This print-out should have 24 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
2.
3.
5
4
3
2
1
0
5
4
3
2
1
0
Power (W)
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
1. P ∝ d and P ∝ ρ only
2. P ∝ ℓ only
4.
3. P ∝ ℓ, P ∝ d and P ∝ ρ
4. P ∝ d only
Power (W)
6.
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
A variable resistor is connected across a
constant voltage source.
Which of the following graphs represents
the power P dissipated by the resistor as a
function of its resistance R?
7.
5
4
3
2
1
0
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
Power (W)
AP EM 1993 MC 67
002 10.0 points
5.
Power (W)
correct
Explanation:
According to Ohm’s law, we know the
power dissipated is
ρℓ
P = I2 R = I2
.
π d2
Therefore, the power is directly proportional
to P ∝ ℓ and P ∝ ρ only.
Power (W)
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
5. P ∝ ℓ and P ∝ ρ only correct
1.
1
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
Power (W)
AP EM 1993 MC 44
001 10.0 points
The power dissipated in a wire carrying a
constant electric current I may be written as
a function of I, the length ℓ of the wire, the
diameter d of the wire, and the resistivity ρ of
the material in the wire.
In this expression, the power P dissipated
is directly proportional to which of the following?
Power (W)
Version 001 – CIRCUITS – holland – (1290)
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
Explanation:
The power dissipated in the resistor has
Version 001 – CIRCUITS – holland – (1290)
2
several expressions
E2
= I2 R ,
R
where the last two are simply derived from the
first equation together with the application of
the Ohm’s law.
Since the resistor is connected to a constant
voltage source E = constant
AP B 1993 MC 20 22
004 (part 1 of 3) 10.0 points
P =EI =
E2
constant
=
,
R
R
tells us that the power is inversely
propor
1
.
tional to the resistance P ∝
R
A battery with an internal resistance is connected to two resistors in series.
16 Ω
20 Ω
Power (W)
X
4Ω
internal
resistance
Y
What is the emf E of the battery?
1. E = 10.8 V
5
4
3
2
1
0
2. E = 12.0 V correct
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
AP EM 1998 MC 42
003 10.0 points
A wire of resistance R dissipates power P
when a current I passes through it. The wire
is replaced by another wire with resistance
3 R.
The power P ′ dissipated by the new wire
when the same current passed through it is
3. E = 13.2 V
4. E = 1.2 V
5. E = 6.0 V
Explanation:
Let :
1. P ′ = 3 P . correct
R1
R2
r
I
= 16 Ω ,
= 20 Ω ,
= 4 Ω , and
= 0.3 A .
R1
2. P ′ = 6 P .
P
.
3
P
4. P ′ = .
9
3. P ′ =
I
X
R2
E
r
internal
resistance
The total resistance of the circuit is
5. P ′ = P .
Explanation:
The power dissipated by a resistor is given
by
P = I2 R .
So if we change R to 3 R, the power dissipated
will be changed correspondingly to
′
E
0.3 A
P =
2
′
2
P = I (R ) = I (3 R) = 3 P .
Rtotal = r + R1 + R2
= 4 Ω + 16 Ω + 20 Ω
= 40 Ω ,
so the emf of the battery is
E = I Rtotal = (0.3 A) (40 Ω)
= 12 V .
Y
Version 001 – CIRCUITS – holland – (1290)
005 (part 2 of 3) 10.0 points
What is the potential difference across the
terminals Y and X of the battery?
1. VY X = 10.8 V correct
3
AP EM 1993 MC 57 58
007 (part 1 of 2) 10.0 points
The switch has been open for a long period
of time.
R2
2. VY X = 12.0 V
C
3. VY X = 1.2 V
4. VY X = 6.0 V
R1
V
S
5. VY X = 13.2 V
Explanation:
The potential difference across the terminals of the battery is
VY X = E − I r
= 12 V − (0.3 A) (4 Ω)
= 10.8 V ,
or
VY X = VXY = I (R1 + R2 )
= (0.3 A)(16 Ω + 20 Ω)
= 10.8 V .
006 (part 3 of 3) 10.0 points
What power Pinternal is dissipated by the 4 Ω
internal resistance of the battery?
1. Pinternal = 1.2 W
2. Pinternal = 4.8 W
3. Pinternal = 0.36 W correct
4. Pinternal = 3.2 W
5. Pinternal = 3.6 W
Explanation:
The power dissipated by the r = 4 Ω internal resistance is
Pinternal = I 2 r = (0.3 A)2 (4 Ω)
= 0.36 W .
Immediately after the switch is closed, the
current supplied by the battery is
V
.
R2
V (R1 + R2 )
2. I0 =
.
R1 R2
V
. correct
3. I0 =
R1
1. I0 =
4. I0 = 0 .
5. I0 =
V
.
R1 + R2
Explanation:
Before the switch is closed, there is no
charge on the capacitor, so the voltage is zero
across the capacitor at this time. Because it
is not possible to change the charge on the
capacitor like a step function (or the current
should be infinitely large), immediately after
the switch is closed, the voltage across the capacitor (and R2 ) is still zero. Therefore, the
voltage across R1 is V ; i.e., think of the capacitor as being a short-circuit for this instant
of time.
So the current supplied by the battery,
which is the same as the current going through
V
R1 , is I0 =
.
R1
008 (part 2 of 2) 10.0 points
A long time after the switch has been closed,
the current I∞ supplied by the battery is
Version 001 – CIRCUITS – holland – (1290)
V
.
R2
4. |EC | = 13 V
2. I∞ = 0 .
5. |EC | = 24 V
V
. correct
R1 + R2
V
.
4. I∞ =
R1
V (R1 + R2 )
.
5. I∞ =
R1 R2
Explanation:
After a long time, the capacitor has been
charged and remained stable. That means
the current going through R1 is the same as
the current going through R2 ; i.e., think of
the capacitor as being a open-circuit for this
time.
So we can write down the equation
3. I∞ =
6. |EC | = 28 V
7. |EC | = 5 V
8. |EC | = 4 V
9. |EC | = 18 V
10. |EC | = 9 V correct
Explanation:
t
a
It
R3
V = I∞ R 1 + I∞ R 2 ,
which gives the current I∞ as
I∞ =
E
b
7Ω
16 µF
S
V
.
R1 + R2
The circuit has been connected as shown in
the figure for a “long” time.
24 V
R4
Ib
AP EM 1993 MC 71
009 (part 1 of 2) 10.0 points
1Ω
b
It
Ib
Let :
5Ω
R2
R1
C
1. I∞ =
4
= 5 Ω,
= 7 Ω,
= 1 Ω,
= 23 Ω , and
= 16 µF .
After a “long time” implies that the capacitor C is fully charged and therefore the
capacitor acts as an open circuit with no current flowing to it. The equivalent circuit is
23 Ω
It
It
a
R1
R3
S
What is the magnitude of the electric potential EC across the capacitor?
R1
R2
R3
R4
C
Ib
R2
b
R4
Ib
1. |EC | = 33 V
Rt = R1 + R2 = 5 Ω + 7 Ω = 12 Ω
2. |EC | = 17 V
Rb = R3 + R4 = 1 Ω + 23 Ω = 24 Ω
E
24 V
It =
=
=2A
Rt
12 Ω
3. |EC | = 19 V
Version 001 – CIRCUITS – holland – (1290)
E
24 V
=1A
=
Rb
24 Ω
Ieq
C
Across R1
Req
Ib =
5
E1 = It R1 = (2 A) (5 Ω) = 10 V .
Across R3
where
E3 = Ib R3 = (1 A) (1 Ω) = 1 V .
Since E1 and E3 are “measured” from the same
point “a”, the potential difference across C
must be
EC = E3 − E1 = 1 V − 10 V = −9 V
Rℓ = R1 + R3 = 5 Ω + 1 Ω = 6 Ω,
Rr = R2 + R4 = 7 Ω + 23 Ω = 30 Ω
and
|EC | = 9 V .
010 (part 2 of 2) 10.0 points
If the battery is disconnected, how long does it
1
Et
=
take for the capacitor to discharge to
E0
e
of its initial voltage?
−1
Req =
1
1
+
Rℓ Rr
=
1
1
+
6 Ω 30 Ω
−1
= 5 Ω.
Therefore the time constant τ is
τ ≡ Req C = (5 Ω) (16 µF) = 80 µs .
1. tEt /E0 = 728 µs
The equation for discharge of the capacitor is
2. tEt /E0 = 510 µs
Qt
= e−t/τ , or
Q0
1
Et
= e−t/τ = .
E0
e
3. tEt /E0 = 476 µs
4. tEt /E0 = 720 µs
Taking the logarithm of both sides, we have
t
1
− = ln
τ
e
t = −τ (− ln e)
5. tEt /E0 = 231 µs
6. tEt /E0 = 660 µs
7. tEt /E0 = 546 µs
= −(80 µs) (−1) = 80 µs .
8. tEt /E0 = 80 µs correct
9. tEt /E0 = 120 µs
AP EM 1998 MC 15
011 (part 1 of 2) 10.0 points
10. tEt /E0 = 414 µs
Explanation:
With the battery removed, the circuit is
Ir
Iℓ
R1
R3
Iℓ
R2
r
X
Y
9Ω
C
ℓ
The following diagram shows part of a
closed electrical circuit.
55 Ω
17 Ω
R4
E
Ir
I
Version 001 – CIRCUITS – holland – (1290)
Find the electric resistance RXY of the part
of the circuit shown between point X and Y .
Since R12 and R3 are connected parallel,
their equivalent resistance RXY is
1
1
R3 + R12
1
+
=
=
RXY
R12 R3
R12 R3
(72 Ω)(9 Ω)
R12 R3
=
= 8Ω .
RXY =
R12 + R3
72 Ω + 9 Ω
1. RXY = 2 Ω
2. RXY = 10 Ω
3. RXY = 9 Ω
012 (part 2 of 2) 10.0 points
When there is a steady current in the circuit,
the amount of charge passing a point per unit
time is
4. RXY = 3 Ω
5. RXY = 8 Ω correct
6. RXY = 5 Ω
1. greater at point X than at point Y .
7. RXY = 42 Ω
2. greater in the 9 Ω resistor than in the 17 Ω
resistor. correct
8. RXY = 36 Ω
3. the same everywhere in the circuit.
9. RXY = 28 Ω
4. greater in the 55 Ω resistor than in the
17 Ω resistor.
10. RXY = 35 Ω
Explanation:
R1
R2
X
Y
R3
E
Let :
6
I
R1 = 55 Ω ,
R2 = 17 Ω ,
R3 = 9 Ω .
and
Since R1 and R2 are in series, their equivalent resistance R12 is
5. greater in the 17 Ω resistor than in the
9 Ω resistor.
Explanation:
The amount of charge passing a point per
unit of time (the same as the current at a
point) is not the same everywhere, but it is
the same at point X as at point Y ; i.e., it is
the same in the 55 Ω resistor as in the 17 Ω
resistor. It is greater in the 9 Ω resistor than
in the 55 Ω or 17 Ω resistor.
From Ohm’s Law E = I R, we have
I12 =
I3 =
R12 = R1 + R2 = 55 Ω + 17 Ω = 72 Ω .
E
E
=
R12
72 Ω
E
E
=
R3
9Ω
I3 > I12 .
R12
X
Y
72 Ω
R3
E
9Ω
I
When there is a steady current in the circuit, the amount of charge passing a point per
unit time is greater in the smaller 9 Ω resistor
than in the larger 17 Ω resistor.
AP EM 1998 MC 36
013 10.0 points
Version 001 – CIRCUITS – holland – (1290)
A resistor R and a capacitor C are connected
in series to a battery of terminal voltage V0 .
Which of the following equations relating
the current I in the circuit and the charge Q
on the capacitor describes this circuit?
1.
Q
−IR=0
C
3. V0 − C
6Ω
5.
4Ω
8Ω
correct
20 Ω
6.
2. V0 + Q C − I 2 R = 0
7
20 Ω
Explanation:
dQ
− I2 R = 0
dt
Q
4. V0 − − I R = 0 correct
C
1 Q2
5. V02 −
− I2 R = 0
2 C
Explanation:
Kirchhoff’s law tells us the voltage drop
across any closed circuit loop is zero. We
know the voltage drop across the resistor is
Q
I R and that across the capacitor is , so the
C
equation describing the circuit is
Q
V0 − − I R = 0
C
Let :
P = 200 W
E = 60 V .
Power is
P =
and
E2
,
Req
so the combination of resistors dissipating
200 W when connected to a 60 V power supply will have the equivalent resistance of
Req =
Consider
(60 V)2
E2
=
= 18 Ω .
P
200 W
R11
R12
AP EM 1998 MC 37
014 10.0 points
Req = R11 + R12 = 6 Ω + 10 Ω = 16 Ω .
Which of the following combinations of resistors would dissipate 200 W when connected
to a 60 V power supply?
Consider
R22
12 Ω
1.
2.
R21
6Ω
12 Ω
6Ω
1
1
1
=
+
Req
R21 R22
10 Ω
so
18 Ω
Req =
18 Ω
3.
Consider
R31
18 Ω
4Ω
4.
1
1
1
+
20 Ω 20 Ω
R32
= 10 Ω .
R33
12 Ω
16 Ω
Req = R31 + R32 + R33
= 6 Ω + 4 Ω + 8 Ω = 18 Ω .
Version 001 – CIRCUITS – holland – (1290)
Consider
R43
4. Ampère’s law
R41
5. Faraday’s law
Explanation:
Kirchhoff’s loop rule
R42
1
1
1
1
=
+
+
Req
R41 R42 R43
so
Req =
Consider
1
= 6 Ω.
1
1
1
+
+
6Ω 4Ω 8Ω
R52
R53
R51
1
1
1
=
+
Req
R51 R52 + R53
Req =
1
1
1
+
16 Ω 4 Ω + 12 Ω
Consider
= 8 Ω.
R63
R61
R62
Req = R61 +
8
1
1
1
+
R62 R63
1
= 12 Ω .
=6Ω+
1
1
+
12 Ω 12 Ω
AP B 1993 MC 14
015 10.0 points
Kirchhoff’s loop rule for circuit analysis is an
expression of which of the following?
1. Ohm’s law
2. Conservation of energy correct
3. Conservation of charge
X
V = V1 + V2 + V3 + · · · = 0
follows from the conservation of energy.
AP B 1993 MC 15 16
016 (part 1 of 2) 10.0 points
Consider the circuit
2 µF
3 µF
c
a
b
5 µF
4 µF
100 V
What is the equivalent capacitance for this
network?
1. Cequivalent =
10
µF
7
2. Cequivalent = 7 µF correct
3. Cequivalent = 14 µF
7
µF
3
3
= µF
2
4. Cequivalent =
5. Cequivalent
Explanation:
Let : C1
C2
C3
C4
EB
= 2 µF ,
= 4 µF ,
= 3 µF ,
= 5 µF , and
= 100 V .
Version 001 – CIRCUITS – holland – (1290)
018
C1
C3
c
a
b
C2
C4
9
10.0 points
Consider resistors R1 and R2 connected in
series
R1
R2
E
EB
C1 and C2 are connected in parallel, so
and in parallel
R1
C12 = C1 + C2 = 6 µF .
C12 and C3 are connected in series, so
1
1
1
C3 + C12
=
+
=
C123
C12 C3
C12 C3
(6 µF) (3 µF)
C12 C3
=
= 2 µF .
C123 =
C3 + C12
6 µF + 3 µF
C123 and C4 are connected in parallel, so
C = C4 + C123 = 7 µF .
017 (part 2 of 2) 10.0 points
What is the charge stored in the 5-µF lowerright capacitor?
1. Q1 = 1, 100 µC
2. Q1 = 360 µC
3. Q1 = 1, 800 µC
4. Q1 = 500 µC correct
E
to a source of emf E that has no internal
resistance.
How does the power dissipated by the resistors in these two cases compare?
1. It is greater for the series connection.
2. It is different for each connection, but one
must know the values of E to know which is
greater.
3. It is different for each connection, but one
must know the values of R1 and R2 to know
which is greater.
4. It is the same for both connections
5. It is greater for the parallel connection.
correct
5. Q1 = 710 µC
Explanation:
Let :
R2
C4 = 5 µF and
EB = 100 V .
The charge stored in a capacitor is given by
Q = C V , so
Q4 = C4 V = (5 µF) (100 V) = 500 µC .
AP B 1993 MC 50
Explanation:
The power dissipated by the resistors is
E2
P =
.
Req
The equivalent resistance for a series connection is
Rs = R1 + R2 .
The equivalent resistance for a parallel connection is
R1 R2
.
Rp =
R1 + R2
Version 001 – CIRCUITS – holland – (1290)
Regardless of the values of R1 and R2 , Rp <
Rs , so more power is dissipated in the parallel
connection.
AP EM 1993 MC 39 40
019 (part 1 of 2) 10.0 points
Consider the system of equivalent capacitors.
2 µF
2 µF
2 µF
2 µF
a
b
2 µF
2 µF
EB
10
C1
C2
C3
C4
a
b
C5
C6
Eab
We have 3 series combinations, each with
1
1
1
2
equivalent capacitance ′ = + = , of
C
C C
C
the capacitors connected in parallel, so
C
2 µF
C C C
+ + =3
=3
2
2
2
2
2
= 3 µF .
Cab =
Find the equivalent capacitance Cab of the
network of capacitors.
1. C = 12 µF
2
µF
3
4
3. C = µF
3
2. C =
4. C = 4 µF
5. C =
1
µF
3
020 (part 2 of 2) 10.0 points
What potential difference must be applied
between points a and b so that the charge
on each plate of each capacitor will have a
magnitude of 6 µC?
1. Vab = 4 V
2. Vab =
3
V
2
6. C = 6 µF
3. Vab = 2 V
7. C = 3 µF correct
4. Vab = 3 V
8. C = 2 µF
5. Vab =
2
V
3
6. Vab = 6 V correct
Explanation:
7. Vab = 9 V
8. Vab = 18 V
Explanation:
Let :
Let :
C1 = C2 = C3 = C4 = C5 = C6
= C = 2 µF .
Q = 6 µC and
EB = 6 V .
Regard the system as an equivalent capacitor with the capacitance Cab . The charge on
Version 001 – CIRCUITS – holland – (1290)
a series set of capacitors is the same: Q12 =
Q34 = Q56 = 6 µC , and the charge on a parallel set of capacitors is the sum of the charges
on each branch: Qtotal = 3 (6 µC) = 18 µC ,
so
Vab =
C1
a
Qtotal
18 µC
= 6 V.
=
Cab
3 µF
Find
3 µF
the equivalent capacitance of the circuit.
4
µF
3
2. C = 2 µF correct
3. C = 1 µF
4. C = 4 µF
2
µF
3
3
6. C = µF
2
5. C =
7. C = 3 µF
Explanation:
C6
c
1
1
2
=
= C
1
1
1
3
3
+
+
2C 2C 2C
2C
2
= (3 µF) = 2 µF .
3
Cad =
EB
022 (part 2 of 2) 10.0 points
What potential difference must be applied
across the capacitor network so that the
charge on each plate of each capacitor will
have a magnitude of 6 µC?
1. Vad = 6 V correct
2. Vad = 2 V
3. Vad =
4
V
3
4. Vad = 3 V
5. Vad = 8 V
2
V
3
3
= V
2
6. Vad =
7. Vad
8. Vad = 4 V
Explanation:
Let :
c
We have 3 parallel combinations, each with
equivalent capacitance C ′ = C + C = 2 C , of
the capacitors connected in series, so
3 µF
1. C =
C4
EB
3 µF
3 µF
b
C5
AP EM 1993 MC 40 41
021 (part 1 of 2) 10.0 points
Consider the following system of equivalent
capacitors.
3 µF
C3
C2
d
3 µF
11
C1 = C2 = C3 = C4 = C5 = C6
= C = 3 µF .
Let :
Q = 6 µC and
EB = 6 V .
Version 001 – CIRCUITS – holland – (1290)
Regard the system as an equivalent capacitor with the capacitance Cad = 2 µF. The
charge on each parallel set of capacitors is
the sum of the charges on each of the capacitors: Q12 = Q34 = Q56 = 12 µC , and
the charge on a series set of capacitors is the
same: Qtotal = 12 µC , so
Vad =
X
9Ω
R1
Y
R3
E
Let :
The following diagram shows part of a
closed electrical circuit.
22 Ω
X
R2
Qtotal
12 µC
=
= 6 V.
Cad
2 µF
AP EM 1998 MC 16
023 (part 1 of 2) 10.0 points
12
I
R1 = 9 Ω ,
R2 = 22 Ω ,
R3 = 99 Ω .
and
Since R2 and R3 are connected parallel,
their equivalent resistance R23 is
1
1
R3 + R2
1
=
+
=
R23
R2 R3
R2 R3
R2 R3
(22 Ω) (99 Ω)
R23 =
=
= 18 Ω .
R2 + R3
22 Ω + 99 Ω
Y
99 Ω
E
X
I
R1
R23
9Ω
18 Ω
E
Find the electric resistance RXY of the part
of the circuit shown between point X and Y .
1. RXY = 15 Ω .
Y
I
Since R1 and R23 are in series, their equivalent resistance RXY is
RXY = R1 + R23 = 9 Ω + 18 Ω = 27 Ω .
2. RXY = 33 Ω .
RXY
X
3. RXY = 17 Ω .
4. RXY = 3 Ω .
Y
27 Ω
E
I
5. RXY = 29 Ω .
6. RXY = 43 Ω .
7. RXY = 7 Ω .
8. RXY = 9 Ω .
9. RXY = 39 Ω .
10. RXY = 27 Ω . correct
Explanation:
024 (part 2 of 2) 10.0 points
When there is a steady current in the circuit,
the amount of charge passing a point per unit
time is
1. greater at point X than at point Y .
2. greater in the 22 Ω resistor than in the
9 Ω resistor.
3. greater in the 99 Ω resistor than in the
9 Ω resistor.
Version 001 – CIRCUITS – holland – (1290)
4. greater in the 22 Ω resistor than in the
99 Ω resistor. correct
5. the same everywhere in the circuit.
Explanation:
The amount of charge passing a point per
unit of time (the same as the current at a
point) is not the same everywhere, but it
is the same at point X as at point Y , and
it is the same in the 9 Ω resistor as in the
combination of the parallel R2 + R3 = 18 Ω
resistors. Therefore the current is less in the
22 Ω resistor or 99 Ω resistor than in the 9 Ω
resistor, and it is greater in the 22 Ω resistor
than in the 9 Ω or 99 Ω resistor.
From Ohm’s Law E = I R, we have
I2 =
E
E
=
R2
22 Ω
E
E
=
R3
99 Ω
I2 > I3 .
I3 =
When there is a steady current in the circuit, the amount of charge passing a point
per unit time is greater in the smaller 22 Ω
resistor than in the larger 99 Ω resistor.
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