1. Simplify this circuit to find the total power absorbed by all resistors. R1 R3 47

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1 Hoework 1. iplify this iruit to find the total power asored y all resistors. R1 R 6 R R 47 Ans: Use series and parallel oinations to siplify iruit. R1R= =94Ω and this resistane is in parallel with R4: 940 R eq (94 0) This is in series with R1 so the total resistane aross the 6 soure is 48.45=58.45Ω. PTOTAL W iplify this iruit to find what fration of s appears aross the 180 oh resistor as o. R1 R s 47 R 47 R6 180 o 470 R4 R5 470 Ans: The iruit an e analyzed as a voltage divider with upper part fored y the su of R1 and R in series = R a = 47=80Ω. The other leg is fored y the oination of R, R4, R5 and R6, whih an e done as follows: R4 and R5 in parallel: 470*470/(470470)=470/ = 5Ω. This resistane is in series with R, so they add: 547=8Ω. This resistane is in parallel with R6, so the equivalent resistane for the lower leg of the divider, R =8*180/(8180)=9.9Ω Now, R a and R for a voltage divider. The fration of s aross R and, hene, aross R6 (sine it is the resistane etween the sae two nodes as R6), is: 9.9/(9.9 80)=0.579.

2 . You have a urrent sensor that has an internal resistane of Ω. It an handle a axiu urrent of 1A, ut you want to easure urrents up to 0A. What value of Rs would allow this? What is the axiu power it ay have to asor. (By the way, this is often done and Rs is alled a shunt.) Rs 0A ENOR Ans: Use urrent division: The axiu urrent through the sensor is: I ENOR R 0 A ( R ) olving for Rs when I ENOR = 1A: 0( R ) 1 ( R ) R 0R 99R 0.1R If 1A is flowing through the sensor, then 99A ust e flowing through Rs when the total urrent is 0A. P I R R R 99 (0.1) W Note: The shunt would get ERY hot. 4. You have a voltage sensor that easures up to, ut you need to easure the voltage etween a wire and ground that an reah up to (1K). The sensor has an internal resistane of Ω. What value of R would you use? What is the axiu power it ay have to asor? (Again, this is often done. R is referred to as a ultiplier.) R ENOR 0 Ans: This is a voltage divider prole.

3 R.01 R.01R.01R 990 R or 99K.01 P R R 9.9W R This resistor would also get soewhat hot. 5. You need to redue the in the prole aove to 0 for a la experient. What iruit would you use and with what values? Whatever iruit you use ay not asor ore than 1 watt. Ans: This alls for a voltage divider of 0.1, ut also the onstraint that the total resistane an asor no ore than 1W. We will use this onstraint first to alulate the total resistane of the divider, R T. Reeer the entire appears aross this resistane. P divider R T 1W R T R T or 1egoh (1 M ) 1 This gives us the total resistane for the forula for the voltage divider: R R divider ( Ra R RT R 0 1 M.1(1 M) R 00 or 0K This eans R a is 00000=900000Ω or 900KΩ

4 6. In the la, you have ore proles. The experient alls for a 50 oh resistor in soe part of the iruit, ut all you have are oxes of 470 and 0 oh resistors. How an you ake a 50 oh resistor with a oination of these values? Ans: Try two 0 oh resistors in parallel (50 ohs) and that in series with a 470: 47050= Below is a sheati of a iruit soeone else designed. The designer got sik efore finishing his design (It happens!). How any watts does R4 asor? Resistors oe in sizes ale to asor 1/8, 1/4, and 1/ watt. What size would you use? The next day they tell you s is now 5. Can you tell the how uh R4 has to asor without doing all the nodal analysis again? How uh ust it asor? Can you use the sae size resistor(in wattage)as efore? If not, what size? R1 R. 4.7 s = R R4 5 Ans: Using nodal analysis with the nodes arked as shown: R1 a R. 4.7 s = R R4 5 The node equations are: a a

5 Whih yield =.959. This eans R4 will asor /5 watts or 0.171W. A 1/4watt resistor should handle this niely. When they hange the speifiation to a voltage soure of 5, you an use the property of linearity to deterine the new value of : (1.667) 1.54 This leads to a new power for R4 of: (1.54) /5=0.476W. The resistor would have to e hanged to at least a 1/watt. However, good engineering pratie ditates it should e twie that or 1W. 8. This is the iruit odel of an atual transistor aplifier (really!). The gain of the aplifier is o s. What is the gain of the aplifier? Note the diretion of I and the dependent urrent soure. I 0 I s o Ans: Using nodal analysis and naing the enter node voltage : I 0I 1I 0 I ustituting this into the first equation to get I :

6 0.91 1I I E I 5 And this goes into the equation for o in ters of the dependent urrent soure: 0 I (4700) O 4.5 o the gain of the iruit is 4.5. The inus sign indiates it will invert the signal. 9. What is in this iruit? A Ans: Using nodal analysis and the nodes as arked: 0.5 a 50 A We an write the following node equations:

7 A a Noting that a and for a super node. The voltage soure gives us one ore piee of inforation: a 0.5 This an e sustituted into the first equation to eliinate a: 0.5 A whih gives: (.01) This an e sustituted into the seond equation to give: Notie what is happening. We will have a strange expression: This an ONLY e true if =0.. What is o? 1A 5 1 o

8 Ans: The one volt voltage soure deterines exlusively what happens to the right of it. This akes the two right resistors a voltage divider and o is given y: O Notie the signs. 11. This is the odel for a iruit using a FET. What is o in ters of s? 1K s K g.1g K o Ans: The voltage soure, s, and the two resistors should e failiar y now as a voltage divider iruit. g in ters of s is: 000 g ( 000) Notie the sign, sine s is upside down. This akes the output dependent urrent soure:. Noting the diretion of the dependent urrent soure and the polarity indiated for o:. O 0( ) A very high gain iruit.

9 1. What is o? 1 A 4 o Ans: Using nodal analysis on the three nodes fro left to right a, and, the nodal equations are: a a 1 a 1 4 These siplify into the syste of three equations: a a a Using Craer s Rule to solve for, whih is also o, gives: o =1.01