Version 001 – Quest 6 Review Circuits – tubman – (IBII201516)
This print-out should have 17 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Four Light Bulbs 02
001 (part 1 of 2) 10.0 points
Four identical light bulbs are connected either in series (circuit 1) or parallel (circuit 2)
to a constant voltage battery with negligible
internal resistance, as shown.
Circuit A
E
1
bulb is
E
E
R= ,
4R
4
so the power of each bulb in circuit 1 is
V =IR=
V2
E2
=
.
R
16 R
In circuit 2, the voltage across each bulb is
identical; namely E. Hence the power of each
bulb in circuit 2 is
E2
P2 =
= 16 P1 ,
R
more than 4 times brighter than the bulbs in
circuit 1.
P1 =
002 (part 2 of 2) 10.0 points
If one of the bulbs in circuit 2 is unscrewed
and removed from its socket, the remaining 3
bulbs
1. are unaffected. correct
2. go out.
3. become dimmer.
4. become brighter.
Circuit B
E
Compared to the individual bulbs in circuit
1, the individual bulbs in circuit 2 are
1. the same brightness.
2. less than
1
as bright.
4
3. more than 4 times brighter. correct
4.
Explanation:
Since the bulbs are parallel, after one of
the bulbs is unscrewed, the voltage across
each remaining bulb is unchanged, and the
brightness is unaffected.
Battery with internal resistance
003 (part 1 of 2) 10.0 points
The internal resistance r of a battery with
emf E is connected to a load resistor with
resistance R.
191 Ω
1
as bright.
4
5. 4 times brighter.
I
A
10 V
14 Ω
B
internal
resistance
Explanation:
In circuit 1, the voltage across each light
Find the potential difference VBA = VA −
VB .
Version 001 – Quest 6 Review Circuits – tubman – (IBII201516)
Explanation:
The power dissipated by the load is
2
E
2
P =i R=
R.
r+R
Correct answer: 9.31707 V.
Explanation:
Let : E = 10 V ,
R = 191 Ω ,
and
r = 14 Ω .
The current through the circuit is
i=
E
,
r+R
so the potential difference VA − VB is
ER
r+R
(10 V) (191 Ω)
=
14 Ω + 191 Ω
= 9.31707 V .
VA − VB = i R =
004 (part 2 of 2) 10.0 points
What is the power P dissipated by the load
resistor R?
E2
r
R
2
E
2. P =
r
R
E2
3. P =
r
1. P =
4. P = (r + R) E 2
5. P =
E2
rR
6. P = R E 2
7. P = r E 2
E2
8. P =
r+R
2
E
9. P =
r
r+R
2
E
R correct
10. P =
r+R
2
Internal Resistance 02
005 (part 1 of 2) 10.0 points
A battery with an emf of 6.2 V and internal
resistance of 1.05 Ω is connected across a load
resistor R.
If the current in the circuit is 1.86 A, what
is the value of R?
Correct answer: 2.28333 Ω.
Explanation:
Let : E = 6.2 V ,
I = 1.86 A ,
Ri = 1.05 Ω .
and
The electromotive force E is given by
E = I (R + Ri )
E
R = − Ri
I
6.2 V
=
− 1.05 Ω
1.86 A
= 2.28333 Ω .
006 (part 2 of 2) 10.0 points
What power is dissipated in the internal resistance of the battery?
Correct answer: 3.63258 W.
Explanation:
The power dissipation due to the internal
resistance is
P = I 2 Ri
= (1.86 A)2 (1.05 Ω)
= 3.63258 W .
Resistance of Nichrome Wire
007 (part 1 of 2) 10.0 points
Version 001 – Quest 6 Review Circuits – tubman – (IBII201516)
Calculate the resistance per unit length
of a nichrome wire, which has a radius of
0.251 mm. The resistivity of nichrome is
1.5 × 10−6 Ω · m.
Explanation:
Let : P
V
t
r
Correct answer: 7.57869 Ω/m.
Explanation:
Let :
ρ = 1.5 × 10−6 Ω · m and
r = 0.251 mm = 0.000251 m .
The cross-sectional area is A = π r 2 , so the
resistance per unit length λ is
λ≡
R
ℓ 1
ρ
1.5 × 10−6 Ω · m
=ρ
=
=
ℓ
A ℓ
π r2
π (0.000251 m)2
= 7.57869 Ω/m .
008 (part 2 of 2) 10.0 points
If a potential difference of 25 V is maintained
across a 9.2 m length of the nichrome wire,
what is the current in the wire?
Correct answer: 0.358557 A.
Explanation:
= 70 W ,
= 120 V ,
= 31 days , and
= 3.7 cents/kW · h .
The power is P = I V . The energy used is
E = P t, and the cost is
C =Er
= P tr
= (70 W) (31 days) (3.7 cents/kW · h)
24 h
1 kW
1 dollar
×
day
1000 W
100 cents
= 1.92696 dollars .
010 (part 2 of 3) 10.0 points
b) What is the resistance of the bulb?
Correct answer: 205.714 ohm.
Explanation:
Since the power of the light bulb is
P =
Let : V = 25 V and
ℓ = 9.2 m .
The resistance is R = λ ℓ , so applying
Ohm’s law,
I=
V
V
25 V
=
=
R
λℓ
(7.57869 Ω/m)(9.2 m)
= 0.358557 A .
V2
,
R
the resistance is
R=
(120 V)2
V2
=
= 205.714 ohm .
P
70 W
011 (part 3 of 3) 10.0 points
c) What is the current in the bulb?
Correct answer: 583.333 mA.
Lightbulb100W
009 (part 1 of 3) 10.0 points
A 70 W light bulb is plugged into a standard
120 V outlet.
a) How much does it cost per month (31
days) to leave the light turned on? Assume
electric energy cost of 3.7 cents/kW · h.
Correct answer: 1.92696 dollars.
3
Explanation:
The current
P
70 W 1000 mA
=
·
V
120 V
1A
= 583.333 mA .
I=
light bulbs in a circuit 02a
Version 001 – Quest 6 Review Circuits – tubman – (IBII201516)
4
012 (part 1 of 3) 10.0 points
Assume the battery is ideal (it has no internal resistance) and connecting wires have
no resistance. Unlike most real bulbs, the resistances of the bulbs in the questions below
do not change as the current through them
changes. Three identical bulbs are in the circuit as shown below in the figure. (The switch
S is initially closed.)
S
C
A
E
B
Which of the following correctly ranks the
bulbs in brightness?
1. Bulb B and C are equally bright, and
each is brighter than A.
2. None of these is correct.
3. All bulbs are equally bright.
4. Bulb A is the brightest, and B and C are
equally bright. correct
5. Bulb A is the brightest, B is next brightest, and C is the least brightest.
Explanation:
Bulb B and C are connected parallel, they
have the same potential difference, thus they
are equally bright. IA = IB + IC , so A the
brighter than B and C.
013 (part 2 of 3) 10.0 points
Which of the following correctly ranks the
current flowing through the bulbs?
1. Bulb B and C have the same current, and
each has more current than A.
2. None of these is correct.
3. Bulb A has the largest current, and C has
the smallest current.
4. Bulb A has the largest current, and B and
C have the same current. correct
5. All bulbs have the same current flowing
through them.
Explanation:
As mentioned in the last part, B and C
have
same potential difference and current
the
V
I=
, while the current of A is the sum
R
of that of B and C.
014 (part 3 of 3) 10.0 points
Which of the following correctly ranks the
potential difference across these bulbs?
1. Bulb B and C have the same potential
difference, and each has more potential difference than A.
2. The potential difference is largest across
A, and smallest across C.
3. Bulb A has the largest potential difference, and B and C have the same potential
difference. correct
4. All bulbs have the same potential across
them.
5. None of these is correct.
Explanation:
B and C have the same potential difference
because they are connected in parallel. Compare the potential differences of A and B (or
C): since IA > IB = IC and V = IR, bulb A
has the largest potential difference.
AP EM 1998 MC 15num
015 (part 1 of 2) 10.0 points
The following diagram shows part of a
closed electrical circuit.
Version 001 – Quest 6 Review Circuits – tubman – (IBII201516)
4Ω
11 Ω
1. the same everywhere in the circuit.
X
Y
2. greater in the 4 Ω resistor than in the 10 Ω
resistor.
10 Ω
E
I
Find the electric resistance RXY of the part
of the circuit shown between point X and Y .
Correct answer: 6 Ω.
R2
X
Y
R3
E
I
Let :
3. greater in the 10 Ω resistor than in the
11 Ω resistor. correct
4. greater at point X than at point Y .
5. greater in the 4 Ω resistor than in the 11 Ω
resistor.
Explanation:
R1
R1 = 4 Ω ,
R2 = 11 Ω ,
R3 = 10 Ω .
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 4 Ω resistor as in the 11 Ω
resistor. It is greater in the 10 Ω resistor than
in the 4 Ω or 11 Ω resistor.
From Ohm’s Law E = I R, we have
and
I12 =
E
E
=
R12
15 Ω
I3 =
E
E
=
R3
10 Ω
Since R1 and R2 are in series, their equivalent resistance R12 is
R12 = R1 + R2 = 4 Ω + 11 Ω = 15 Ω .
R12
X
I3 > I12 .
Y
15 Ω
R3
E
10 Ω
I
Since R12 and R3 are connected parallel,
their equivalent resistance RXY is
1
RXY
RXY
5
1
1
R3 + R12
+
=
R12 R3
R12 R3
R12 R3
(15 Ω) (10 Ω)
=
=
= 6Ω .
R12 + R3
15 Ω + 10 Ω
When there is a steady current in the circuit, the amount of charge passing a point
per unit time is greater in the smaller 10 Ω
resistor than in the larger 11 Ω resistor.
Current in Two Loop Circuit b
017 10.0 points
9.6 Ω
=
016 (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
I3
3.9 V
6.9 Ω
I2
2.9 Ω
1V
0.7 Ω
I1
6.1 V
Version 001 – Quest 6 Review Circuits – tubman – (IBII201516)
6
E1 + E2 RD Find the current I1 in the 0.7 Ω resistor
=0−1 at the bottom of the circuit between the two
E3
RD power supplies.
E1 + E2 0 + (−1) E3
RC Correct answer: 0.634432 A.
= − [(E1 + E2 ) RD − E3 RD ]
Explanation:
− [RC (E1 + E2 ) − 0]
RD
= RD (E3 − E1 − E2 ) − RC (E1 + E2 )
= (9.6 Ω) (3.9 V − 6.1 V − 1 V)
I3
E3
RC
−(6.9 Ω) (6.1 V + 1 V)
= −79.71 V Ω .
I2
E2
E1
Expanding along the first column, the deRB
RA
nominator is
I1
1
1
−1 At a junction (Conservation of Charge)
0 RD D = RA + RB
0
RC RD I1 + I2 − I3 = 0 .
(1)
0 RD Kirchhoff’s law on the bottom loop gives
= 1 RC RD (RA + RB ) I1 + RD I3 = E1 + E2 . (2)
1
−1 +0
− (RA + RB ) RC RD Kirchhoff’s law on the top loop gives
= 0 − RC RD − (RA + RB ) (RD + RC )
R C I2 + R D I3 = E 3 .
(3)
= (6.9 Ω) (9.6 Ω)
−(0.7 Ω + 2.9 Ω) (9.6 Ω + 6.9 Ω)
= −125.64 Ω2 , and
Let : RA = 0.7 Ω ,
RB = 2.9 Ω ,
−79.71 V Ω
D1
=
= 0.634432 A .
I1 =
RC = 6.9 Ω ,
D
−125.64 Ω2
RD = 9.6 Ω ,
E1 = 6.1 V ,
E2 = 1 V , and
E3 = 3.9 V .
Using determinants,
0
1
−1 E 1 + E2 0 RD E3
RC RD I1 = 1
1
−1
RA + RB
0 RD 0
RC RD Expanding along the first row, the numerator is
0
1
−1 D1 = E1 + E2 0 RD E3
RC RD