Download Electricity Review Sheet Solutions

Electricity Review Sheet
1
Consider the following statements given below and
determine if the charge of Object A is:
A. positive
B. negative
C. neutral
I. Object A is charged by friction using animal fur. Animal
fur has a weaker electron affinity than object A.
II. Object A is charged by contact using a negatively
charged object.
III. Object A is charged by induction using a
positively charged object.
IV. Object A is attracted to a negatively charged
object.
2
A glass rod is given a positive charge by rubbing it
with silk. The rod has become positive by
(A) losing electrons
(C) losing protons
(B) gaining electrons
(D) gaining protons
3
A positively charged rod is held near the knob of a
neutral electroscope. Which diagram best represents the
distribution of charge on the electroscope?
4
9
The diagram below
represents the electric field
surrounding two charged
spheres, A and B. What is
the sign of the charge of
each sphere?
(A) Sphere A is positive and sphere B is negative.
(B) Sphere A is negative and sphere B is positive.
(C) Both spheres are positive.
(D) Both spheres are negative.
5
What is the magnitude of the electrostatic force
between two electrons separated by a distance of 1.00
× 10–8 meter?
(A) 2.56 × 10–22 N
(C) 2.30 × 10–12 N
–20
(B) 2.30 × 10 N
(D) 1.44 × 10–1 N
6
A negatively charged comb has –8.0 C of charge on
it. How many excess electrons are on it?
(A) 5 × 1013
(C) 2 × 10-14
19
(B) 5 × 10
(D) 5 × 10-25
7
The electrical resistance of a metallic conductor is
inversely proportional to its
(A) temperature
(C) length
(B) cross-sectional area (D) resistivity
8
What is the current in a 100-ohm resistor connected
to a 0.40-volt source of potential difference?
(A) 250 mA
(C) 2.5 mA
(B) 40 mA
(D) 4.0 mA
The potential difference between two points in an
electric field is 10 volts. The energy to move a
positive charge of 2 × 10-4 C from a point at the
lower potential to a point at the higher potential is:
(A) 5 × 104 J
(C) 2 × 10-3 J
(B) 2 × 10-2 J
(D) 2 × 10-5 J
10 At a distance 10 cm from a point charge, the electric
field is 3600 N/C and points toward the charge. The
amount of this point charge is
(A) 40 microcoulombs
(B) 4 microcoulombs
(C) 4 nanocoulombs
(D) 4 megacoulombs
11 The diagram to the right
shows a point, P, located
between two oppositely
charged parallel plates.
The electric field between the plates
(A) points up, and is varying (non-uniform)
(B) points down, and is varying (non-uniform)
(C) points up, and is constant (uniform)
(D) points down, and is constant (uniform)
12 The diagram to the right
represents an electric circuit
consisting of a 12-volt battery, a
3.0-ohm resistor, R1, and a
variable resistor, R2.
At what value must the variable resistor be set to
produce a current of 2.0 ampere through R1?
(A) 6.0 Ω
(C) 3.0 Ω
(B) 9.0 Ω
(D) 12 Ω
13 An electric iron operating at 120 volts draws 10
amperes of current. How much heat energy is
delivered by the iron in 30 seconds?
(A) 3.0 × 102 J
(C) 3.6 × 103 J
(B) 1.2 × 103 J
(D) 3.6 × 104 J
14 A 20-ohm resistor and a 30ohm resistor are connected in
parallel to a 12-volt battery as
shown. As shown to the right,
an ammeter is connected.
What is the equivalent resistance of the circuit?
(A) 10 Ω
(C) 25 Ω
(B) 12 Ω
(D) 50 Ω
15 In the previous question, what is the current reading
of the ammeter?
(A) 1.0 A
(C) 0.40 A
(B) 0.60 A
(D) 0.20 A
16
In the previous question, what is the power of the
30-ohm resistor?
(A) 4.8 W
(C) 30 W
(B) 12 W
(D) 75 W
17 Two resistors are connected to a source of voltage as
shown in the diagram below.
At which position should an ammeter be placed to
measure the current passing only through resistor R1
and a voltmeter be placed to measure the potential
difference across R2, respectively?
(A) 2 and 1
(C) 3 and 1
(B) 3 and 2
(D) 4 and 1
21 The schematic diagram to the
right shows and ammeter A, and
three resistors. Solve the
following with equations, then
check with tables:
a Calculate resistance R1.
b Calculate current in R2.
c Calculate current in R3.
d Calculate power in R1.
22 The wiring diagram shown below shows four light
bulbs and a battery arranged in a combination circuit.
The battery is 6 volts, and each light bulb is 12 .
18 Which combination of resistors has the smallest
equivalent resistance?
Problem Solving: Do the following problems on separate
paper. Show the general equation, substitutions,
calculations, final results with units.
19 The drawing below shows the atomic nucleus of two
atoms each with 8 protons, and a nearby electron.
2.4 x 10-14
electron
a
b
c
23 The schematic diagram shown below shows five
resistors and a battery are arranged in a combination
circuit. Solve with tables.
5.9 x 10-15
protons
Find the force between the two groups of protons.
Find the net force on the electron.
Find the electric field at the electron’s location.
20 Two resistors are wired in series with a 12-volt
battery. The first resistor has 10 volts across it, and
the second resistor has 0.5 amps through it. Solve the
following with equations, then check with tables:
a
b
c
d
e
a Draw the schematic diagram for the circuit.
b Calculate the circuit’s equivalent resistance.
Use tables to find:
c Calculate the potential difference of each resistor.
d Calculate the current through each resistor.
e Calculate the power consumed by each resistor.
Draw and label the circuit schematic diagram.
Calculate the potential difference across R2.
Calculate the resistance R1.
Calculate the combined resistance of R1 and R2.
Calculate the amount of power R1 consumes.
a
b
Calculate the equivalent resistance, total current,
and total power for this circuit.
HONORS: Calculate the current, voltage, and
power of all five resistors.
24 A +3.0 nanocoulumb charge is located 1.0 m to the
left of a –1.0 nanocoulumb charge.
a
b
c
Draw the field lines around these two charges.
Calculate the electric field at the midpoint
between the two charges.
HONORS: Where should a third charge of +2 nC
be placed so that it is in static equilibrium?
Answers
1 I-B, II-B, III-B, IV-A or C 2 A 3 D 4 B 5 C 6 A 7 B 8 D 9 C 10 C 11 D 12 C 13 D 14 B 15 B 16 A 17 C 18 D
19a 424 N b 5.26 N, right c 3.29 × 1019 N/C, left 20b 2 v c 20 Ω d 24 Ω e 5 W 21a 200 Ω b 0.5 A c 0.67 A d 72 W
22b 20 Ω c VA–VD : 3.6, 1.2, 1.2, 2.4 V d IA–ID: 0.3, 0.1, 0.1, 0.2 A e PA– PD : 1.08, 0.12, 0.12, 0.48 W
23a 12 Ω, 3 A, 108 W b I1–I5: 1.8, 1.8, 1.2, 3, 3 AV1–V5 : 10.8, 7.2, 18, 12, 6 V; P1– P5 : 19.44, 12.96, 21.6, 36, 18 W
Electricity Review Sheet Solutions
1. answers: I – B, II – B, III - B, IV – A or C
I. Friction charging causes electrons to move from
the animal fur to Object A, which becomes negative.
II. Conduction charging causes excess electrons in
the negatively charged object to be shared with
Object A, so object A becomes negatively charged.
III. Induction charging causes Object A to be
oppositely charge, so using a positive object induces
a negative charge on Object A.
IV. Object A can be attracted to a negatively charged
object if it oppositely charged positive, but also may
be attracted by polarization if it is neutral.
resistance (much like a drinking straw has less
resistance than a narrow stirring straw.)
2. answer: A
9. answer: C
Ordinarily only electrons can be moved from
one body to another (movement of protons
requires high-energy collisions in accelerators!).
So a glass rod acquires a positive charge, when
rubbed with silk, by losing electrons.
DV =
8. answer: D
Using the definition of resistance as the ratio of
voltage divided by current:
DV
DV 0.40 volt
R=
Þ I=
=
= 0.004 A
I
R
100 W
Converting amps to milliamps:
1 mA
0.004 A ´ -3 = 4.0 mA
10 A
4. answer: B
By definition, electric field lines point towards
negative charges and away from positive
charges (because a test charge is always
assumed to be positive). So sphere A must be
negative and sphere B must be positive to create
the field lines shown in the drawing.
work
2 ´ 10 -4
work = 2 ´ 10 -3 J
E=
kq
(9 ´ 10 9 )(q)
3600
=
q = 4 ´ 10 -9 C = 4 nC
2
2
r
0.1
11. answer: D
The electric field inside this parallel plate
capacitor is constant (except at the left and right
ends). The field points down because a positive
test charge is used to define the direction of an
electric field, and the positive test charge would
be repelled by the positive charges above, and
attracted by the negative charges below it.
12. answer: C
DV = I(R1 + R2 )
12 = 2(3 + R2 )
R2 = 3 W
13. answer: D
PE = P(t) = IDV(t) = 10(120)(30) = 3.6 ´ 10 4 J
14. answer: B
5. answer: C
Fe =
10 =
10. answer: C
3. answer: D
The positively charged rod near the electroscope
will attract some electrons into the knob (top) of
the electroscope, leaving the electroscope
polarized, but still neutral. The correct
representation has negative signs in the knob,
and positive signs in the leaves (bottom).
PE work
=
q
q
-19 2
kq1q2 9 ´ 10 (1.6 ´ 10 )
=
r2
(1.00 ´ 10 -8 )2
9
1
1
1
1
1
1
= +
+
=
+
Req R1 R2 R3 20 30
Fe = 2.30 ´ 10 -12 N
15. answer: B
6. answer: A
I=
number of electrons =
charge
fundamental charge
DV 12
=
= 0.60 A
R
20
16. answer: A
DV 2 12 2
=
= 4.8 W
R
30
q
-8 ´ 10 -6
n= =
= 5 ´ 1013
-19
e -1.6 ´ 10
P=
7. answer: B
17. answer: C
Electrical resistance depends directly on
temperature, length, and resistivity. It depends
inversely on cross-sectional area. That is, the
wider the diameter of resistor the less the
Req = 12 W
The ammeter must be placed in series with
resistor #1 within the branch, so at position 3.
The voltmeter must be placed in parallel with
resistor #2, so at position 1.
18. answer: D
21a.
circuit A: Req = 2 + 2 = 4 W
-1
DV1 = I1R1
-1 -1
120 = 0.60R1
circuit B: Req = (2 + 2 ) = 1 W
21b.
circuit C: Req = 1+ 1 = 2 W
DV2 = I 2 R2
-1
-1 -1
circuit D: Req = (1 + 1 ) =
1
2
W
120 = I 2 (240)
I 2 = 0.5 A
21c.
P3 = I 3DV3
smallest resistance is circuit D
80 = I 3 (120)
I 3 = 0.67 A
21d.
19a.
Fe =
R1 = 200 W
kq1q2 (9 ´10 )(8 ´1.6 ´10
=
r2
(5.9 ´10 -15 )2
9
-19 2
)
P = I1DV1 = 0.6(120) = 72 W
= 424 N
19b.
R1
R
200
I
0.6
V
120
P
72
kq1q2 (9 ´ 10 9 )(-1.6 ´ 10 -19 )(8 ´ 1.6 ´ 10 -19 )
=
r2
(2.4 ´ 10 -14 )2
Fe = -3.2 N (to the right)
R2
240
0.5
120
60
R3
180
0.67
120
80
Req
67.9
1.767
120
212
Fe =
-19
kq1q3 (9 ´ 10 )(-1.6 ´ 10 )(8 ´ 1.6 ´ 10
=
r2
(2.4 ´ 10 -14 + 5.9 ´ 10 -15 )2
Fe = -2.06 N (to the right)
9
Fe =
-19
)
22a.
SF = 3.2 + 2.06 = 5.26 N (to the right)
19c.
E=
Fe
5.26
=
= 3.29 ´ 1019 N/C, to the left
-19
q0 (1.6 ´ 10 )
I2 = 0.5 A
Req = RABC + RD = 8 +12 = 20 W
22c,d,e.
12 V
Series:
20b.
DV = DV1 + DV2
12 = 10 + DV2
DV2 = 2 v
20c.
I = I1 = I 2 = 0.5 A
DV1 = I1 R1
10 = 0.5R1
12 = 0.5Req
Req = 24 W
20e.
P2 = I 2 DV2 = 0.5(10) = 5 W
table method:
R
R1
4
RB
R
12
I
0.1
V
1.2
P
0.12
RC
12
0.1
1.2
0.12
RBC
24
0.1
2.4
0.24
RBC
R
24
I
0.1
V
2.4
P
0.24
RD
12
0.2
2.4
0.48
RBCD
8
0.3
2.4
0.72
I
0.3
V
2.4
P
0.72
Parallel:
R1 = 20 W
20d.
DV = IReq
RBC = RB + RC = 12 +12 = 24 W
RABC = (RBC -1 + RA -1 )-1 = (24 -1 +12 -1 )-1 = 8 W
20a.
ΔV1 = 10 v
22b.
Series:
I
0.5
V
2
P
1
RBCD
R
8
RA
12
0.3
3.6
1.08
Req
20
0.3
6
1.80
R2
20
0.5
10
5
Req
24
0.5
12
6
Series:
23a.
R12 = R1 + R2 = 6 + 4 = 10 W
R123 = (R12 + R3 ) = (10 + 15 ) = 6 W
R1
R
6
Req = R123 + R4 + R5 = 6 + 4 + 2 = 12 W
R2
4
1.8
7.2
12.96
R12
10
1.8
18
32.4
R12
R
10
I
1.8
V
18
P
32.4
R3
15
1.2
18
21.6
R123
6
3
18
54
I
3
V
18
P
54
-1
-1 -1
-1
-1 -1
DV 36
I=
=
= 3A
Req 12
P=
I
1.8
V
10.8
P
19.44
Parallel:
DV 2 36 2
=
= 108 W
Req
12
23b.
I = I4 = I5 = 3 A
Series:
DV5 = I 5 R5 = 3(2) = 6 v
P5 = I 5 DV5 = 3(6) = 18 W
R123
R
6
DV4 = I 4 R4 = 3(4) = 12 v
R4
4
3
12
36
R5
2
3
6
18
Req
12
3
36
108
P4 = I 4 DV4 = 3(12) = 36 W
DV3 = DV123 = DV - DV45 = 36 - 18 = 18 v
DV3 = I 3 R3
18 = I 3 (15)
I 3 = 1.2 A
24a.
P3 = I 3 DV3 = 1.2(18) = 21.6 W
I1 = I 2 = I - I 3 = 3 - 1.2 = 1.8 A
DV2 = I 2 R2 = 1.8(4) = 7.2 v
P2 = I 2 DV2 = 1.8(7.2) = 12.96 W
DV1 = I1 R1 = 1.8(6) = 10.8 v
P1 = I1DV1 = 1.8(10.8) = 19.44 W
24b.
q1 = +3 nC ´
10 -9 C
= +3 ´ 10 -9 C
1 nC
Then complete the third table for the final series circuit.
Only the known values are shown in bold. All others are
found by multiplication or division.
q2 = –1 nC ´
10 -9 C
= –1 ´ 10 -9 C
1 nC
Then complete the second table. Notice the bottom row is
simply taken from the table below it.
r12
Using the table method, work through each resistor column
from the first table, to the second, then to the third.
Finally, complete the first table. Again, notice the bottom
row is simply taken from the table below it
E1 =
kq1
E2 =
kq2
r12
=
=
9.0 ´ 10 9 (3 ´ 10 -9 )
= 108 N/C, right
( 0.5 m )2
9.0 ´ 10 9 (1´ 10 -9 )
( 0.5 m )2
= 36 N/C, right
ETOTAL = E1 + E2 = 108 N/C + 36 N/C = 144 N C, right
24c.
k(3 nC)(2 nC) k(1 nC)(2 nC)
=
Þ r = 1.37 m
(1+ r)2
(r)2