, `UNITS AND ELECTRICAL QUANTITIES
1. Determine the magnitude of the voltage required to accelerate an electron to a kinetic
energy of 8×10−15 Joules?
a. 50000 b. 5000 c. 500 d. 50
SOLUTION:
−15
W 8 × 10
E= Q = 1.6 ×10
−9 = 50, 000 Volts
2. Determine the maximum number of electrons that can exist in the third shell of an atom?
a. 15 b. 16 c. 18 d. 19
SOLUTION:
N = 2 n2 = 2(3)2
N = 18 electrons
3. A car battery supplies a current of 50 A to the starter motor. How much charge passes
through the starter in ½ minute?
a. 15 b. 150 c. 1500 d. 15000
SOLUTION:
60 s
Q = IT = 50 (0.5 min × 1min ) = 1500 c
4. How many circular mils does a round copper rod of 0.25-inch diameter has?
a. 63,500 b. 62,500 c. 64000 d. 65000
SOLUTION:
Remember:
1000 mils 1 inch = 1000 mils
d = 0.25 in × = 250 mils
1∈¿ ¿
1 MCM = 1000 cm
A = d 2 = (250)2 = 62, 500 cm
5. What is the size in square millimeter (mm 2) is the cable of 250 mm size?
a. 126.67mm² b. 127.50mm² c. 125. 76mm² d. 129.50mm²
, SOLUTION:
250 MCM 250,000 CM
2 2
d (12.7)
A = d2 d = √ A A= =
4 4
d = √ 250,000 = 500 mils A = 126.67 mm 2
1 inch 25.4 mm
d = 500 mils × × d = 12.7 mm
1000 mils 1inch
6. A copper rod of 10 m long has a diameter with one inch. If the resistivity of copper is
10.371 circular mil per foot, what is the resistance
a. 3.4×10^-3 ohm b. 3.4×10^-4 ohm c. 3.5×10^-6 ohm d. 3.6×10^-4 ohm
SOLUTION:
3.281 ft
L = 10 m × = 32.81 ft. A = d 2 = (1000)2 = 1,000,000 cm
1m
pL ( 10.371 ) (32.81)
d = 2 inch = 1000 mils R = = = 3.4 ×
A 1,000,000
10
−4
7. A one- meter rod of 2 cm diameter is drawn until its resistance is times the initial
resistance. Its length afterwards is?
a. 15m b. 10m c. 20m d. 5m
SOLUTION:
√
R2 L2 2 R
= ( ¿ L2 = L1 ( 2 )
R1 L1 R1
L2=1(
√ 100 R1
R1
) = 10 m
8. A kilometer of wire having a diameter of 11.7 mm and a resistance of 0.031 ohm is drawn
down so that its diameter is 5.0 mm. what does its resistance become?
a. 0.95ohm b. 0.93ohm c. 0.97ohm d. 0.91ohm
SOLUTION:
R2 d1 4 d1 4
= ( ¿ R2 = R1 ( ¿
R1 d2 d2
, 11.7 4
R2 = 0.031 ( ¿ = 0.93
5
9. If a current of 2A flows through a point in a wire for 30 seconds, how many columns pass
through the point in the wire?
a. 70 coulombs b. 50 coulombs c. 60 coulombs d. 40 coulombs
SOLUTION:
coulombs
Q=It=2 × 30 sec
sec
Q = 60 coulombs
10. What quantity of charge must be delivered by a battery with potential difference of 100 V
to do 500 J of work?
a. 5coulombs b. 6 coulombs c. 7 coulombs d. 8coulombs
SOLUTION:
W 500
Q= = Q = 5 coulombs
E 100
11. The resistance of 120 meters of wire is 12 ohms. What is its conductance?
a. 0.0855 mho b. 0.0875 mho c. 0.0886 mho d. 0.0833 mho
SOLUTION:
1 1
G= = G = 0.0833 mho
R 12
12. A radio receiver draws 0.9 A at 110 V. if the jet is used 3 hours per day, how much energy
does it consume in 10 days?
a. 2.94kWh b. 2.79 kWh c. 2.97kWh d. 2.49kWh
SOLUTION:
P = EI = 110(0.9) = 0.099 kW
h
W = Pt = 0.099 (3 × 10 day)
day
W = 2.97 kWh
1. Determine the magnitude of the voltage required to accelerate an electron to a kinetic
energy of 8×10−15 Joules?
a. 50000 b. 5000 c. 500 d. 50
SOLUTION:
−15
W 8 × 10
E= Q = 1.6 ×10
−9 = 50, 000 Volts
2. Determine the maximum number of electrons that can exist in the third shell of an atom?
a. 15 b. 16 c. 18 d. 19
SOLUTION:
N = 2 n2 = 2(3)2
N = 18 electrons
3. A car battery supplies a current of 50 A to the starter motor. How much charge passes
through the starter in ½ minute?
a. 15 b. 150 c. 1500 d. 15000
SOLUTION:
60 s
Q = IT = 50 (0.5 min × 1min ) = 1500 c
4. How many circular mils does a round copper rod of 0.25-inch diameter has?
a. 63,500 b. 62,500 c. 64000 d. 65000
SOLUTION:
Remember:
1000 mils 1 inch = 1000 mils
d = 0.25 in × = 250 mils
1∈¿ ¿
1 MCM = 1000 cm
A = d 2 = (250)2 = 62, 500 cm
5. What is the size in square millimeter (mm 2) is the cable of 250 mm size?
a. 126.67mm² b. 127.50mm² c. 125. 76mm² d. 129.50mm²
, SOLUTION:
250 MCM 250,000 CM
2 2
d (12.7)
A = d2 d = √ A A= =
4 4
d = √ 250,000 = 500 mils A = 126.67 mm 2
1 inch 25.4 mm
d = 500 mils × × d = 12.7 mm
1000 mils 1inch
6. A copper rod of 10 m long has a diameter with one inch. If the resistivity of copper is
10.371 circular mil per foot, what is the resistance
a. 3.4×10^-3 ohm b. 3.4×10^-4 ohm c. 3.5×10^-6 ohm d. 3.6×10^-4 ohm
SOLUTION:
3.281 ft
L = 10 m × = 32.81 ft. A = d 2 = (1000)2 = 1,000,000 cm
1m
pL ( 10.371 ) (32.81)
d = 2 inch = 1000 mils R = = = 3.4 ×
A 1,000,000
10
−4
7. A one- meter rod of 2 cm diameter is drawn until its resistance is times the initial
resistance. Its length afterwards is?
a. 15m b. 10m c. 20m d. 5m
SOLUTION:
√
R2 L2 2 R
= ( ¿ L2 = L1 ( 2 )
R1 L1 R1
L2=1(
√ 100 R1
R1
) = 10 m
8. A kilometer of wire having a diameter of 11.7 mm and a resistance of 0.031 ohm is drawn
down so that its diameter is 5.0 mm. what does its resistance become?
a. 0.95ohm b. 0.93ohm c. 0.97ohm d. 0.91ohm
SOLUTION:
R2 d1 4 d1 4
= ( ¿ R2 = R1 ( ¿
R1 d2 d2
, 11.7 4
R2 = 0.031 ( ¿ = 0.93
5
9. If a current of 2A flows through a point in a wire for 30 seconds, how many columns pass
through the point in the wire?
a. 70 coulombs b. 50 coulombs c. 60 coulombs d. 40 coulombs
SOLUTION:
coulombs
Q=It=2 × 30 sec
sec
Q = 60 coulombs
10. What quantity of charge must be delivered by a battery with potential difference of 100 V
to do 500 J of work?
a. 5coulombs b. 6 coulombs c. 7 coulombs d. 8coulombs
SOLUTION:
W 500
Q= = Q = 5 coulombs
E 100
11. The resistance of 120 meters of wire is 12 ohms. What is its conductance?
a. 0.0855 mho b. 0.0875 mho c. 0.0886 mho d. 0.0833 mho
SOLUTION:
1 1
G= = G = 0.0833 mho
R 12
12. A radio receiver draws 0.9 A at 110 V. if the jet is used 3 hours per day, how much energy
does it consume in 10 days?
a. 2.94kWh b. 2.79 kWh c. 2.97kWh d. 2.49kWh
SOLUTION:
P = EI = 110(0.9) = 0.099 kW
h
W = Pt = 0.099 (3 × 10 day)
day
W = 2.97 kWh
