8. The voltage drop across resistor R3 in Fig. 68 is
XXXXXXXXXXV. (3) I V XXXXXXXXXXV XXXXXXXXXXV XXXXXXXXXXV XXXXXXXXXXV.
9. The power P4 dissipated in resistor R4 of Fig. 68 is
XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW.
10. The total power delivered by the battery in Fig. 68 to the resistors is
XXXXXXXXXXW. " XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW.
11. In the design of the FET amplifier circuit of Fig. 69 select a value for resistor R, that will bias the gate I respect to the source. Find the value of R,. IG = 0 ( I XXXXXXXXXX: XXXXXXXXXXSI (2) 2 kit (4) 5 VI
P. The power dissipated by resistor R, in Fig. 69 is (ic (1) 450pW XXXXXXXXXXW. (2) 450Af XXXXXXXXXXW.
, it is necessary to .5 V negative with
(5) 10 kc XXXXXXXXXXkf.2
XXXXXXXXXXW XXXXXXXXXXW.
13. Figure 70 shows a bridge circuit such as might be encountered in the design of a force-measurement system. The current drawn from the battery is (Hint: Fig. 49)
XXXXXXXXXXA XXXXXXXXXXA.
XXXXXXXXXXA XXXXXXXXXXA XXXXXXXXXXA XXXXXXXXXXA.
14. The voltage across resistor R5 in Fig. 70 is (1) —5 V XXXXXXXXXXV. (2) — I V XXXXXXXXXXV.
20. The source resistor in Fig. 72 has a value of 4000 S2. As an experi-ment, we wish to reduce the resistance between the source and ground to 3000E2 by connecting another resistor, R1f in parallel with resistor R. The value of resistor Rs that will reduce the total resistance to
XXXXXXXXXXis (Hint: Topic 2 App. Ex. #4) (I) 30005.-? XXXXXXXXXX, XXXXXXXXXX4) 10,000 Ct XXXXXXXXXX,000 2.
21. Figure 73 shows a milliammeter with a shunt resistor Rs connected to increase its full-scale range to 0.5 A. The value of the shunt resistor is XXXXXXXXXXn XXXXXXXXXXSI XXXXXXXXXXf XXXXXXXXXXSI
22. In the design of a direct-coupled FET amplifier such as that shown in Fig. 74 it is necessary to calculate values of resistors that will provide proper operating potentials for the FET's. In Fig. 74, R2 is (Hint: First solve the loop containing D and S of FET,, G and S of FET: and R2.)
XXXXXXXXXXS XXXXXXXXXXa XXXXXXXXXXn XXXXXXXXXXa
23. The value of resistor R., in Fig. 74 is
XXXXXXXXXXa XXXXXXXXXXn XXXXXXXXXXn XXXXXXXXXXa XXXXXXXXXXa (6) 38,600 n.
24. The value of resistor R, in Fig. 74 is
XXXXXXXXXX CI XXXXXXXXXXC XXXXXXXXXX5500 fl XXXXXXXXXXThe value of resistor R5 in Fig. 74 is XXXXXXXXXXa XXXXXXXXXX6870 a XXXXXXXXXX3666 a XXXXXXXXXXa
26. The resistor in the circuit of Fig. 74 that dissipates the most power is (1) R2. (3) R4. (5) R6. (2) R3. (4) R5. (6) R7.
27. The power dissipated in FET2 in Fig. 74 is
(1) 0.018W XXXXXXXXXX036W XXXXXXXXXX8W XXXXXXXXXXW XXXXXXXXXXW XXXXXXXXXXW.
28. 'What resistance would be measured between terminals a and b of the circuit show-n in Fig. 75? (Hint: Convert the 2 T networks to 2 n: networks)
XXXXXXXXXXS XXXXXXXXXXS XXXXXXXXXX XXXXXXXXXX
29. Refer to Fig. 76. Some of the voltages an$ currents are indicated on the diagram, and also the conventional current direction through some of the elements. Select from the list below the law or laws that will enable you to find the current through R. If more than one selection is cor-rect, pick that one which lists the minimum number of laws necessary. (1) Ohm's law (2) Kirchhoff's voltage law (3) Kirchhoff's current law (4) Kirchhoff's voltage law and Ohm's law (5) Ohm's law, Kirchhoff's voltage law, and Kirchhoff's current law.
R,
6
24 V 3 A
Fig. 76
30. The current through resistor R5 of Fig. 76 is
(1) 1 A XXXXXXXXXXA.
(3) 3 A XXXXXXXXXXA.
5 A--=2 v R7
4 n
(5) 7 A. (6) None of the above.
31. The direction of current through resistor Rs of Fig. 76 is (1) from A to D. (2) from D to A. (3) There isn't enough information given to answer this question.