PHYSICS TOPICAL:
Electric Circuits
Test 1
Time: 22 Minutes*
Number of Questions: 17
* The timing restrictions for the science topical tests are optional. If you
are using this test for the sole purpose of content reinforcement, you
may want to disregard the time limit.
MCAT
DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage
preceding each group of questions. Study the passage,
then select the single best answer to each question in the
group. Some of the questions are not based on a
descriptive passage; you must also select the best answer
to these questions. If you are unsure of the best answer,
eliminate the choices that you know are incorrect, then
select an answer from the choices that remain. Indicate
your selection by blackening the corresponding circle on
your answer sheet. A periodic table is provided below for
your use with the questions.
PERIODIC TABLE OF THE ELEMENTS
1
H
1.0
2
He
4.0
3
Li
6.9
4
Be
9.0
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
F
19.0
10
Ne
20.2
11
Na
23.0
12
Mg
24.3
13
Al
27.0
14
Si
28.1
15
P
31.0
16
S
32.1
17
Cl
35.5
18
Ar
39.9
19
K
39.1
20
Ca
40.1
21
Sc
45.0
22
Ti
47.9
23
V
50.9
24
Cr
52.0
25
Mn
54.9
26
Fe
55.8
27
Co
58.9
28
Ni
58.7
29
Cu
63.5
30
Zn
65.4
31
Ga
69.7
32
Ge
72.6
33
As
74.9
34
Se
79.0
35
Br
79.9
36
Kr
83.8
37
Rb
85.5
38
Sr
87.6
39
Y
88.9
40
Zr
91.2
41
Nb
92.9
42
Mo
95.9
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3
55
Cs
132.9
56
Ba
137.3
57
La *
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222)
87
Fr
(223)
88
Ra
226.0
89
Ac †
227.0
104
Rf
(261)
105
Ha
(262)
106
Unh
(263)
107
Uns
(262)
108
Uno
(265)
109
Une
(267)
*
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
Lu
175.0
†
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(260)
GO ON TO THE NEXT PAGE.
2
as developed by
Electric Circuits Test 1
Passage I (Questions 1–6)
Figure 1 shows a schematic diagram for a device
known as a potentiometer. The potentiometer is useful for
measuring an unknown electromotive force (emf) ε x in
terms of a known emf ε 0 . In addition to these two, a third
emf ε is used to produce the current i. The three emfs have
internal resistances rx, r0, and r, respectively.
The central circuit element of the potentiometer is a
long resistor, having total resistance R , with a sliding
contact at Point D. The resistivity ρ of this variable
resistor is constant. Therefore, the resistance between
Points A and D is given by α R, where α is the distance
from A to D divided by the distance from A to E. A very
sensitive ammeter called a galvanometer is connected
between Points C and D in order to measure the current
through that section of the circuit.
The operation of the potentiometer is conducted in
two steps. First, the unknown emf ε x is connected in the
circuit between Points B and C, and the sliding contact at
Point D is adjusted so that the galvanometer reads
ix = 0. Using the loop theorem for the current loop ABCD,
it can be shown that the current i satisfies the equation i =
ε x /(α xR). In the second step, the unknown emf ε x is
replaced by the known emf ε 0 , and the sliding contact at
Point D is again adjusted so that the galvanometer reads
zero. Applying the loop theorem again leads to the result
that the current i is also given by i = ε 0/(α 0R). Equating
the two expressions for i gives for the unknown emf ε x =
ε 0 α x /α 0 .
C
B
εx or ε0 rx or r0
ix or i0
Galvanometer
αR
(1 – α )R
A
E
D
i
F
G
ε
r
Figure 1
1 . The internal resistances rx and r0 can be neglected when
calculating the current i because:
A . they are small relative to R.
B . they are nearly equal and will cancel each other.
C . they have no effect when the galvanometer reads
zero.
D . they cancel against the internal resistance of the
third emf ε.
2 . The current i is assumed to be the same when ε 0 is
connected as it is when ε x is connected. If the
galvanometer reads zero in both cases, which of the
following is also a correct expression for i?
A.
B.
C.
D.
ε/(r + α xR)
ε /R
ε/(r + R)
ε/(r + α 0R)
3 . In order for the galvanometer to be a good ammeter:
A . its internal resistance should be high
dissipate a lot of energy.
B . its internal resistance should be high
dissipate very little energy.
C . its internal resistance should be low
dissipate a lot of energy.
D . its internal resistance should be low
dissipate very little energy.
and it should
and it should
and it should
and it should
4 . Consider the example where ε 0 = 6 V. If the length
AD is equal to 0.2 m when ε x is connected and equal
to 0.4 m when ε 0 is connected, what is the value of
εx?
A.
B.
C.
D.
2V
3V
6V
12 V
5 . If the distance from Point A to Point E is doubled
while keeping the distance from Point A to Point D
constant, then the resistance αR would:
A.
B.
C.
D.
be reduced by a factor of 1/4.
be reduced by a factor of 1/2.
remain the same.
also double.
GO ON TO THE NEXT PAGE.
KAPLAN
3
MCAT
6 . In addition to the loop theorem, the junction theorem
is also needed in order to calculate the current i. The
junction theorem states that the algebraic sum of the
currents at any junction must be zero. On what
physical principle is the junction theorem based?
A.
B.
C.
D.
Conservation of electric charge
Conservation of the momentum of the electron
Conservation of energy
Newton’s third law
GO ON TO THE NEXT PAGE.
4
as developed by
Electric Circuits Test 1
Passage II (Questions 7–12)
A
RC circuits are used in many applications where
time-dependent currents are involved. The circuit shown in
Figure 1 consists of a resistor, having resistance R,
connected in series with a capacitor, having capacitance C,
and an emf ε . A short circuit is included that makes it
possible to exclude the voltage source ε from the circuit by
throwing the switch S to Position B. In this position, the
capacitor completely discharges through the resistor.
When the switch is thrown to Position A, the emf
produces a current i that passes through the resistor and
deposits positive charge on one plate of the capacitor. An
equal amount of positive charge leaves the other plate and
flows back to the voltage source. It can be shown that the
charge q on the positive plate of the capacitor as a function
of time satisfies the equation q = C ε(1 – e–t/RC), where t =
0 corresponds to the instant that the switch is thrown to
Position A. Once the capacitor is sufficiently charged, the
switch can be thrown to Position B and it will discharge
through the resistor. As the capacitor discharges, q satisfies
the equation q = Cεe–t/RC, where t = 0 now corresponds to
the instant that the switch is thrown to Position B.
A
B
A
B
A
VC
(a)
VR
(b)
VR + VC
R
(c)
S
B
ε
C
Figure 1
A simple model of a time-dependent current can be
produced by switching S between Positions A and B at
regular time intervals. Figure 2a shows a cathode-ray
oscilloscope display of the voltage V C across the capacitor
as a function of time; Figure 2b shows the voltage V R
across the resistor as a function of time; and Figure 2c
shows the sum V C + V R as a function of time. On the
displays shown in Figure 2, the time t = 0 corresponds to
the instant that the switch is thrown to position A. Each
horizontal division on the oscilloscope screen equals 0.1 s,
and each vertical division equals 1.0 V. (Note: The internal
resistance of the emf ε is considered to be negligible.)
7 . When the switch S is in Position A, at what value of
time t does the charge q on the capacitor reach its
maximum value?
A.
B.
C.
D.
t=0
t = 1/RC
t = RC
t=∞
8 . How would the graph of Vc versus time change if
another resistor were added in series to the original
resistor?
A.
B.
C.
D.
The upper limit of the voltage would decrease.
The upper limit of the voltage would increase.
It would take less time for the voltage to plateau.
It would take longer for the voltage to plateau.
GO ON TO THE NEXT PAGE.
KAPLAN
5
MCAT
9 . When the switch is in Position B, which of the
following is the correct expression for the voltage VC
across the capacitor as a function of time t?
A.
B.
C.
D.
ε(1 – e–t/RC)
εe–t/RC
εet/RC
ε(1 – et/RC)
1 0 . Another capacitor, having capacitance C, is added in
series with the original capacitor and resistor such that
the two capacitors are adjacent. If the switch is thrown
to Position A and remains there indefinitely, then as t
→ ∞, the charge on each capacitor approaches:
A.
B.
C.
D.
C ε /2.
2C ε .
2 ε /C.
ε/(2C).
1 1 . What is the magnitude of the emf ε?
A.
B.
C.
D.
0.5 V
10V
l.5 V
2.0 V
1 2 . Why does the capacitor discharge through the resistor
when the switch S is in Position B?
A . Because the charge on the positive plate of the
capacitor is at a higher potential than the charge
on the negative plate
B . Because the charge on the negative plate of the
capacitor is at a higher potential than the charge
on the positive plate
C . Because the emf draws a current that has to pass
through the resistor
D . Because the voltage VC across the capacitor equals
zero
GO ON TO THE NEXT PAGE.
6
as developed by
Electric Circuits Test 1
Questions 13 through 17 are NOT
based on a descriptive passage.
1 3 . If the distance between the plates of an isolated,
charged parallel-plate capacitor is increased, which of
the following will also occur?
I. The capacitance will decrease.
II. The charge on the plates will decrease.
III. The potential difference between the plates
will increase.
A.
B.
C.
D.
1 6 . An AC voltage source generates a sinusoidal current
with amplitude equal to 2 A. If all the current passes
through a l kΩ resistor, what is the average power
dissipated by the resistor?
A . 2 kW
B . 2 kW
C . 2 2 kW
D . 4 kW
1 7 . Both the length and the radius of a cylindrical wire are
reduced to half of their original value. How does the
resistance of the wire change?
I only
II only
I and III only
II and III only
1 4 . An emf is connected in parallel to a 150 Ω resistor and
a light bulb with a resistance of 750 Ω as shown in
the circuit below. If the emf produces a total current of
0.2 A, what is the total power dissipated by the
resistor and the light bulb?
A.
B.
C.
D.
It is reduced to half of its original value.
It remains the same.
It increases by a factor of 2.
It increases by a factor of 4.
i = 0.2 A
ε
A.
B.
C.
D.
150Ω
750Ω
5.0 W
6.0 W
30 W
36 W
1 5 . A parallel-plate capacitor has a capacitance of 1 nF in a
vacuum. If the capacitance increases to 2 nF when a
dielectric material is inserted between the plates, what
is the dielectric constant of the material?
A.
B.
C.
D.
1
2
4
8
END OF TEST
KAPLAN
7
MCAT
ANSWER KEY:
1.
C
2.
C
3.
D
4.
B
5.
C
8
6.
7.
8.
9.
10.
A
D
D
B
A
11.
12.
13.
14.
15.
D
A
C
A
B
16.
17.
B
C
as developed by
Electric Circuits Test 1
EXPLANATIONS
Passage I (Questions 1–6)
1.
C
In order to answer this question, it is important to have a good understanding of what is going on in the circuit shown
in Figure 1. In this circuit each emf is drawn as an ideal emf connected in series with a resistor representing the internal
resistance of the emf. The internal resistance of an emf is actually just the intrinsic electrical resistance of its components.
As explained in the third paragraph of the passage, the loop theorem is used on the current loop ABCD to find the
expression for the current i. We need not know how exactly the relation given is obtained, but we need to make use of the fact
that the galvanometer reads zero. The galvanometer, as stated in the passage, is a very sensitive ammeter: in other words, it
measures current. The fact that to determine i, we adjust the potentiometer so that the galvanometer reads zero means that when
the emf ε x is connected in the loop ABCD, the current ix is zero, and when the emf ε 0 is connected, the current i 0 is zero.
Therefore, there is no current passing through the resistors representing the internal resistances. By Ohm’s law, the voltage drop
across a resistor is equal to IR, where I is the current flowing through the resistor and R its resistance. Therefore, there is no
voltage drop due to the internal resistances since there is no current flowing through the components of the emf. So, when the
galvanometer reads zero, the internal resistances have no effect on the calculation of current i.
2.
C
At first glance, this question seems to require some manipulation of the expressions for the current i given in the
passage. But in fact, all of the answer choices contain the emf ε instead of either ε x or ε 0 . The current i can be determined by
applying Ohm’s law. Since the galvanometer reads zero, there is no current through the upper portion of the circuit. The circuit
can therefore be simplified to one in which we have a battery with emf ε in series with two resistors: r and R. The voltage drop
across the two resistors added together must be equal to the voltage supplied by the battery (Kirchhoff’s law):
ε = iR + ir = i(r + R)
ε
i=
(r + R)
3.
D
The internal resistance of the galvanometer should be low so that it dissipates little energy. An ammeter is connected
in series in order to measure the current. It would not be desirable for the presence of the ammeter to actually change the current
by a significant amount. If the ammeter has a high internal resistance, it will act as a large resistance in series to whatever else
is in the circuit, and thus it will affect the net current through the circuit. More concretely, suppose we would like to measure
the current through the circuit on the left hand side below:
R
R
Ram
V
V
By introducing an ammeter into the circuit, however, we have in essence added a small resistance, Ram, that arises
because of the internal resistance of the ammeter. (See right hand side.) From Ohm’s law, we could see that the current has
changed from i = V/R to i’ = V/(R + Ram). In other words, the act of measuring the current has caused the current to decrease.
This is obviously undesirable. In an ideal case, the internal resistance of the ammeter would be zero, so the ammeter just gives a
reading of the current without changing it in any way. In general, then, we would like to minimize the internal resistance of the
ammeter so the change in current is negligible. We can therefore eliminate choices A and B. Furthermore, if the presence of the
ammeter is not to perturb the circuit too much, it will dissipate little energy. Choice D is therefore the correct answer.
KAPLAN
9
MCAT
4.
B
This question is a straightforward application of the last equation given in the passage. The equation expresses an
αx
unknown emf εx in terms of a known emf ε0. Specifically, εx = ε0
:
α0
α x (ADx/AE) ADx 0.2 1
=
=
=
=
α 0 (AD0/AE) AD0 0.4 2
The unknown emf is thus half the known emf, or 3 V. Note that the variables αx and α o are calculated from measured
distances in the experiment. Each α is the ratio of two distances: that of AD to that of AE. We do not know the value of AE,
the total length of the resistor, but since we are dividing the two α’s, this factor cancels.
5.
C
The resistance would remain the same. This question asks about the variable resistance αR. It is important to have a
conceptual understanding of what this expression is really saying: α, as described in the passage, is the distance AD divided by
the distance of AE: in other words, it is the fraction of the length of the resistor that is “used” in the loop ABCD. R is the total
resistance, and so the variable resistance that presents itself to the loop ABCD is a fraction of the total resistance, with the
fraction being equal to that of the length that is “sampled.” This should make good sense intuitively: if, for example, point D is
halfway between points A and E, the length of the resistor that is part of the loop is 1/2 that of the total length, and the
resistance is 1/2 R. Conversely, if we slide the contact so that points D and E coincide, the loop ABCD sees the entire resistor,
and so the resistance is R. (α = 1.)
The resistivity ρ of the variable resistor is said to be constant in the passage. If we double the length AE such that AE’
= 2AE, then, we have doubled the length of the resistor and thus the total resistance will be doubled as well, since the resistance
can be written as R = ρL/A, where L is the length and A the cross-sectional area. But since we have kept the contact point D at
the same position, α’ is now AD/AE’ = AD/2AE = half the old value of α. The resistance seen by the loop ABCD is then:
α’R’ =
α
2R = αR
2
In other words, it has not changed. This should be expected physically without any calculations actually: the resistance
between points A and D should depend only on the length of AD. Without altering the resistivity, extending the part of the
resistor that is outside of the loop in question cannot possibly affect its value.
6.
A
Current is charge over time. Physically, it represents a flow of electric charge and can loosely be interpreted as an
incompressible fluid of positive charge. The junction theorem, as described in the question stem, states that the algebraic sum of
the currents at any junction must be zero. In other words, given a complicated junction such as the following:
i1
i3
i2
i5
i4
we know that the net current flowing in has to equal the net current flowing out:
i 1 + i2 + i3 = i4 + i5
In other words, the amount of charge going in per unit time must equal the amount of charge exiting the junction per
unit time. This is therefore a result of the conservation of charge.
10
as developed by
Electric Circuits Test 1
Passage II (Questions 7–12)
7.
D
This question requires you to analyze the formula in the passage that gives the charge on the capacitor as a function of
time. Since there are two formulas, though, we have to make sure we use the correct one. When the switch is in position A the
capacitor is being charged by the emf, and when the switch is in position B the capacitor discharges across the resistor. This
question asks about the case where the switch is in position A; the applicable formula is therefore q = Cε(1 – e–t/RC), where q is
the charge, C is the capacitance, ε is the emf, t is the time, and R is the resistance of the resistor. The time t = 0 corresponds to
the instant that the switch is thrown to position A. Physically, since the capacitor is being charged by the emf, we would
expect that the charge on the capacitor is increasing as time passes. Therefore, the maximum charge will occur at maximum
time t, which is as it approaches infinity. Let us verify that the mathematical formula is consistent with this rather intuitive
approach.
At t = 0, the exponential term is e0 which is 1. The charge q is therefore Cε(1 – 1) = 0. This makes sense as the
capacitor is uncharged at t = 0. Let us examine closer the exponential term for t > 0. Since t, R and C are all positive quantities,
the exponential term is e to a negative number. As t increases, e is being raised to a number that is increasingly negative, which
means that the entire exponential term is becoming smaller and smaller, approaching zero. This term, however, is being
subtracted from a constant 1; so as the exponential term gets closer and closer to zero, (1 – e–t/RC) is increasing and getting
closer and closer to one. (Mathematically, we say that the quantity (1 – e–t/RC) is asymptotically approaching one, since it can
never increase beyond one as the value of the exponential term never gets below zero.)
1 – e –t/RC
e –t/RC
1
1
t
0
0
t
The above graphs illustrate the behavior of the exponential term itself and of the quantity (1 – e–t/RC). There is also a
multiplicative factor of Cε. As t approaches infinity, then, the charge approaches the value of Cε × 1 = Cε. The capacitor
becomes fully charged and the potential difference (or voltage drop) across the capacitor plates is equal to the emf. There is no
potential drop across the resistor because no current flows through the circuit anymore at that point, so according to Ohm’s law,
the voltage drop is V = iR = 0 × R = 0.
8.
D
By adding another resistor in series we have increased the overall resistance of the circuit. Qualitatively, the current
through the circuit at any time t would be smaller because of the higher resistance. The flow of charge to the capacitor is
therefore slowed, and so it will take longer for the capacitor to charge up to a particular level. Choices A and B are incorrect
because given enough time, the charge on a plate of the capacitor will approach the value of Cε as discussed above in the
explanation to #7.
9.
B
We are told that when the switch is in position B, the charge on the capacitor is described by the equation q = Cεe–t/RC,
where t = 0 is the time when the switch is thrown to position B. From the definition of capacitance, we have:
C=
q
q
, or V =
V
C
In this case, then, the voltage across the capacitor is:
Vc =
Cεe–t/RC
= εe–t/RC
C
At time t = 0 the capacitor is fully charged, and the voltage across it is equal to the emf ε. As time passes, the
capacitor discharges, and the voltage approaches zero as the time approaches infinity.
KAPLAN
11
MCAT
10.
A
First of all, we should recognize that the charge on each capacitor is the same. We could have inferred this from the
answer choices which make no distinction between the capacitors, or we could have reasoned it out from conservation of charge.
Consider two capacitors are connected in series as shown below, with the plates labeled A, B, E, F:
C1
A
B
C2
E
F
As the charge flows, let us say that the voltage source deposits positive charge on plate A and absorbs positive charge
from plate F. This connects the circuit, and leaves net negative charge on plate F. Note that the negative plate of C1, plate B, is
connected to the positive plate of C2, plate E. Since there can be no creation of net charge, the charges on these two plates have
to be equal in magnitude (but opposite in sign). From the workings of a capacitor, we know that this equality of charge
magnitude is manifested in the other plates as well (A and F):
C1
+q
-q
C2
+q
-q
We can think of this charge as being stored on an effective capacitor that has the same capacitance as this series
arrangement. Capacitors in series add in the same way as resistors in parallel do; i.e.:
1
1
1
1 + 1
2
=
+
=
=
Ceff C
C
C
C
Ceff =
C
2
As t goes to infinity, the capacitors will be fully charged. Using the equation q = Cε(1 – e–t/RC) given in the passage
and discussed in the explanation to #7, the charge approaches:
q = Ceffε =
Cε
2
where we have used the effective capacitance. In other words, the same amount of charge that was originally on one capacitor is
now spread over two capacitors.
11.
D
The information needed for the answer is not found directly in the text of the passage but is contained in Figure 2. Each
of the three plots that make up Figure 2 is a graph of voltage as a function of time, but none gives the emf ε directly. Consider
the circuit in Figure 1 when the switch is in position A. Using Kirchhoff’s law, we know that the emf ε must be equal to the
sum of the voltage drops across the resistor and the capacitor (VR and V C respectively). So we do have exactly what we need
after all: the sum (VR + VC) is plotted as a function of time in (c), and its value when the switch is in position A must be equal
to the magnitude of the emf. Looking at the section of the graph that corresponds to position A, we see that the emf equals two
12
as developed by
Electric Circuits Test 1
divisions in the vertical direction. We are told in the passage that each vertical division equals 1.0 V. Therefore, the emf must be
2 × 1.0 = 2.0 V.
12.
A
The circuit when the switch is at position B consists only of a capacitor in series with a resistor. The emf is no longer
a part of the circuit and has no effect. The charge that is initially stored in the capacitor at high potential will flow to a lower
potential. This is analogous to water flowing downhill until it reaches a point of minimum gravitational potential energy.
The positive terminal of a battery is at a higher potential than the negative terminal: positive charge hence flows from
the positive to the negative terminal aroung the circuit. This applies to capacitor plates as well: The plate with positive charge
is at a higher potential than the plate with the negative charge. Without an emf to provide the energy to separate the charges,
charge will flow from the positive to the negative plate if a path exists. In this case, the path is through the resistor:
i
R
C
+q
–q
As the charge flows, the positive and negative charges cancel each other until each plate is neutral (no more separation
of charge). The capacitor is then completely discharged, and the two plates are at the same potential. Choice A is consistent
with this analysis.
Choice B is incorrect because as just described, the reverse is true.
Choice C is incorrect because the emf is no longer part of the circuit when the switch is at position B. A current does
exist through the resistor, but it has nothing to do with the emf.
Choice D is also incorrect: The voltage across the capacitor is not zero since the plates are at different potentials. (The
voltage across the capacitor is just the potential difference across the plates.) This quantity is zero only when the capacitor is
fully discharged.
Independent Questions
13.
C
Capacitance in general is defined as C = Q/V, but for parallel-plate capacitors, we have an equation that gives the
capacitance in terms of its geometry:
C = ε0
A
d
where ε0 is the permittivity constant, A the area of the overlap of the plates, and d the distance between them. If the separation
is increased, therefore, the capacitance will decrease. Statement I is therefore correct. We are told that the capacitor is isolated;
there is thus no pathway for the charge on one to flow to the other and be neutralized, nor is there any external circuitry (emf,
for example) that can generate more charge separation on the plates. The amount of charge on each plate therefore has to remain
the same; so statement II is incorrect. To evaluate whether statement III is true, we need to resort to the more general definition
of capacitance given above: C = Q/V. Since we know that the capacitance changes, the right hand side of the equation must also
change. The charge, however, is constant. Therefore the voltage across it has to change. Statement III is hence also correct in
addition to statement I.
14.
A
The key to answering this question is to remember that the power dissipated by a resistor equals i 2 R, where i is the
current through the resistor and R its resistance. The bulb acts only as a second resistor in this question. (It dissipates electrical
energy in the form of light and heat instead of just heat like normal resistors.) Since the two resistors are in parallel, the
effective resistance is:
1
1
1
1 × 5 + 1
6
=
+
=
=
Reff 150 Ω
750 Ω
750Ω
750Ω
KAPLAN
13
MCAT
750Ω
= 125Ω
6
The current through is given as 0.2 A, and so the power dissipated is:
Reff =
P = (0.2)2 × 125 = 0.04 × 125 = 5.0 W
which is choice A. Choice B is the incorrect result one would have obtained by using the 150-Ω resistor alone; choice C is the
incorrect result one would have obtained by using the bulb alone. Choice D is the result of adding the resistors in series rather
than in parallel.
15.
B
The dielectric constant κ of a material inserted in between the plates of a capacitor affects the capacitance as follows:
C’ = κC
where C’ is the new capacitance and C the capacitance if the material were absent. In this case, the capacitance increases from 1
nF without the dielectric to 2 nF with the dielectric. The dielectric constant must therefore be 2.
16.
B
This question is about a sinusoidal current, which generally has the form:
i = Imaxsinωt
where Imax is the maximum current (or amplitude of the oscillation in current), and ω the angular frequency. The graph of the
current versus time is a sine curve (hence the name sinusoidal):
i
Imax
t
2 /ω
In this case, Imax is given as 2 A. The resistor dissipating power has a resistance of R = 1kΩ = 1000Ω. In the case of a
DC current the power dissipated by the resistor is i 2 R, but we are now interested in the time average of the power over one
period since the current is changing. R is constant, so we need to determine the average of i2 over one period. This is the square
of the root-mean-square current. The root-mean-square current, irms, is related to the maximum current via:
irms =
Imax
2
The square of irms is the average of the square of the actual current over one period, which is what we are after:
2
2
I

2 I
i 2 = ( i rms ) =  max  = max
 2 
2
The average power is thus Imax2R/2 = 22 × 1000/2 = 2000 W = 2 kW.
14
as developed by
Electric Circuits Test 1
17.
C
We need to relate the resistance of a wire to its length and cross-sectional area. The relevant equation is:
R=
ρL
A
where R is the resistance, ρ is the resistivity which is a property of the material out of which the resistor is made, L is the
length and A is the cross-sectional area. For a cylindrical wire the cross section is a circle, and A is thus r 2 where r is the
radius. If both L and r are halved, the new resistance R’ is:
R’ =
ρ(L/2) 1/2 ρL
=
= 2R
(r/2)2
1/4 r2
The resistance has thus been doubled. One can obtain this answer without laborious plug-and-chug by noting that the
length appears in the numerator, while the radius appears in the denominator squared. If we change both in the same direction by
the same factor, the two work against each other, but the radius effect wins out because of the square.
KAPLAN
15