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3. a) Find the Norton equivalent seen by the 0.3 A current source .

3. a) Find the Norton equivalent seen by the 0.3 A current source . b) Find the power delivered by the 0.3 A current source . . If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit.

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3. a) Find the Norton equivalent seen by the 0.3 A current source .

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  1. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. • If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit. • If not sure about dependent or independent sources, then for both Theveninand Norton (a 1st or b 1st OK): • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem). • find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM; MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes). Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if it is a completely different problem. • Obtain REQ= VOC /ISC. Go to 4. • For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just follow 2 above. • Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.

  2. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. + A - B • If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit. • If not sure about dependent or independent sources, then for both Theveninand Norton (a 1st or b 1st OK): • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem). • find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM; MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes). Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if it is a completely different problem. • Obtain REQ= VOC /ISC. Go to 4. • For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just follow 2 above. • Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.

  3. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. + A - B • If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit. • If not sure about dependent or independent sources, then for both Theveninand Norton (a 1st or b 1st OK): • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem). • find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM; MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes). Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if it is a completely different problem. • Obtain REQ= VOC /ISC. Go to 4. • For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just follow 2 above. • Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.

  4. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. • If not sure about dependent or independent sources, then for both Thevenin and Norton: • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM, or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem). • Identify the node, choose a reference. • Identify node voltage that is already known (given) • Apply KCL to each unknown node. If no voltage source directly attached to a node: • Draw current vector away from node • If a current flows in a resistor, apply Ohm’s law (VX-VY)/R • If a current is to a current source, write the current source with proper polarity • Add all the currents and let = 0 • If a direct voltage source attached to a node: • An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. • If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or • An unknown current can be introduced to be solved later. • Assemble all the equations and identified additional unknown besides node voltage. • Solve the equations • Use the known node voltage to derive other quantities asked by the problem

  5. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. C • If not sure about dependent or independent sources, then for both Thevenin and Norton: • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM, or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem). • Identify the node, choose a reference. • Identify node voltage that is already known (given) • Apply KCL to each unknown node. If no voltage source directly attached to a node: • Draw current vector away from node • If a current flows in a resistor, apply Ohm’s law (VX-VY)/R • If a current is to a current source, write the current source with proper polarity • Add all the currents and let = 0 • If a direct voltage source attached to a node: • An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. • If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or • An unknown current can be introduced to be solved later. • Assemble all the equations and identified additional unknown besides node voltage. • Solve the equations • Use the known node voltage to derive other quantities asked by the problem

  6. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. C • If not sure about dependent or independent sources, then for both Thevenin and Norton: • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM, or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem).

  7. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. C • If not sure about dependent or independent sources, then for both Thevenin and Norton: • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM, or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem).

  8. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. + A - B • If “as seen by a device,” then remove the device (not part of the EQ), otherwise, it is part of the circuit. • If not sure about dependent or independent sources, then for both Theveninand Norton (a 1st or b 1st OK): • find open circuit voltage VOC: the voltage across the 2 points of interest. Use KVL, NVM or MCM; NVM tends to give voltage directly, (which might be a bit more convenient, but depending on the problem). • find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM; MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes). Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if it is a completely different problem. • Obtain REQ= VOC /ISC. Go to 4. • For If you are sure that there are NO dependent sources, and if you think finding REQ is easy, you can zero all the independent sources (voltage-> shorted wire, current-> cut open wire). If you don’t think it is easy, just follow 2 above. • Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor.

  9. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. C • If not sure about dependent or independent sources, then for both Theveninand Norton: • find short circuit ISC: connect a wire across the 2 points and find its current. Use KCL, NVM or MCM; MCM tends to give current directly, but may NOT be convenient if too many equations (many meshes). Most critical: this circuit has NOTHING to do with circuit in #2a. It is a different circuit and solve it as if it is a completely different problem.

  10. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. • If not sure about dependent or independent sources, then for both Theveninand Norton: • Obtain REQ= VOC / ISC. Go to 4.

  11. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. • If not sure about dependent or independent sources, then for both Theveninand Norton: • Obtain REQ= VOC / ISC. Go to 4.

  12. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor. A INo RNo B

  13. 3. a) Find the Norton equivalent seen by the 0.3 A current source. b) Find the power delivered by the 0.3 A current source. Can we do an independent verification of the power (delivered or absorbed) by the 0.3 A source?

  14. 3. b) Find the power delivered by the 0.3 A current source. • Identify the node, choose a reference. • Identify node voltage that is already known (given) • Apply KCL to each unknown node. If no voltage source directly attached to a node: • Draw current vector away from node • If a current flows in a resistor, apply Ohm’s law (VX-VY)/R • If a current is to a current source, write the current source with proper polarity • Add all the currents and let = 0 • If a direct voltage source attached to a node: • An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. • If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or • An unknown current can be introduced to be solved later. • Assemble all the equations and identified additional unknown besides node voltage. • Solve the equations • Use the known node voltage to derive other quantities asked by the problem

  15. 3. b) Find the power delivered by the 0.3 A current source. C A B • Identify the node, choose a reference. • Identify node voltage that is already known (given) • Apply KCL to each unknown node. If no voltage source directly attached to a node: • Draw current vector away from node • If a current flows in a resistor, apply Ohm’s law (VX-VY)/R • If a current is to a current source, write the current source with proper polarity • Add all the currents and let = 0 • If a direct voltage source attached to a node: • An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. • If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or • An unknown current can be introduced to be solved later. • Assemble all the equations and identified additional unknown besides node voltage. • Solve the equations • Use the known node voltage to derive other quantities asked by the problem

  16. 3. b) Find the power delivered by the 0.3 A current source. C A B • Identify the node, choose a reference. • Identify node voltage that is already known (given) • Apply KCL to each unknown node. If no voltage source directly attached to a node: • Draw current vector away from node • If a current flows in a resistor, apply Ohm’s law (VX-VY)/R • If a current is to a current source, write the current source with proper polarity • Add all the currents and let = 0 • If a direct voltage source attached to a node: • An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. • If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or • An unknown current can be introduced to be solved later. • Assemble all the equations and identified additional unknown besides node voltage. • Solve the equations • Use the known node voltage to derive other quantities asked by the problem

  17. 3. b) Find the power delivered by the 0.3 A current source. C A B • Identify the node, choose a reference. • Identify node voltage that is already known (given) • Apply KCL to each unknown node. If no voltage source directly attached to a node: • Draw current vector away from node • If a current flows in a resistor, apply Ohm’s law (VX-VY)/R • If a current is to a current source, write the current source with proper polarity • Add all the currents and let = 0 • If a direct voltage source attached to a node: • An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. • If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or • An unknown current can be introduced to be solved later. • Assemble all the equations and identified additional unknown besides node voltage. • Solve the equations • Use the known node voltage to derive other quantities asked by the problem

  18. C A B

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