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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. R = 5[] 1 R3 = 33[] vS1 = 18[V] + - R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] R5 = 12[] R4 = 56[] vS2 = 7[V] + 1. 2. 3. 4. - 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 Thevenin and Norton (a 1st or b 1st OK): a. 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). b. 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. c. 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. 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. R = 5[] 1 R3 = 33[] vS1 = 18[V] + - R4 = 56[] R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] - + B A R5 = 12[] vS2 = 7[V] + 1. 2. 3. 4. - 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 Thevenin and Norton (a 1st or b 1st OK): a. 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). b. 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. c. 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. 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. R = 5[] 1 R3 = 33[] vS1 = 18[V] + - R4 = 56[] R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] - + B A R5 = 12[] vS2 = 7[V] + 1. 2. 3. 4. - 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 Thevenin and Norton (a 1st or b 1st OK): a. 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). b. 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. c. 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. 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. 2. If not sure about dependent or independent sources, then for both Thevenin and Norton: a. 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). 1. Identify the node, choose a reference. 2. Identify node voltage that is already known (given) 3. Apply KCL to each unknown node. If no voltage source directly attached to a node: a. Draw current vector away from node b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R c. If a current is to a current source, write the current source with proper polarity d. Add all the currents and let = 0 4. If a direct voltage source attached to a node: a. An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. b. If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or c. An unknown current can be introduced to be solved later. 5. Assemble all the equations and identified additional unknown besides node voltage. 6. Solve the equations 7. Use the known node voltage to derive other quantities asked by the problem 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 2. If not sure about dependent or independent sources, then for both Thevenin and Norton: a. 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). 1. Identify the node, choose a reference. 2. Identify node voltage that is already known (given) 3. Apply KCL to each unknown node. If no voltage source directly attached to a node: a. Draw current vector away from node b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R c. If a current is to a current source, write the current source with proper polarity d. Add all the currents and let = 0 4. If a direct voltage source attached to a node: a. An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. b. If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or c. An unknown current can be introduced to be solved later. 5. Assemble all the equations and identified additional unknown besides node voltage. 6. Solve the equations 7. Use the known node voltage to derive other quantities asked by the problem 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 2. If not sure about dependent or independent sources, then for both Thevenin and Norton: a. 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). 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 2. R1i1 vS 1 i1 R2i1 R5i1 If not sure about dependent or independent sources, then for both Thevenin and Norton: a. 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). vS 2 i1 R1 R2 R5 vS 1 vS 2 0 v A R5i1 vS 2 vS1 R1i1 R2i1 v A 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. R = 5[] 1 R3 = 33[] vS1 = 18[V] + - R4 = 56[] R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] I SC - B + A R5 = 12[] vS2 = 7[V] + 1. 2. 3. 4. - 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 Thevenin and Norton (a 1st or b 1st OK): a. 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). b. 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. c. 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. 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 2. If not sure about dependent or independent sources, then for both Thevenin and Norton: b. 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. 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. 2. If not sure about dependent or independent sources, then for both Thevenin and Norton: c. Obtain REQ= VOC / ISC. Go to 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. 2. If not sure about dependent or independent sources, then for both Thevenin and Norton: c. Obtain REQ= VOC / ISC. Go to 4. v A vS 1 0 vS 1 R2 R1 R2 R1 v A ( vS 2 ) 0 vS 2 R5 R5 iSC vS 1 v S 2 0 R2 R1 R5 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. 4. Draw TEQ or NEQ with proper source with correct polarity and equivalent resistor. A INo iS 2 0.3 A RNo iRNo B 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. R = 5[] 1 R3 = 33[] + 3. vS1 = 18[V] - R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] R5 = 12[] R4 = 56[] vS2 = 7[V] + - Can we do an independent verification of the power (delivered or absorbed) by the 0.3 A source? 3. b) Find the power delivered by the 0.3 A current source. R1 = 5[] R3 = 33[] vS1 = 18[V] + - R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] R5 = 12[] R4 = 56[] vS2 = 7[V] + - 1. Identify the node, choose a reference. 2. Identify node voltage that is already known (given) 3. Apply KCL to each unknown node. If no voltage source directly attached to a node: a. Draw current vector away from node b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R c. If a current is to a current source, write the current source with proper polarity d. Add all the currents and let = 0 4. If a direct voltage source attached to a node: a. An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. b. If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or c. An unknown current can be introduced to be solved later. 5. Assemble all the equations and identified additional unknown besides node voltage. 6. Solve the equations 7. Use the known node voltage to derive other quantities asked by the problem 3. b) Find the power delivered by the 0.3 A current source. C R1 = 5[] R3 = 33[] vS1 = 18[V] + - R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] A R4 = 56[] B R5 = 12[] vS2 = 7[V] + - 1. Identify the node, choose a reference. 2. Identify node voltage that is already known (given) 3. Apply KCL to each unknown node. If no voltage source directly attached to a node: a. Draw current vector away from node b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R c. If a current is to a current source, write the current source with proper polarity d. Add all the currents and let = 0 4. If a direct voltage source attached to a node: a. An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. b. If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or c. An unknown current can be introduced to be solved later. 5. Assemble all the equations and identified additional unknown besides node voltage. 6. Solve the equations 7. Use the known node voltage to derive other quantities asked by the problem 3. b) Find the power delivered by the 0.3 A current source. C R1 = 5[] R3 = 33[] vS1 = 18[V] + - R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] A R4 = 56[] B R5 = 12[] vS2 = 7[V] + - 1. Identify the node, choose a reference. 2. Identify node voltage that is already known (given) 3. Apply KCL to each unknown node. If no voltage source directly attached to a node: a. Draw current vector away from node b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R c. If a current is to a current source, write the current source with proper polarity d. Add all the currents and let = 0 4. If a direct voltage source attached to a node: a. An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. b. If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or c. An unknown current can be introduced to be solved later. 5. Assemble all the equations and identified additional unknown besides node voltage. 6. Solve the equations 7. Use the known node voltage to derive other quantities asked by the problem 3. b) Find the power delivered by the 0.3 A current source. C R1 = 5[] R3 = 33[] vS1 = 18[V] + - R2 = 9[] iS1 = 15[A] iS2 = 0.3[A] A R4 = 56[] B R5 = 12[] vS2 = 7[V] + - 1. Identify the node, choose a reference. 2. Identify node voltage that is already known (given) 3. Apply KCL to each unknown node. If no voltage source directly attached to a node: a. Draw current vector away from node b. If a current flows in a resistor, apply Ohm’s law (VX-VY)/R c. If a current is to a current source, write the current source with proper polarity d. Add all the currents and let = 0 4. If a direct voltage source attached to a node: a. An equation can be: VX-VY=VS where VY is the node at the other side of the voltage source. b. If the source is in series with a resistor the other terminal (and no branching), Norton equivalent circuit can be applied; or c. An unknown current can be introduced to be solved later. 5. Assemble all the equations and identified additional unknown besides node voltage. 6. Solve the equations 7. Use the known node voltage to derive other quantities asked by the problem C R1 = 5[] R3 = 33[] vS1 = 18[V] R2 = 9[] iS1 = 15[A] + - iS2 = 0.3[A] A B R4 = 56[] R5 = 12[] vS2 = 7[V] + - R1 , R2 , R3 , R5 , vS1 , vS2 , iS2 Solve vA vA vS1 R1 R2 iS2 vA 5., 9 , 33 , 12., 18, 7., 0.3 ; vS2 R5 0 , vA 2.6 Power delivered (active convention) by the 0.3 A source: P v A iS2 . v A 0.78 2.6