Download 150HOMEWORK1SOLUTIONS Assignment Page

1.
CONST 150 HOMEWORK#1 SOLUTIONS
FIND CURRENT AND POWER USED BY R1
SHOW ALL WORK
R1= 10KΩ
10V
USE OHM’S LAW TO FIND CURRENT
V  IR FLOWING THRU R1
SOLVE FOR I
V
10V
10V
1.0 101V
I  


R 10 K 10,000 1.0 10 4 
 1.0 101 4 A  1.0 10 3 A
FIND POWER USED BY R1
WE CAN SOLVE BY USING EITHER:
I  1mA
P  IV OR P  I 2 R
FIRST USING:
P  IV
P  1mA  10V
P  0.001A  10V
P  1 10 3 A  1.0  101V
P  1 10 31  1 10  2 W
P  0.01W  10mW
OR USING:
P  I 2R


P  110 A1.0 10 
3
2
P  110 A 1.0 104 
6
4
P  1106 4W  1102W
P  0.01W  10mW
2.
CONST 150 HOMEWORK#1 SOLUTIONS
FIND VOLTAGE AND POWER USED BY R1
SHOW ALL WORK I=5mA
R1
10V
Since this is a series circuit with only one load R1
The source (10V) and load R1 voltages are the same.
To solve for the power used by R1.
USE P  IV
P  5mA 10V
P  5 10 3 A 1.0 101V
P  5 10  2W
P  50mW
3.
10V
CONST 150 HOMEWORK#1 SOLUTIONS
R1= 50Ω
FIND TOTAL POWER USED BY THE 50Ω RESISTOR
SHOW ALL WORK. GIVEN I=200mA
You can solve this problem 2 ways, by using either
P=IV or P=I²R
USING P  IV
P  200mA 10V
P  2 10  23 A 1.0 101V
P  2 10
11
W
P  2.0 10 W
0
P  I 2R
P  200mA2  50


P  4 10 A 5.0 10 
3
2
P  2 10 10 A  5.0 101 
2
2
1
P  2.0 1W
P  20 10 21W  2.0 101 10 1W
P  2W
P  2 100W  2W
4.
CONST 150 HOMEWORK#1 SOLUTIONS
FIND THE POWER USED BY THIS 8Ω SPEAKER
WHICH IS POWERED BY A 20VDC BATTERY
SHOW ALL WORK
V2
USE P 
FROM THE FORMULA WHEEL
R
2

20V 
P
8
400V
P
8
P  50W
CONST 150 HOMEWORK#1 SOLUTIONS
5.
SHOW ALL WORK
•Three resistors are connected to a 10-V
battery as shown in the diagram above. What
is the current through the 2.0 Ω resistor?
A.0.25 A
B.0.50 A
C.1.0 A
D.2.0 A
E.4.0 A
SINCE THIS IS A SERIES CIRCUIT,
RT  4  4  2
10V
RT  10
10Ω
VT  I T RT
VT
IT 
RT
10V
IT 
10
I T  1A
IN SERIES CIRCUIT CURRENT IS THE SAME IN EACH ELEMENT.
JUST SOLVE FOR IT USING OHM’S LAW AND WE HAVE THE ANSWER.
CORRECT ANSWER IS C
6.
CONST 150 HOMEWORK#1 SOLUTIONS
R1
R2
R3
1.A 100 Ω, 120 Ω, and 150 Ω resistor are connected
to a 9-V battery in the circuit shown above. Which of
the three resistors dissipates the most power?
A.the 100 Ω resistor
B.the 120 Ω resistor
C.the 150 Ω resistor
D.both the 120 Ω and 150 Ω
E.all dissipate the same power
6. SOLUTION FOR PROBLEM 6
RT
9V
SINCE THE 2 RESISTORS ARE IN PARALLEL (120Ω and 150Ω)
SOLVE FOR THE COMBINED RESISTANCE USING ;
R2,3 
R2  R3
R2  R3

120 150
R2,3  67
120  150
RT  R1  R2,3
RT  100  67  167
SOLVE FOR IT:
VT
9V
VT  I T RT , I T 

 0.05 A  50mA
RT 167
V1  IT R1  0.05 A100  5V
VT  V1  V2,3
V2,3  VT  V1  9V  5V  4V
2
2
V
5V
P100  100 
 0.25W
R1
100
2
P120 
2
V120
4V

 0.13W
R1
120
2
V150
4V 2
P150 

 0.11W
R1
150
100Ω RESISTOR USES THE MOST POWER 0.25WATTS
CORRECT ANSWER IS A
CONST 150 HOMEWORK#1 SOULTIONS
7.
CONST 150 HOMEWORK#1 SOLUTIONS
IT=I1=I2+I3
FIND I2,I3 ,RT SHOW WORK
7. SOLUTION FOR PROBLEM 7
TREAT THIS CIRCUIT AS SERIES- PARALLEL.
FIRST FIND COMBINED RESISTANCE FOR THE 2 RESISTORS IN PARALLEL,
R2,3
R R
 2 3
R2  R3
4 1.33 5.33 2


 1
4  1.33 5.33
12V
IN A SERIES CIRCUIT ALL THE RESISTORS
HAVE THE SAME CURRENT. SO,
IT  I1  I 2  I 3   4 A
R1
2
R2,3
1
USING OHM ' S LAW SOLVE FOR V2 ,V 2,3
V1  I T  R1  4 A  2  8V
VT  V1  V2,3
V2,3  VT  V1  12V  8V  4V
4V
 1A
4
4V
I3 
 3A
1.33
I T  I 2  I 3 USING KIRKOFFS CURRENT LAW
FIND RT, THEN USE OHM’S
LAW TO FIND IT
I2 
I T  1A  3 A  4 A
CONST 150 HOMEWORK#1 SOULTIONS
RT  R1  R2,3
OR 
RT  2  1  3
VT 12V
IT 

 4A
RT
3
8. CONST 150 HOMEWORK#1 SOULTIONS
HOW MUCH WILL IT COST TO RUN A 100W INCANDESCENT LIGHT BULB
FOR 30 DAYS ? YOU ARE BILLED AT 3¢ PER KILOWATT-HOUR.
30DAYSX24HRS=720 HRSX100W=72,000WATT-HOUR
72kWhr  72000Whr  0.03
72000  0.03

 2.16 $
kWhr
1000
HOW MUCH WILL IT COST TO RUN A EQUIVALENT 10W LED LIGHT BULB FOR 30 DAYS ?
30DAYSX24HRS=720 HRSX10W=7,200WATT-HOUR
72kWhr  7200Whr  0.03
7200  0.03

 21.6
kWhr
1000
¢