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resistance?
10. The total resistance in Figure 4-68 is 20 KN. What is the value of Rg?
FIGURE 4-68
RI
R2
WW
4.7 kΩ 1.0 ΚΩ
R; Ž
2.2 kΩ
Rg RA
0W W
3.9 k 2
11. Determine the resistance between each of the following sets of pins on the PC board in
Figure 4-65.
(a) pin 1 and pin 8 (b) pin 2 and pin 3
(c) pin 4 and pin 7 (d) pin 5 and pin 6
12. If all the resistors in Figure 4-65 are connected in series, what is the total resistance?
SECTION 4-3
Current in a Series Circuit
13. What is the current through each of four resistors in a series circuit if the source voltage is 12 V
and the total resistance is 120 12?
14. The current from the source in Figure 4-69 is 5 mA. How much current does each milliamme-
ter in the circuit indicate?
FIGURE 4-69
+
R1
mA
mA
+
+
R2
w
Vs =
-
mA
+
w
R3
SECTION 4-4 Application of Ohm's Law
15. What is the current in each circuit of Figure 470? Show how to connect an ammeter in each case.
R
2.2 kΩ
1.0 M.2
RI
+
+
R2
5.5 =
R2
5.6 ΚΩ
16V =
w
2.2 ΜΩ
R3
1.0 k 2
560 k2
R2
(a)
(b)
16. Determine the voltage across each resistor in Figure 4_70.
SECTION 4-6 Kirchhoff's Voltage Law
21. The following voltage drops are measured across each of three resistors in series: 5.5 V, 8.2 V,
and 12.3 V. What is the value of the source voltage to which these resistors are connected?
22. Five resistors are in series with a 20 V source. The voltage drops across four of the resistors are
1.5 V, 5.5 V, 3 V, and 6 V. How much voltage is across the fifth resistor?
23. Determine the unspecified voltage drop(s) in each circuit of Figure 4–71. Show how to connect
a voltmeter to measure each unknown voltage drop.
2V
V2
3.2 V
+
W w
8V
M
R
w
R
+
15 V
v
Holi
1V
Vs = 특
w
2R
0.5 V
+
1.5 V
4R
3R
w
(a)
(b)
SECTION 4-7 Voltage Dividers
24. The total resistance of a series circuit is 500 12. What percentage of the total voltage appears
across a 22 12 resistor in the series circuit?
25. Find the voltage between A and B in each voltage divider of Figure 4–72.
FIGURE 4-72
1.0 kΩ
100 Ω
+
+
OA
12 v =
-OA
8v
Hole
2.2 kΩ
47 22
ww
3.3 k 12
OB
OB
101 =
(a)
(b)
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