Version 001 – Circuits II – tubman – (12125)
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Light Bulb 01
001 (part 1 of 3) 10.0 points
A 100-W light bulb is plugged into a standard
120 V outlet.
How much does it cost per month (31 days)
to leave the light turned on? Assume electric
energy cost of 6 cents/kWh
Correct answer: $4.464.
Explanation:
Let :
1
Explanation:
V =IR
V
120 V 1000 mA
I=
=
×
R
144 Ω
A
= 833.333 mA .
Resistor in a Circuit
004 (part 1 of 2) 10.0 points
A 12.6 Ω resistor is connected to a 10.2 V
battery.
What is the current in the circuit?
Correct answer: 0.809524 A.
C = $0.06/kWh ,
P = 100 W , and
t = 31 d .
Explanation:
Let : R = 12.6 Ω and
V = 10.2 V .
Voltage is
V =IR
V
10.2 V
I=
=
= 0.809524 A .
R
12.6 Ω
C = ($0.06/kWh)(100 W)(31 d)
1 kW
24 h
×
1000 W
1d
= $4.464 .
005 (part 2 of 2) 10.0 points
How much thermal energy is produced in
24.7 min?
002 (part 2 of 3) 10.0 points
What is the resistance of the bulb?
Correct answer: 12237.1 J.
Correct answer: 144 Ω.
Explanation:
Explanation:
Let :
Let : t = 24.7 min .
V = 120 V .
Power is
E
t
E =Pt=IV t
P =
V2
R
V2
(120 V )2
R=
=
= 144 Ω .
P
100 W
P =
003 (part 3 of 3) 10.0 points
What is the current in the bulb?
Correct answer: 833.333 mA.
60 s
= (0.809524 A)(10.2 V)(24.7 min)
min
= 12237.1 J .
Car Battery
006 (part 1 of 2) 10.0 points
A 12 V automobile battery is connected to an
Version 001 – Circuits II – tubman – (12125)
electric starter motor. The current through
the motor is 244 A.
How much power does the motor dissipate?
009 (part 2 of 5) 10.0 points
What is the resistance of a single light?
Correct answer: 2928 W.
Correct answer: 12.8401 Ω.
Explanation:
Power is
P = I V = (244 A) (12 V)
2
Explanation:
The total resistance is n times the resistance of one bulb, so
= 2928 W .
007 (part 2 of 2) 10.0 points
How much energy does the battery deliver to
the motor during a 19.5 s period?
010 (part 3 of 5) 10.0 points
How much power is dissipated in a single
light?
Correct answer: 57096 J.
Explanation:
Let :
V = 12 V ,
I = 244 A ,
t = 19.5 s .
Correct answer: 2.63636 W.
and
Energy is
E = P · T = (I V ) t
= (2928 W) (19.5 s)
= 57096 J .
Christmas Lights
008 (part 1 of 5) 10.0 points
A string of 22 identical Christmas tree lights
are connected in series to a 128 V source. The
string dissipates 58 W.
What is the equivalent resistance of the
light string?
Correct answer: 282.483 Ω.
Explanation:
n = 22 ,
V = 128 V ,
P = 58 W .
Power is defined by
Explanation:
The total power is n times the power of one
bulb, so
P = n P1
P
58 W
P1 =
=
= 2.63636 W .
n
22
011 (part 4 of 5) 10.0 points
One of the bulbs quits burning. The string
has a wire that shorts out the bulb filament
when it quits burning, dropping the resistance
of that bulb to zero. All the rest of the bulbs
remain burning.
What is the resistance of the light string
now?
Correct answer: 269.643 Ω.
Let :
and
V2
R
(128 V)2
V2
=
= 282.483 Ω .
R=
P
58 W
P =
R = n R1
R
282.483 Ω
R1 =
=
= 12.8401 Ω .
n
22
Explanation:
We now have n − 1 bulbs, each with a
resistance of R1 , so
Rnew = (n − 1) R1
= (22 − 1) 12.8401 Ω
= 269.643 Ω .
Version 001 – Circuits II – tubman – (12125)
3
The resistors are in parallel, so
012 (part 5 of 5) 10.0 points
How much power is dissipated by the string
now?
Correct answer: 60.7619 W.
Explanation:
V2
Rnew
V2
=
(n − 1) R1
(128 V)2
=
(22 − 1) (12.8401 Ω)
Pnew =
= 60.7619 W .
Electric Circuit 01
013 10.0 points
If the different parts of an electric circuit are
found on separate branches of the circuit, the
circuit is called
1. a parallel circuit. correct
2. a series circuit.
3. a transistor circuit.
4. an open circuit.
Explanation:
1
1
1
1
=
+
+
Req
R1 R2 R3
−1
1
1
1
Req =
+
+
R1 R2 R3
−1
1
1
1
+
+
=
2 Ω 9 Ω 17 Ω
= 1.49268 Ω .
015 (part 2 of 2) 10.0 points
b) Determine the current in the circuit.
Correct answer: 8.03922 A.
Explanation:
Let :
∆V = 12 V .
∆V = I R
∆V
12 V
I=
=
= 8.03922 A .
Req
1.49268 Ω
Resistors in Parallel
016 (part 1 of 3) 10.0 points
A 18.3 Ω and a 23.8 Ω resistor are connected
parallel. A difference in potential of 40.7 V is
applied to the combination.
What is the equivalent resistance of the
parallel circuit?
Correct answer: 10.3454 Ω.
Explanation:
Holt SF 20Rev 18
014 (part 1 of 2) 10.0 points
A 2 Ω resistor, a 9 Ω resistor, and a 17 Ω
resistor are connected in parallel across a 12 V
battery.
Determine the equivalent resistance for the
circuit.
Correct answer: 1.49268 Ω.
Explanation:
Let : R1 = 2 Ω ,
R2 = 9 Ω , and
R3 = 17 Ω .
Let : R1 = 18.3 Ω ,
R2 = 23.8 Ω , and
V = 40.7 V .
R1 and R2 are in parallel, so
1
1
1
R2 + R1
=
+
=
R
R1 R2
R1 R2
R1 R2
R1 + R2
(18.3 Ω)(23.8 Ω)
=
18.3 Ω + 23.8 Ω
R=
= 10.3454 Ω .
Version 001 – Circuits II – tubman – (12125)
017 (part 2 of 3) 10.0 points
What is the current in the circuit?
Correct answer: 3.93413 A.
Explanation:
By Ohm’s Law, V = I R.
I=
V
40.7 V
=
= 3.93413 A .
R
10.3454 Ω
018 (part 3 of 3) 10.0 points
How much current passes through the 18.3 Ω
resistor?
Correct answer: 2.22404 A.
Explanation:
In a parallel circuit, the voltages are the
same, so the current through R1 is
I1 =
V
40.7 V
=
= 2.22404 A .
R1
18.3 Ω
4