5) The source voltage in Figure 6-67 is 100V. How much voltage does each of the meters read?
Each meter will read 100V because the resistors are in parallel.
11) In the circuit of Figure 6-69, determine the resistance R2, R3, R4
Es
R1
47ohm
R2 R3 R4
5.03mA
1mA 2.14mA
470uA
Figure 6-69
HINT:
It is sometimes helpful to build a chart listing all the circuit givens in a problem. It is easier to see what information that is available to help solve the circuit’s unknowns. Do this starting out and it will help you see what to look for so you can use Ohm’s and Watt’s Laws to solve the given unknowns.
Resistance
R
T
=
Component
Es
R1
R2
Voltage Current
5.03mA
1mA
2.14mA
47Ω
Power
R3
R4
470µA
We need to determine Es for the parallel combination of Figure 6-69. Looking at the stated values in the problem you should see that the values of R
1 we can use Ohm’s Law to solve for V
R1 because they are in parallel.
Es
=
V
R 1
=
I
R 1
×
R
1 and I
R1
are given in this problem, so
, which is the same voltage as Es, V
R2
, V
R3
, and V
R4
VR 1 1 mA 47 47 mV
Now add these values to the chart as more knowns.

Component
Es
R1
R2
Voltage
47mV
47mV
47mV
Current
5.03mA
1mA
2.14mA
Resistance
R
T
=
47Ω
Power
R3
R4
47mV
47mV
470µA
It’s now easy to see how to use Ohm’s Law to solve for the rest of the problems unknowns.
R
2
=
V
I
R 2
⇒
R 2
47 mV
2.14
mA
⇒
21.96
Ω
I
R
3
I
R 4
=
I
T
−
I
R 1
−
I
R 2
−
I
R 3
R 4
=
V
I
R 3
⇒
R 3
=
5.03
mA
47
470
− mV
1
µ
A mA
⇒
−
100
Ω
2.14
mA
−
470
µ
A
⇒
1.42
mA
R
4
=
V
I
R 4
⇒
R 4
47 mV
1.42
mA
⇒
33.10
Ω
Don’t forget to add these values to the chart.
15) Find the total resistance for each of the following groups of parallel resistors:
1
=
1
+
1
R R R
T 1 2
R
T
=
R R
2
R
1
+
R
2
(a)
(b)
560Ω, 1kΩ = 358.97Ω
47Ω, 56Ω = 25.55Ω
(c) 1.5kΩ, 2.2kΩ, 10kΩ = 818.86Ω
(d) 1MΩ, 470kΩ, 1kΩ, 2.7MΩ = 996.51 Ω
20) If the total resistance in Figure 6-71 is 389.2Ω, what is the value or R
2
?
R1
680ohm
R2
1
=
1
−
1
R 2 RT R 1
Figure 6-71
R
1
=
1
Ω
−
1
2 389.2
680
Ω
R 2
=
910.1
Ω
⇒
1.098785
mS

21 What is the total resistance between point A and ground in Figure 6-72 for the following conditions?
SW1 SW2
R1
510kohm
R2
470kohm
R3
910kohm
(a) SW1 and SW2 open = 510kΩ
(b) SW1 closed, SW2 open = 244.59kΩ
(c) SW1 open, SW2 closed = 510kΩ
(d) SW1 and SW2 closed = 192.78kΩ
Figure 6-72
(a) R1 will be the only resistor seen by the voltage source.
(b) R1 and R2 will be connected in parallel as the load seen by the voltage source a.
Use the formula
1
𝑅𝑇
=
1
𝑅1
+
1
𝑅2
to solve for the total resistance
(c) R1, R2, and R3 will be connected in parallel as the load seen by the voltage sources a.
Use the formula
1
𝑅𝑇
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
to solve for the total resistance

22) What is the total current in each circuit of Figure 6-73?
Es
10V
R1
33ohm
R2
33ohm
R3
27ohm
Es
25V
R1
1kohm
R2
4.7kohm
R3
560ohm
Figure 6-73(a)
R
T
I
T
=
10.24
Ω
=
Es
⇒
R
T
10 V
10.24
Ω
⇒
976.56
mA
Figure 6-73(b)
R
T
=
333.50
Ω
I
T
=
Es
⇒
R
T
25 V
333.50
Ω
⇒
74.96
mA

26) Find the values of the unspecified labeled quantities in each circuit of Figure 6-74.
500mA
R3
10V
100V
R1
1kohm
R2
680ohm
R1 R2
I1 I2 I3
I
R 1
=
I
T
−
( I
R 2
+
I
Figure 6-74(b)
R 3
) I
R 2
I
R 2
Figure 6-74(a)
=
I
T
−
I
R 1
=
150 mA
−
100 mA
⇒
50 mA
R
1
=
10 V
100 mA
=
100
Ω
R
2
=
10 V
50 mA
=
200
Ω
I
R 2
=
V
R 2
R
2
⇒
100 V
680
Ω
=
147.06
mA
I
R 3
I
R 1
=
V
R 3
⇒
R
3
=
500 mA
100
−
1 k
V
Ω
=
100
(147.06
mA mA
+
100 mA )
=
252.94
mA
R
1
=
100 V
=
395.35
Ω
252.94
mA
Using Current Division Rule
I
R 3
=
V
R 3
=
100 mA
R
3
I
R 2
=
R
3
×
I
R 3
⇒
R
2
1 k
Ω
680
Ω
×
100 mA
=
147.06
mA
I
R 1
= found as above
29) Find the values of the unspecified quantities in Figure 6-77 on page 219 of the textbook.

I
T
250mA
50m
A
I
2
I
3
Figure 6-77
I
R 2
+
I
R 3
=
250 mA
I
R 3
I
R 2
=
V
R 3
⇒
R
3
=
250 mA
100
1.2
− k
V
Ω
=
83.33
83.33
mA
= mA
166.67
I
T
=
I
R 1
+
I
R 2
+
I
R 3
⇒
300 mA mA
R
1
=
100 V
50 mA
=
2 k
Ω
R
2
=
100 V
=
599.99
Ω
166.67
mA
30) Determine the current through R
L
in each of the circuits of Figure 6-78.
(a) 3mA (I
1
(b) 10μA (I
1
+ I
2
– I
2
= I
L
= I
)
L
)
(c) 0.5A = 500mA (I
1
+ I
3
– I
2
= I
L
38) Five parallel resistors each handle 250mW. What is the total power?
P
T
=5x250mW = 1.25W