ACT_Electricity_1_Team072

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University of Cincinnati, Main Campus *

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1100

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Electrical Engineering

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Apr 3, 2024

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pdf

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Task 1: Arrangement 1: It is a circuit 3 circuits Arrangement 2: It is a circuit 6 circuits Arrangement 3: It has a circuit It currently has one circuit To fix it we must connect the negative end of the power source to the rest of the circuit to close the loop Arrangement 4: It doesn't have a circuit To fix it we must tie the ends of everything that doesn't have two sides connected
Task 2: Determine the power supplied by each of the elements of the circuit below. P1: (-1A*6V)= -6W P2: (2A*3V) = 6W P3A: (-3A*5V) = -15W P4A: (-4A*3V) = -12W P2A: (2A*8V) = 16W P5V: (2A*5V) = 10W P3A: (-3A*5V) = -15W Which elements are supplying power and which are absorbing power? P1, P3A and P4A are all absorbing power while P2 supplies it. Is it possible for these values to be present in a real circuit? No because the values don't add up to zero.
Task 3: Part A: Calculate the current, I, through the resistor, R, assuming R = 6.8 ohms. Be sure to select appropriate units for current 2V/6.8 Ω = 1.76 A Repeat (a) assuming R = 4.7 kilo-ohms 12V/8.1 kΩ = 2.55 mA Repeat assuming R = 8.1 milli-ohms 12V/8.1MΩ = 1.481µA
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Part B: If a refrigerator draws 2.2A at 120V, what is the resistance of the refrigerator. 120V/2.2A = 54.54 Ω A 9V DC battery can provide 45mA of current. What is the power rating of the battery? 0.405W = 9V * 0.045A The current through 10 kilo-ohms resistor 1.5 mA. What is the power dissipated by the resistor? 10,000Ω*.0015A = 15mV 15mV*1.5mA= 22.5mW If the voltage across a resistor is doubled, what happens to the power dissipated by the resistor? Verify your answer by doing calculations. If voltage is doubles then, power is quadrupled. P = v^2/R 12V^2/2Ω = 72W 24V^2/2Ω = 288W 288W/72W= 4
Task 4: Given the circuit below, determine each of the unknown quantities shown along with the power supplied or absorbed in each element. If you know them do not use any series or parallel combinations/properties to perform these calculations Note: the 2A current applies to both the voltage source and the 10 ohm resistor and the 4V voltage applies both the 5 ohm and R resistors V10) 2A*10Ω= 20V I5Ω) 4V/5Ω = .8A