Transcript It is sometimes difficult to find the polarity of an

Chapter 20
Super Review
Circuits
1. What is the
potential difference
across a resistor of
6 Ω that carries a
current of 3 A?
V = IR
V=3x6
V = 18 V
2. A light bulb has a
resistance of 36 Ω
when operating at
12 V. What is the
current?
V = IR
12 = I x 36
1/3 A = I
3. If a 60 W light
bulb is connected
to a 120 V circuit,
what is the
current?
P = IV
60 = I x 120
0.5 A = I
4. A 1000 W heater
has a current of
10 A. What is
the potential
difference?
P = IV
1000 = 10 x V
100 V = V
5. Three resistors are
3 Ω, 4 Ω, and 5 Ω.
If connected in
series, what is
the equivalent
resistance?
R S = R1 + R2 + R3
RS = 3 + 4 + 5
RS = 12 Ω
6. Two resistors,
4 Ω and 8 Ω are
connected in parallel.
What is the
equivalent
resistance?
1/RP = 1/R1 + 1/R2
1/RP = 1/4 + 1/8
1/RP = 0.25 + 0.125
1/RP = 0.375
RP = 2.67 Ω
7. What is the equivalent
resistance of the circuit
shown above?
RS = R1 + R2
RS = 5 + 3
RS = 8 Ω
8. What is the equivalent
resistance of the circuit
shown above?
1/RP = 1/R1 + 1/R2
1/RP = 1/12 + 1/6
1/RP = 0.0833 + 0.167
1/RP = 0.25
RP = 4.0 Ω
9. What is the power
output of the circuit
shown above?
R S = R 1 + R 2 RS = 5 + 3
RS = 8 Ω
V = IR
12 = I x 8
I = 1.5 A
P = IV
P = 1.5 x 12
P = 18 W
10. What is the power
output of the circuit
shown above?
1/RP = 1/12 + 1/6
RP = 4.0 Ω
V = IR
8 = I x 4.0
I=2A
P = IV
P=2x8
P = 16 W
11. A potential
difference of 120 V
is placed across a
resistor with a current
of 0.50 A. What is the
resistance of this
resistor?
V = IR
120 = 0.5 x R
240 Ω = R
12. A 6 Ω resistor
has 7 A of
current. What is
the potential
difference?
V = IR
V=7x6
V = 42 V
13. Three resistors
of 5 Ω, 7 Ω, and
9 Ω are connected
in series. What is
their equivalent
resistance?
R S = R1 + R2 + R3
RS = 5 + 7 + 9
RS = 21 Ω
14. Two resistors, both
20 Ω, are connected in
parallel. What is the
equivalent resistance?
1/RP = 1/R1 + 1/R2
1/RP = 1/20 + 1/20
1/RP = 2/20 = 1/10
RP = 10 Ω
Hint: When two resistors of
the same resistance are
connected in parallel, the
equivalent resistance is ½
of one of the resistances.
In the previous example
with two 20 Ω resistors,
½ of 20 Ω is 10 Ω.
1/RP = 1/R1 + 1/R2 + 1/R3
1/RP = 1/3 + 1/4 + 1/5
1/RP = 0.33 + 0.25 + 0.2
1/RP = 0.78
RP = 1.28 Ω