ECET 110 Week 5 Homework
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CHAPTER 9
1a. Using the superposition theorem, determine the current through the 12 Ohm resistor of fig. 9.119. we will use superposition to analyze this circuit. First short E2 and find the current thru the 12 Ohms resistor and then Short E1 and keep E2 in the circuit and find the same current. Then algebraically add the two currents (add if in the same directions and subtract if opposite direction
Short V2
12Ohms resistor—IR3
IR1 and IR3 are the same value
IR3=4A
Short V1
12 Ohms resistor—IR3
IR3=5A
Subtract the values due to opposing directions
IR3=IR3(V1)+IR3(V2)=5A-4A=1A
1b. Convert both voltage sources to current sources and recalculate the current to the 12 Ohm resistor.
Subtract the two current sources (because they are opposite directions) and the result will be one current source. You can find individual currents using current divider rule
1c. How do the results of parts a and b compare?
The answers are to be the same.
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ECET 110 Week 5 Homework
Click Below Link To Purchase
www.foxtutor.com/product/ecet-110-week-5-homework
CHAPTER 9
1a. Using the superposition theorem, determine the current through the 12 Ohm resistor of fig. 9.119. we will use superposition to analyze this circuit. First short E2 and find the current thru the 12 Ohms resistor and then Short E1 and keep E2 in the circuit and find the same current. Then algebraically add the two currents (add if in the same directions and subtract if opposite direction
Short V2
12Ohms resistor—IR3
IR1 and IR3 are the same value
IR3=4A
Short V1
12 Ohms resistor—IR3
IR3=5A
Subtract the values due to opposing directions
IR3=IR3(V1)+IR3(V2)=5A-4A=1A
1b. Convert both voltage sources to current sources and recalculate the current to the 12 Ohm resistor.
Subtract the two current sources (because they are opposite directions) and the result will be one current source. You can find individual currents using current divider rule
1c. How do the results of parts a and b compare?
The answers are to be the same.