SCI-111 Assignment #6b Solutions
Text Problems:
Chapter Number
6
Questions for Thought
5, 7, 8, 10, 11
Group B Problems
11 to 17
Problems
11. An electric motor draws a current of 11.5 A in a 240 V circuit. What is the power of this
motor?
This is a straight forward electric power question: P=V∙I.
P  V  I  240 V 11.5 A  2,760 Watts  2.763 kW
The motor produces 2.763 kW of power.
12. A swimming pool requiring a 2.0 hp motor to filter and circulate the water runs for 18
hours a day. What is the monthly electric cost for running this pool pump if electricity
costs $0.10 per kWh?
This problem is one of those interesting electric cost problems. It is simply a multiplication of
(Power x Time x Cost)
Total Cost  Power   Time   Cost per kWh 

 0.746 kW 

  2.0 hp  
1
hp



Dollars
 80.57
Month
 
hr
days
  18

30

month
  day
 
Dollars 
   0.10

kWh 
 
The pool pump costs $80.57 to run per month.
13. Is it possible for two people to simultaneously operate 1,300 W hair dryers on the same 120
V circuit without tripping a 15 A circuit breaker? Explain.
This is an electric power problem where you should use P=V∙I and solve for I. This current will
be the current drawn by the hair driers, If this total current is less than 15 A, then the circuit
breaker will not trip. Otherwise the breaker will trip.
P V  I
I 
P 2 1300 Watts

 21.7 A
V
120 Volts
No, you cannot run both hair dryers at the same time because they will draw a combined current
in excess of the 15 amps the circuit breaker allows.
14. A step-up transformer has a primary coil of 100 loops and a secondary coil of 1,500 loops.
If the primary coil is supplied with a household current of 120 V and 15 A,
a. What voltage is produced in the secondary circuit?
Clearly a transformer problem. We only have the transformer equation
VP VS

and
NP NS
conservation of energy (Power in = power out).
VP V S

NP NS
VS  N S 
VP
120 V
 1,500 Loops 
 1,800 V
NP
100 Loops
The step-up transformer has 1,800 volts on the secondary.
b. What current flows in the secondary circuit?
I like to solve this problem using conservation of energy, rather than just plugging into
another formula. The power on the primary side of the transformer must equal the power
on the secondary side. On the primary side, the power is given by
P = V∙I = 120 V∙15 A = 1,800 W
So, on the secondary side
P V I
I 
P 1,800 Watts

1A
V 1,800 Volts
The secondary loop draws only 1 Amps. (Note: This is why they use high voltage in
transmission lines; they draw less current for a given power delivered to the home.
Transmission line power losses are proportional to the current in the lines.)
15. The step-down transformer in a local neighborhood reduces the voltage from a 7,200 V
line to 120 V.
a. If there are 125 loops on the secondary, how many are on the primary coil?
V
V
Clearly a transformer problem. We only have the transformer equation P  S and
NP NS
conservation of energy (Power in = power out).
VP V S

NP NS
N P  VP 
NS
125 Loops
 7,200 V 
 7,500 Loops
VS
120 V
The step-down transformer has 7,500 loops on the secondary. (Note: The step-down
transformer steps the voltage down, but not the number of turns.)
b. What current does the transformer draw from the line if the current in the
secondary is 36 A?
I like to solve this problem using conservation of energy, rather than just plugging into
another formula. The power on the primary side of the transformer must equal the power
on the secondary side. On the secondary side, the power is given by
P = V∙I = 120 V∙36 A = 4,320 W
So, on the primary side
P V  I
I 
P 4,320 Watts

 0.6 A
V 7,200 Volts
The primary loop draws only 0.6 Amps. (Note: This is why they use high voltage in
transmission lines; they draw less current for a given power delivered to the home.
Transmission line power losses are proportional to the current in the lines.)
c. What are the power input and output?
I have already answered this in part (b). The power on both sides of the transformer is
4,320 Watts.
16. A step-down transformer connected to a 120 V electric generator has 30 loops on the
primary for each loop in the secondary.
a. What is the voltage of the secondary?
Clearly yet another transformer problem. We only have the transformer equation
VP VS

and conservation of energy (Power in = power out).
NP NS
VP V S

NP NS
VS  N S 
N
VP
1
 VP  S  120 V   4 V
NP
NP
30
The secondary coil produces 4 volts on this step down transformer.
b. If the transformer has a 90 A current in the primary, what is the current in the
secondary?
I like to solve this problem using conservation of energy, rather than just plugging into
another formula. The power on the primary side of the transformer must equal the power
on the secondary side. On the primary side, the power is given by
P = V∙I = 120 V∙90 A = 10,800 W
So, on the secondary side
P V I
I 
P 10,800 Watts

 2,700 A
V
4 Volts
Yikes!!! That’s a lot of amps. The secondary would draw 90 amps of current.
c. What are the power input and output?
I have already answered this in part (b). The power on both sides of the transformer is
10,800 Watts.
17. What is the power of an 8 Ohm bulb when three such bulbs are connected in a 12 volt
series circuit?
If two bulbs are connected in series, then their resistances add. The total resistance of
these two bulbs in the circuit will be 24 Ohms. Now we use Ohm’s Law V=I∙R and the
Power formula P=V∙I to solve the problem.
V V 2 12 V 


 6 Watts.
R R 24 Ohm
2
Combine the two equations to get P  V 
The three bulbs in series draw 6 Watts of power.