Download Some Electricity Problems Answers

Some Electricity Problems Answers
1. Write the three rules governing voltage, current, and resistance
relationships in a series circuit containing a battery and three resistors.
(1)
V = V1 + V2 + V3
(2)
I = I 1 = I2 = I3
(3)
Req = R1 + R2 + R3
2. Write the three rules governing voltage, current, and resistance
relationships in a parallel circuit containing a battery and
three resistors.
(1)
V = V1 = V2 = V3
(2)
I = I 1 + I2 + I3
(3)
1
Req
=
1
R1
+
1
R2
+
1
R3
3. Draw a schematic diagram of a circuit containing a 6.0-V
battery hooked in series with three resistors of 2-Ω, 2-Ω,
and 4-Ω, respectively.
2Ω
2Ω
6V
4Ω
3. Draw a schematic diagram of a circuit containing a 6.0-V
battery hooked in series with three resistors of 2-Ω, 2-Ω,
and 4-Ω, respectively.
(a) Determine the equivalent resistance.
2Ω
2Ω
6V
6V
4Ω
Req = R1 + R2 + R3 = (2 Ω) + (2 Ω) + (4 Ω)
Req = 8 Ω
Req
3. Draw a schematic diagram of a circuit containing a 6.0-V
battery hooked in series with three resistors of 2-Ω, 2-Ω,
and 4-Ω, respectively.
(b) Determine the current I emerging from the battery.
2Ω
Use simpler circuit
IT
IT
6V
2Ω
6V
4Ω
V = IR
I = 0.75 A
I =
V
Req
=
6V
8Ω
8Ω
(c) Determine the voltage drops V1, V2, and V3.
Current is the same everywhere
V1
= (0.75 A)(2 Ω)
2Ω
0.75 A
6V
V1 = I1 R1 = I R1
V1 = 1.5 V
I1
0.75 A
I2
2Ω
I3
4Ω
V3
Note that V1 + V2 + V3 = 6 V
V2
V2 = I2 R2 = I R2
= (0.75 A)(2 Ω)
V2 = 1.5 V
V3 = I3 R3 = I R3
= (0.75 A)(4 Ω)
V3 = 3.0 V
4. Draw a schematic diagram of a circuit containing a 6.0-V
battery hooked in parallel with three resistors of resistances
2.0-Ω, 2.0-Ω, and 4.0-Ω, respectively.
6V
2Ω
2Ω
4Ω
4. Draw a schematic diagram of a circuit containing a 6.0-V
battery hooked in parallel with three resistors of resistances
2.0-Ω, 2.0-Ω, and 4.0-Ω, respectively.
(a) Determine the equivalent resistance Req.
2Ω
6V
1
Req
=
=
1
R1
2
4
+
+
Req = 0.80 Ω
2Ω
1
R2
2
4
+
4Ω
1
=
R3
+
1
4
6V
1
2
=
+
5
4
1
Req
+
2
=
1
4
1
Req
4. Draw a schematic diagram of a circuit containing a 6.0-V
battery hooked in parallel with three resistors of resistances
2.0-Ω, 2.0-Ω, and 4.0-Ω, respectively.
(b) Determine the current emerging from the battery.
IT
IT
6V
V = IR
2Ω
2Ω
I =
4Ω
V
Req
I = 7.5 A
=
6V
6V
0.80 Ω
0.80 Ω
(c) Determine the branch currents I1, I2, and I3.
IT
I1
6V
6V
I1 =
I2 =
I3 =
V1
R1
V2
R2
V3
R3
=
=
=
I2
6V
V
R1
V
R2
V
R3
=
=
=
I3
6V
6V
2Ω
6V
2Ω
6V
4Ω
Voltage is the same
everywhere
= I1 = 3.0 A
= I2 = 3.0 A
= I3 = 1.5 A
Note that I1 + I2 + I3 = 7.5 A, the total current from (b)
5. Consider the complex circuit below.
(a) How many resistors does each electron pass through on
its way through the circuit?
12 Ω
120 V
48 Ω 30 Ω
60 Ω
example of one way
through circuit
24 Ω
Each electron goes through 3 resistors
5. (b) Determine the equivalent resistance of the circuit.
12 Ω
120 V
48 Ω 30 Ω
12
60 Ω
R’
R’’
24 Ω
The 30-Ω and 60-Ω resistor are in parallel, and can be combined into
a single resistor, call it R’
Likewise, the 48-Ω and 24-Ω resistor are in parallel, and can be
combined into a single resistor, call it R’’
5. (b) Determine the equivalent resistance of the circuit.
12 Ω
12
48 Ω 30 Ω
120 V
60 Ω
R’
R’’
24 Ω
1
R’
1
R’’
=
=
1
30
1
24
+
+
1
60
1
48
=
=
2
60
2
48
+
+
1
60
1
48
=
=
3
60
3
48
R’ = 20 Ω
R’’ = 16 Ω
5. (b) Determine the equivalent resistance of the circuit.
12 Ω
120 V
48 Ω 30 Ω
12
60 Ω
20
16
24 Ω
Resistors is series add
Req = 12 + 20 + 16 Ω
Req = 48 Ω
120 V
Req
5. (c) Determine the current emerging from the power source.
12 Ω
12
I
48 Ω 30 Ω
120 V
60 Ω
20
16
24 Ω
Use simpler circuit that has only one resistor
I =
V
Req
I = 2.5 A
=
120 V
48 Ω
I
120 V
48 Ω
5. (d) Determine V1 and I2.
12 Ω
2.5 A
120 V
12
V1
48 Ω 30 Ω
I2
2.5 A
60 Ω
20
16
24 Ω
V1 = I1 R1
I1 = ?
Every electron that comes out of
battery goes through R1 (see part
(a)), so I1 = 2.5 A
V1 = I1 R1 = (2.5 A)(12 Ω)
V1 = 30 V
2.5 A
120 V
48 Ω
5. (d) Determine V1 and I2.
12 Ω
12
I2
2.5 A
48 Ω 30 Ω
120 V
2.5 A
V2
20 V2
16
24 Ω
I2 =
V2
R2
50 V
=
30 Ω
I2 = 1.7 A
V2 = ?
Use first equivalent circuit
that is in series
V2 = I R’
= (2.5 A)(20 Ω)
= 50 V
6. Consider the complex circuit below.
(a) How many resistors does each electron go through
on its way around this circuit?
10 Ω
20 Ω
12 Ω
10 Ω
24 V
60 Ω
20 Ω
An example of one
way around this
circuit
40 Ω
Each electron goes through four resistors
6. Consider the complex circuit below.
(b) Determine the equivalent resistance of the circuit.
10 Ω
20 Ω
12 Ω
R’
12
10 Ω
24 V
60 Ω
40 Ω
20
20 Ω
R’’
The 10-Ω, 20-Ω, and 10-Ω resistors are in parallel, and can be
combined into a single resistor, call it R’
Likewise, the 60-Ω and 40-Ω resistor are in parallel, and can be
combined into a single resistor, call it R’’
6. Consider the complex circuit below.
(b) Determine the equivalent resistance of the circuit.
10 Ω
20 Ω
R’
12 Ω
12
10 Ω
24 V
20
20 Ω
60 Ω
40 Ω
1
R’
=
1
R’’
1
10
=
+
1
60
1
20
+
R’’
+
1
40
1
10
=
=
2+1+2
20
2 + 3
120
=
=
5
20
5
120
R’ = 4 Ω
R’’ = 24 Ω
6. Consider the complex circuit below.
(b) Determine the equivalent resistance of the circuit.
10 Ω
20 Ω
4
12 Ω
12
10 Ω
24 V
60 Ω
20
20 Ω
40 Ω
24
Req = 4 + 12 + 20 + 24 Ω
Req = 60 Ω
24 V
Req
(b) Determine the current emerging from the power source.
10 Ω
20 Ω
I
4
12 Ω
12
10 Ω
24 V
20
20 Ω
60 Ω
40 Ω
I =
V
Req
24
=
I = 0.40 A
24 V
60 Ω
I
24 V
60 Ω
(c) Determine V1 and I2.
10 Ω
V1
20 Ω
0.40 A
24 V
4
12 Ω
12
10 Ω
60 Ω
20
20 Ω
I2
40 Ω
V1 = I1 R1
24
I1 = ?
Every electron that comes out of
battery goes through R1 (see part
(a)), so I1 = 0.40 A
V1 = I1 R1 = (0.40 A)(12 Ω)
V1 = 4.8 V
I
24 V
60 Ω
(c) Determine V1 and I2.
10 Ω
V1
20 Ω
12 Ω
4
12
10 Ω
0.40 A
24 V
60 Ω
I2
40 Ω
24
V2
V2
I2 =
V2
R2
9.6 V
=
60 Ω
I2 = 0.16 A
20
20 Ω
V2 = ?
Use first equivalent circuit
that is in series
V2 = I R’
= (0.40 A)(24 Ω)
= 9.6 V