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Version 001 – Circuits I – tubman – (2016winter1)
This print-out should have 14 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Circuits 03
001 10.0 points
When electrons move back and forth, reversing their direction regularly, the current is
called
1. direct current.
2. a short circuit.
Let : I = 13.0 × 10−3 A
∆Q = 15.0 C .
1
and
∆Q
∆t
15 C
∆Q
=
= 1153.85 s .
∆t =
I
0.013 A
I=
3. alternating current. correct
4. induced current.
Explanation:
Meters 02
002 10.0 points
An instrument used to detect the current in a
circuit is called
1. an ammeter. correct
2. a motor.
3. a voltmeter.
Hewitt CP9 23 03
004 10.0 points
What is drift velocity?
1. The lowest speed of an electron in a metal
2. The speed of an electric field
3. The average speed of atoms in a liquid
4. The average speed of electrons in a conductor in an electric field correct
5. The highest speed of an electron in a
metal
4. a transformer.
5. an ohmmeter.
6. an electroscope.
7. a generator.
Explanation:
Holt SF 19A 01
003 10.0 points
If the current in a wire of a CD player is 13.0
mA, how long would it take for 15.0 C of
charge to pass a point in this wire?
Correct answer: 1153.85 s.
Explanation:
Explanation:
When an electric field is applied to a conductor, the electrons continue their random
motions while simultaneously being nudged
by this field. The collisions interrupt the motion of the electrons along the field lines. The
average speed at which they migrate along a
wire is known as drift velocity.
Comparisons in Conductors 01
005 (part 1 of 2) 10.0 points
Consider two conductors 1 and 2 made of the
same ohmic material; i.e., ρ1 = ρ2 . Denote
the length by ℓ, the cross sectional area by A.
The same voltage V is applied across the ends
of both conductors and the field E is inside of
the conductor.
Version 001 – Circuits I – tubman – (2016winter1)
~E 1
~E 2
I1
V1
I2
3.
V2
4.
b
b
ℓ1
r1
ℓ2
5.
r2
If A2 = 2 A1 , ℓ2 = 2 ℓ1 and V2 = V1 , find
E2
of the electric fields.
the ratio
E1
E2
1.
=1
E1
E2
2.
=8
E1
1
E2
=
3.
E1
4
E2
1
4.
=
E1
8
E2
1
5.
= correct
E1
2
E2
=4
6.
E1
E2
7.
=2
E1
1
E2
=
8.
E1
12
E2
1
9.
=
E1
16
E2
1
10.
=
E1
3
Explanation:
V
1
E=
∝ when V1 = V2 , so
ℓ
ℓ
6.
7.
8.
9.
10.
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
2
1
4
1
=
2
1
=
3
1
=
12
=
=8
=4
= 1 correct
=
1
8
Explanation:
V
1
ℓ
ℓ
I =
∝
and R = ρ
∝
when
R
R
A
A
V1 = V2 and ρ1 = ρ2 , so
R1
ℓ 1 A2
ℓ1 (2 A1)
I2
=
=
=
= 1.
I1
R2
ℓ 2 A1
(2 ℓ1 ) A1
Tipler PSE5 25 43
007 10.0 points
A carbon rod with a radius of 1 mm is used to
make a resistor.
What length of the carbon rod should be
used to make a 15.8 Ω resistor? The resistivity
of this material is 7.7 × 10−5 Ω · m .
Correct answer: 644.638 mm.
Explanation:
E2
ℓ1
ℓ1
1
=
=
= .
E1
ℓ2
2 ℓ1
2
006 (part 2 of 2) 10.0 points
I2
of the currents.
Find the ratio
I1
I2
1.
=2
I1
I2
1
2.
=
I1
16
Let : r = 1 mm = 0.001 m ,
ρ = 7.7 × 10−5 Ω · m ,
R = 15.8 Ω .
and
The cross-sectional area of the rod is
A = π r 2 = π(0.001 m)2
= 3.14159 × 10−6 m2
and
Version 001 – Circuits I – tubman – (2016winter1)
3
The resistance is
ρℓ
A
AR
ℓ=
ρ
(3.14159 × 10−6 m2 )(15.8 Ω) 103 mm
=
·
7.7 × 10−5 Ω · m
1m
R=
= 644.638 mm .
Serway CP 17 16
008 10.0 points
A length of copper wire has a resistance 22 Ω.
The wire is cut into three pieces of equal
length, which are then connected as parallel
lengths between points A and B.
What resistance will this new “wire” of
L0
length
have between points A and B?
3
R=
Current in Tungsten Wire
010 10.0 points
A 0.98 V potential difference is maintained
across a 1.8 m length of tungsten wire that
has a cross-sectional area of 0.49 mm2 .
What is the current in the wire? The resistivity of the tungsten is 5.6 × 10−8 Ω · m .
Correct answer: 4.76389 A.
Explanation:
Let :
Correct answer: 2.44444 Ω.
Explanation:
Let : R0 = 22 Ω .
L0
and crossThe new wire has length L =
3
sectional area A = 3 A0 , so its resistance is
L0
ρ
ρL
1 ρ L0
3
R=
=
=
A
3 A0
9 A0
R0
22 Ω
=
=
= 2.44444 Ω .
9
9
Current in a TV
009 10.0 points
A typical color television draws about 2.6 A
when connected to a 106 V source.
What is the effective resistance of the T.V.
set?
Correct answer: 40.7692 Ω.
Explanation:
Let :
V
106 V
=
= 40.7692 Ω .
I
2.6 A
V = 0.98 V ,
A = 0.49 mm2 = 4.9 × 10−7 m2 ,
ℓ = 1.8 m , and
ρ = 5.6 × 10−8 Ω · m .
The resistance is
V
ρℓ
=
I
A
VA
(0.98 V)(4.9 × 10−7 m2 )
I=
=
ρℓ
(5.6 × 10−8 Ω · m)(1.8 m)
R=
= 4.76389 A .
Ohms Law 01
011 10.0 points
According to Ohm’s Law, if the resistance in
a circuit is 50 Ω and the voltage is 7.1 V, what
will be the current flow in the circuit?
1. 57.1 A
2. 355 A
3. 0.142 A correct
4. 7.04225 A
I = 2.6 A and
V = 106 V .
5. 42.9 A
Explanation:
Version 001 – Circuits I – tubman – (2016winter1)
Let :
R = 50 Ω and
V = 7.1 V .
From Ohm’s Law, ∆V = I R, so
I=
V
7.1 V
=
= 0.142 A .
R
50 Ω
Electric Shock
012 (part 1 of 3) 10.0 points
The damage caused by electric shock depends
on the current flowing through the body;
1 mA can be felt and 5 mA is painful. Above
15 mA, a person loses muscle control, and 70
mA can be fatal. A person with dry skin has a
resistance from one arm to the other of about
1.1 × 105 Ω. When skin is wet, the resistance
drops to about 5800 Ω.
What is the minimum voltage placed across
the arms that would produce a current that
could be felt by a person with dry skin?
Correct answer: 110 V.
Explanation:
Let : Imin = 1 mA and
Rdry = 1.1 × 105 Ω .
The minimum voltage depends on the minimum current for a given resistance, so
Vmin = Imin Rdry
= (1 mA)
1A
1000 mA
(1.1 × 105 Ω)
= 110 V .
013 (part 2 of 3) 10.0 points
For the same electric potential what would be
the current if the person had wet skin?
Correct answer: 18.9655 mA.
Explanation:
Let : Rwet = 5800 Ω .
110 V
Vmin
=
I=
Rwet
5800 Ω
4
1000 mA
1A
= 18.9655 mA .
014 (part 3 of 3) 10.0 points
What would be the minimum voltage that
would produce a current that could be felt
when the skin is wet?
Correct answer: 5.8 V.
Explanation:
V1 = Imin Rwet
= (1 mA)
= 5.8 V .
1A
(5800 Ω)
1000 mA