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...

1 answer below »

8. The voltage drop across resistor R3 in Fig. 68 is
(1) 0.5 V. (3) I V. (5) 2.5 V. (2) 0.2 V. -4--(4) 2 V. (6) 3 V.
9. The power P4 dissipated in resistor R4 of Fig. 68 is
(1) 0.125 W. -4-(3) 1.125 W. (5) 2.0 W. (2) 0.25 W. (4) 1.50 W. (6) 2.5 W.
10. The total power delivered by the battery in Fig. 68 to the resistors is
(1) 1.4 W. " (3) 3.6 W. (5) 4.8 W. 1(2) 2.8 W. (4) 4.2 W. (6) 5.2 W.
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 ) 15001:2 (3) 4500 SI (2) 2 kit (4) 5 VI
P. The power dissipated by resistor R, in Fig. 69 is (ic (1) 450pW. (3) 0.05 W. (2) 450Af. (4) 0.055 W.
, it is necessary to .5 V negative with
(5) 10 kc2 (6) 15 kf.2
0) (5) 0.00575 W. (6) 0.075 W.
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)
141) 0.1 A. (2) 0.2 A.
(3) 0.5 A. (5) 2 A. (4) 1 A. (6) 3 A.
14. The voltage across resistor R5 in Fig. 70 is (1) —5 V. (3) 0 V. (2) — I V. (4) 1 V.




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
30001-1 is (Hint: Topic 2 App. Ex. #4) (I) 30005.-?.. (3) 70002. (5) 12,000 2. (2) 5000 2. (4) 10,000 Ct. (6) 14,000 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 (1) 0.101 n (3). 0.404 SI (2) 0.25 f2 (4) 0.505 SI
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.)
(1) 1333 S2. (3) 4000 a (2) 2667 n. (4) 6666 a
23. The value of resistor R., in Fig. 74 is
(1) 3167 a (3) 3750 n. (5) 9500 n. (2) 4750 a (4) 7250 a (6) 38,600 n.




24. The value of resistor R, in Fig. 74 is
(1) 1375 0. (3) 2750 CI. (5) 6870 C2. (2) 1833 0. (4) 5500 fl. (6) 7250 0. 25. The value of resistor R5 in Fig. 74 is (1) 1375 a (3) 2750 0. (5) 6870 a (2) 1833 0. (4) 3666 a (6) 7250 a
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. (3) 0.036W. (5) 1.8W. (2) 0.027 W. (4) 0.72 W. (6) 3.6 W.
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)
(1) 0.0 S2 (5) 1.0 (2) 0.5 S2 (6) 1.70 (3) 0.662 (4) 0.785




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. (2) 2 A.
(3) 3 A. (4) 6 A.
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.

Answered Same DayDec 20, 2021

Answer To: 8. The voltage drop across resistor R3 in Fig. 68 is XXXXXXXXXXV. (3) I V XXXXXXXXXXV XXXXXXXXXXV...

Robert answered on Dec 20 2021
119 Votes
Solution 11:
Here we have a self bias configuration:



Hence part 4
Solution 12:

Power dissipated is:


( )
Hence part 1
Solution 14:
Solution 21:
To increase the range of the milli ammeter we place a shunt resistance across it.
The resistance placed in parallel carries the remaining current. Hence the current carried by resistance
is:



Hence option 4
Solution 22:
Solving loop containing DS of FET1 and GS of FET2:
To calculate the current:
Total current must be: 6mA


Hence option 3
Solution 23:
Current across R3 is:




Current in resistance...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here