = 
4
5
 3
1

5

2
6
3
2
2
 36t

4
3

5
 2t 3

 13t 2  84t  136/3 = 45.3 J.

 3
4
P1.3.7
The voltage drop v V across a certain device, and the current i A through it, in
the direction of voltage drop, are related by:
i  8  2v 2 ,
0v 2V
i  0,
v  0 and v  2 V
(a) Determine the power absorbed by the load when v  1 V and when v  2 V;
(b) at what value of v is the instantaneous power a maximum? (c) If v (t )  2e t V,
t  0 s, what is the total charge that passes through the device from t = 0 to
t  2 s?
Solution: (a) 0  v  2 ; p = vi = v(8 – 2v2) = –2v3 + 8v, and p = 0, v  2 .
At v = 1 V; p = 6 W; at v = 2 V, p = 0.
(b)
dp
2 3
dp
 0  6v 2  8  0  v 
V.
 6v 2  8 , 0  v  2 ; pmax 
dv
3
dv

2
(c) v (t )  2e t  i  8  8e 2t ; q  idt 
0
 8  8e dt 8t  4e 
2
0
 2t
 2t 2
0
 12.07 C.
7-1