Week 3 Assignment Gough

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Grantham University *

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Electrical Engineering

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Jan 9, 2024

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Kaila Gough G0021 Grantham University 10-01-2023
6. The following resistors (one each) are connected in a series circuit: 1.0 Ω, 2.2 Ω, 5.6 Ω, 12 Ω, and 22Ω. Determine the total resistance. Solve this problem using Rt = R1 + R2 + R3 formula Total resistance equals 42.8Ω 8. Calculate RT for each circuit of Figure 5-66. To solve for total resistance simply add each of the resistor amounts within the circuit. 1.0 + 5.6 + 2.2 = 8.8Ω 10. Six 56Ω resistors, eight 100Ω resistors, and two 22Ω resistors are all connected in series. What is the total resistance? To find the total amount of resistance you must multiply the resistance amounts by the number of resistors there are and then add each of the totals. Formula looks like this: Rt = 6 x 56 + 8 x 100 + 2 x 22 Rt = 336 + 800 + 44 Rt = 1180 ohms 12. You have the following resistor values available to you in the lab in unlimited quantities: 10Ω, 100Ω, 470Ω, 560Ω, 680Ω, 1.0kΩ, 2.2kΩ, and 5.6kΩ. All the other standard values are out of stock. A project that you are working on requires an 18kΩ resistance. What combinations of the available values would you use in series to achieve this total resistance? 10+10+10+10+10+10+10+10+10+10+100+100+100+100+100+100+100+100+100+100+470+470+470+47 0+470+470+470+470+560+560+560+560+560+560+560+560+680+680+680+680 1.0k x 9 2.2k x 4 5.6k x 1 16. The current from the source in Figure 5-69 is 5mA. How much current does each milliammeter in the circuit indicate? To find the current through each resistor we can use formula: I = V/R We know that the voltage source vS is 10v., and the resistors R1 and R2 have values of 100 ohms and 200 ohms respectively. Therefore, the current through jR1 is: I1 = Vs/R1 = 10/100 = 0.1A = 100mA Current through R2 is: I2 = Vs/R2 = 10/200 = 0.05A = 50mA
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