Week 5 Assignment Gough - Copy

docx

School

Grantham University *

*We aren’t endorsed by this school

Course

192

Subject

Electrical Engineering

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by hemi300

Kaila Gough G00215407 Grantham University Week 5 Assignment 10/12/2023

Chapter7 2. Visualize and draw the following series-parallel circuits: a. A parallel combination of three branches, each containing two series resistors b. A series combination of three parallel circuits, each containing two resistors 4. For each circuit in Figure 7-62, identify the series and parallel relationships of the resistors viewed from the source. a. V1 is in series with R2 and R4, R1 is in parallel with R2, R2 is also in parallel with R3. b. 7-62-1 R1 is in series with R4, R1 and R4 are in parallel with R2 and R3, R2 and R3 are in series. The V source is in series with all resistors in the circuit. c. R1 is in series with R2 R4 R8 R7 and R5, although R2 is also in parallel with R3 and R6 as well as the V source.

8. A certain circuit is composed of two parallel resistors. The total resistance is 667Ω. One of the resistors is 1.0kΩ. What is the other resistor? 1000 x 667/1000-667 = 2001ohm 10. Repeat problem 9 for each circuit in Figure 7-62. 9.5 and 2.0 a. 9.5 x 2.0/9.5+2.0 = 1.72Mohms b. 2.0 x 9.5 / 2.0 + 9.5 = 1.67Mohms c. Adding up all of the series resistors we get, 18.6kohm for the parallel resistors we get 4k ohms. Adding both of those values gives us a total resistance value of 22.6kohm. 12. Determine the current through each resistor in each circuit in Figure 7-62, then calculate each voltage drop. IR1 = 0.76mA I= I x R2R3R4/Re(R1+R2+R3+R4) iR2 = 0.35mA I = 3 x 10^-3 x 1x3.3 x 6.2/0.333 x (1 + 2.2 + 3.3 + 6.2) = 0.35mA IR3 = 0.24mA 3 x 10^-3 x 1x2.2x6.2/0.333 x (1+ 2.2 + 3.3 + 6.2) = 0.24mA IR4 = 0.12mA 3 x 10^-3 x 1 x 2.2 x 3.3 / 0.333 x (1 +2.2 +3.3 +6.2) =0.12mA

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

14. Determine the resistance between A and B in Figure 7-66 with the source removed. 100/2.1 = 4.762 Ω 16. Determine the voltage at each node with respect to ground in Figure 7-67. A. 50/560 = 0.0892 B. 50/100 = 0.5 C. 50/1000 = 0.05 D. 50/100 = 0.5 0.5 – 0.0892/560k = 0.0007335 + 0.5 – 0.05 / 100k = 0.0045 + 0.05 – 0.5 / 1000k = 0.00045 = 0 0.0007335 + - 0.0045 + 0.00045 = 0 0.0047835 = 0 47.8v Not so sure about this question. 18. Determine the resistance of the circuit in Figure 7-67 as seen from the voltage source. 56k + 560k + 100k + 1000k + 100k + 1000k + 1/1000 = 2816.001 560.00k Ω 20.Determine the voltage, V AB, in Figure 7-68. 4.1v 24. Determine the value of each resistor in Figure 7-72. R1 = 10 Ω R2 = 10 Ω R3 = 20 Ω R4 = 20 Ω R5 = 5 Ω R6 = 0.5 Ω

26. A 12V battery output is divided down to obtain two output voltages. Three 3.3kΩ reisstors are used to provide the two taps. Determine the output voltages. If a 10kΩ load is connected to the higher of the two outputs, what will its loaded value be? Vout = Vin R2/R1 + R2 = 4v Vout = 12 33.3 + 33.3/33.3 +33.3 +33..3 = 8v Req = 33.3 x 10/ 33.3+10 = 7.5kΩ Vout2L = 12 33.3 + 7.5 / 33.3 + 33.3 + 7.5 = 6.67v 28. In Figure 7-73, determine the output voltage with no load across the output terminals. With a 100kΩ load connected from A to B, what is the output voltage? Vout = (22 x 100k) / (10k + 100k) Vout = 2200000/110000 Vout = 20V The output voltage across the 100k ohm load is 20v. 30. In Figure 7-73, determine the continuous current drawn from the source with no load across the output terminals. With a 33kΩ load, what is the current drain? 1/5.6 + ½.7 + 1/33 1/Req = 0.1786 +0.3704 + 0.0303 = 0.5793 Req = 1.726kΩ Rtotal = R1 + Req Rtotal = 10 + 1.726 Rtotal = 11.726 kΩ I = V/R I = 22/11.726 = 1.88mA Therefore, the drain with a 33-load connected from A to B is 1.88mA. This is higher than the current drawn with no load because the load reduces the total resistance of the circuit.

32. The voltage divider in Figure 7-74 has a switched load. Determine the voltage at each tap (V1, V2, and V3) for each position of the switch. Using the voltage divider formula to determine the position voltages in the switch. We have: 120v = 12/68000(10 + 68000) = 120*68000/(10 + 68000) = 104.6153 I wasn’t feeling confident on this question either but I didn’t want to waste too much time on it. 0.Determine the output voltage for the unbalanced bridge in Figure 7-82 for a temperature of 60° C. The temperature resistance characteristic for the thermistor is shown in Figure 7-59. Vout = Vin * (Rx/R3 + Rx – R2/R1 + R2) The values of R1, R2, and R3 are given as 100Ω, 200Ω, and 300Ω. Rx value depends on the temperature. This is due to the fact that it is a thermistor with a nonlinear resistance tempersature characteristic. The formula is set up as follows: Vout = 10 * (150/300 + 150 – 200/ 100 + 200) Vout = 10 * (0.333 - 0.4) Vout = -0.667V The output voltage for the unbalanced bridge for a temperature of 60 degrees Celsius is –0.667V. 54. Look at the meters in Figure 7-86 and determine if there is a fault in the circuit. If there is a fault, identify it. To determine the fault we can set up the following equations: Vout = IR2/V(R1 + R2) = V1R2 / (R1 + R2) Vout = 150(12000+ 12000) =150*12000/12000 + 12000 = 75v Vout = 150 (2200 +5600) = 150*5600/2200 + 5600 = 5600 / 2200 = 42.3v Our second formula solution did indeed match the voltage from the multimeter in the figure that reads 42.3v so that means those resistors are not open or faulted in any way. Which makes the fault between R2 and R1 because our answer was 75v. The multimeter read 81.8v which is higher and that means the fault betwee n R2 and R3 is a open fault. .

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version