Chapter 18
Electric Current and Resistance.
Home Work Solutions
18.1
Problem 18.33
Two identical 100-Ω resistors are joined together as shown in Figure (18.33).
(a) What is the current through each when switch S is open?
(b) What is it when S is closed?
S
9.0 V
100 Ω
100 Ω
Figure 18.33:
Solution
• Given: The circuit shown in Figure (18.33)
• Required: The current through each resistor when the switch is open and when it is
closed.
2
CHAPTER 18. ELECTRIC CURRENT AND RESISTANCE. HOME WORK
SOLUTIONS
(a) when the switch is open, the 100-Ω resistor to the right of the switch is not included
in the circuit and the current I2 through it is zero. The current in the circuit is then
determined by one 100-Ω resistor and the 9.0 V battery. Using Ohm’s law we can get
the current through the resistor to the left of the switch:
V
R
9.0
=
100
= 0.09 A
= 90.0 mA
I1 =
(b) When the switch is closed the two 100Ω resistors are included in the circuit. The
two resistors are connected in parallel and can be replaced by their parallel equivalent,
Rp = R1 R2 /(R1 + R2 ). When the two resistors are equal (R1 = R2 = R), their parallel
equivalent Rp = 12 R = 50Ω. Now using Ohm’s law we can get the current Itotal that is
produced by the battery:
V
Rp
9.0
=
50
= 0.18 A
= 180 mA
Itotal =
Since the two resistors and the battery are connected in parallel, the potential differences
across them are equal. So the currents through the resistors are:
I1 = I2
V
=
R
9.0
=
100
= 0.09 A
= 90.0 mA
18.2. PROBLEM 18.50
18.2
3
Problem 18.50
Calculate the effective capacitance of the net work shown in Figure (18.43). C1 = C2 = C5 =
C6 = 20µF , and C3 = C4 = 5µF .
C1
C3
C2
C4
C6
C5
Figure 18.43:
Solution
• Given: The circuit in Figure (18.43).
• Required: The effective capacitance Ce .
C1 and C2 are connected in parallel, and so are C3 and C4 , we can then replace each pair with
its parallel equivalent; Cp1 and Cp2 respectively. Now, Cp1 , Cp2 , C5 , and C6 are connected in
4
CHAPTER 18. ELECTRIC CURRENT AND RESISTANCE. HOME WORK
SOLUTIONS
series, we then have:
Cp1 =
=
=
Cp2 =
=
1
=
Ce
Ce
C1 + C2
20 + 20
40 µF
5+5
10 µF
1
1
1
1
+
+
+
Cp1 Cp2 C5 C6
1
1
1
1
=
+
+
+
40 10 20 20
= 0.225
= 4.44 µF
Where Ce is the equivalent capacitance of the network shown in Figure (18.43).
18.3. PROBLEM 18.71
18.3
5
Problem 18.71
Find the equivalent resistance of the network shown in Figure (18.49). R1 = 6.8Ω, R2 =
R3 = 5.6Ω, R4 = 2.2Ω, R5 = R6 = R7 = R8 = 10Ω, and R9 = 1.8Ω
R2
R1
R3
R4
R8
R7
R9
R5
R6
Figure 18.49:
Solution
• Given: The circuit shown
• Required: The equivalent resistance, Re .
R10
R1
R4
R12
R9
R11
6
CHAPTER 18. ELECTRIC CURRENT AND RESISTANCE. HOME WORK
SOLUTIONS
R2 and R3 are connected in parallel, R5 and R6 are connected in series, and R7 and R8
are connected in series. Let the equivalent resistance of each pair be R10 , R11 , and R12 ,
respectively. The circuit then reduces to that shown in the Figure above. R11 and R12 are
connected in parallel, let their parallel equivalent be R13 . The circuit now reduces to that
shown in the Figure below:
R10
R1
R4
R9
R13
Now, R1 , R10 , and R4 are connected in series. R9 and R13 are also connected in series. Let
R14 to be the series equivalent of R1 , R10 , and R4 . Let R15 be the series equivalent of R9
and R13 , so the circuit now reduces to:
R14
R15
18.3. PROBLEM 18.71
7
R14 and R15 are connected in parallel, their parallel equivalent Re is the equivalent resistance
of the whole network. We can now calculate:
R10 =
=
R11
R12
=
=
=
=
=
R13 =
=
R14
R15
=
=
=
=
=
Re =
=
=
R2 R3
R2 + R3
1
× 5.6
2
2.8 Ω
R5 + R6
20 Ω
R7 + R8
20 Ω
R11 R1 2
R11 + R12
1
× 20
2
10 Ω
R1 + R10 + R4
11.8 Ω
R9 + R13
11.8 Ω
R14 R15
R14 + R15
1
× 11.8
2
5.9 Ω
CHAPTER 18. ELECTRIC CURRENT AND RESISTANCE. HOME WORK
SOLUTIONS
8
18.4
Problem 18.78
Find the currents I1 , I2 and I3 in the network of Figure (18.54).
1.5 V
a
47 Ω
I1
1.5 V
b
I2
c
I3
10 Ω
15 Ω
150 Ω
3.0 V
f
e
d
Figure 18.54:
Solution
• Given: The circuit shown in Figure (18.54).
• Required: The currents I1 , I2 , and I3 .
Applying Kircchoff’s current rule at junction “b” or “e” we get:
I2 = I1 + I3
(18.1)
We now apply Kircchoff’s voltage rule to loop “ abefa”:
0 = 1.5 − 47I1 + 10I3 − 3 − 15I1
= −62I1 + 10I3 − 1.5
(18.2)
18.4. PROBLEM 18.78
9
Applying Kircchoff’s voltage rule to loop “acdfa” we get:
0 = 1.5 − 47I1 + 1.5 − 150I2 − 3 − 15I1
= −62I1 − 150I2 + 3.0
(18.3)
Equations 18.1, 18.2, and 18.3 contains the three unknowns: I1 , I2 , and I3 . We now solve
these equations. Substituting for I2 from 18.1 into 18.3 we get:
0 = −62I1 − 150(I1 + I3 ) + 3.0
= −62I1 − 150I1 − 150I3 + 3.0
= −212I1 − 150I3 + 3.0
(18.4)
Multiplying Equation 18.2 by 15 we get:
0 = −15 × 62I1 + 15 × 10I3 − 15 × 1.5
= −930I1 + 150I3 − 22.5
(18.5)
Adding Equation 18.4 to Equation 18.5 we get:
0 = −1142I1 − 19.5
19.5
I1 = −
1142
= −17.1 mA
(18.6)
Using equation 18.6 in Equation 18.2 we get:
62 × 19.5
+ 10I3 − 1.5
1142
= 44.1 mA
0 =
I3
(18.7)
Substituting Equation 18.6 and Equation 18.7 into Equation 18.1 we get:
I2 = 44.1 − 17.1
= 27.0 mA
(18.8)
Equation 18.6 shows that the direction of I1 assumed in Figure (18.54) is wrong. the correct
direction of I1 is going out of junction “b”.