Download Problem Set 3 with solutions

Problem Set 3 Solution.
1.
A toaster has a resistance of 0.014kS and is plugged into a 120-V outlet. What
is the current in kA?
Step 1. Conversion
Given: 0.014kS
Path: kS 6 S
Target: ?S
C. F.:
Computation:
Step 2. Solving
Using Ohm’s Law
Step 3. Conversion
Given: 8.57A
Path: A 6 kA
Target: ?kA
C. F.:
Computation:
2.
In the arctic, electric socks are useful because they have an electric wire woven
into the socks. Usually, a 9-V battery powers the circuit in each of the socks. The
current in the wire is 110mA. Determine the resistance of the wire in mS.
Step 1. Conversion
Given: 110mA
Path: mA 6 A
C. F.:
Computation:
Step 2. Solving
Using Ohm’s Law
Target: ?A
Step 3. Conversion
Given: 81.2S
Path: S 6 mS
Target: ?mS
C. F.:
Computation:
3.
A light bulb that is plugged into 0.120-kV outlet has a resistance of 0.192kS.
a.
What is the current through the bulb in mA?
Step 1. Conversion
Given: 0.120kV
Path: kV 6 V
Target: ?V
C. F.:
Computation:
Given: 0.192kS
Path: kS 6 S
Target: ?S
C. F.:
Computation:
Step 2. Solving
Using Ohm’s Law
Step 3. Conversion
Given: 0.625A
Path: A 6 mA
C. F.:
Computation:
Target: ?mA
b.
How much energy the bulb consumes in 3 hours?
Step 1.
Given: 3hrs Target: ?s
Path: hr 6 s
C. F.:
Computation:
Step 2.
Let E be the energy consumed by the bulb in 3 hrs
4.
An electric blanket that is connected to 120-V outlet consumes 0.140kW
a.
What is the resistance of the blanket in kS?
Step 1.
Given: 0.140kW
Target: ?W
Path: kW6 W
C. F.:
Computation:
Step 2.
Using Ohm’s Law and the expression for power
Step 3.
Given: 103S Target: ?kS
Path: S6 kS
C. F.:
Computation:
b.
What is the current through the blanket in kA?
Step 1. Is not necessary
Step 2.
Using the Ohm’s Law
Step 3.
Given: 1.17A
Path: A6 kA
Target: ?kA
C. F.:
Computation:
5.
A 0.050kS resistor carries a current 25 A.
a.
How much power in mW is used by this resistor?
Step 1.
Given: 0.050kS
Target: ?S
Path: kS6 S
C. F.:
Computation:
Step 2.
Using the Ohm’s Law and the expression for the power
Step 3.
Given: 31250W
Path: W6 mW
Target: ?mW
C. F.:
Computation:
b.
What is the potential difference in kV across this resistor?
Step 1. is not necessary
Step 2. Using the Ohm’s Law
Step 3.
Step 3.
Given: 1250V
Path: V6 kV
C. F.:
Computation:
Target: ?kV