Analysis with Dependent Sources Topic 6 Analysis with Dependent Sources (2.5)

Look for interim objectives + - 500v 5Ω 20Ω 5iΔ iΔ vo io a b c Find vo in the circuit Look for interim objectives If we know io then we can find vo by Ohm’s Law By KVL: 2 If we know iΔ then we can find 5 iΔ (and also io). Eliminating io: 3 So & 4 By KCL: 1 into 4 5 So 1 By Ohm’s Law: 6 4/30/2019 Analysis with Dependent Sources Where’s the other 20 volts?

Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed. Assessment 2.9 30i1 a b c d e f Let’s label all the nodes + - 5v 54kΩ 6kΩ i1 v 30i1 1v 1.8kΩ 8v i1 30i1 i2 i1 i1 runs through the 5v and 1v sources as well as the 54kΩ resistor Similarly, 30i1 runs through the 1.8kΩ resistor and the 8v source. By KCL at node c: So By KVL around loop a-b-c-f-a: 4/30/2019 Analysis with Dependent Sources

Analysis with Dependent Sources Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed. Assessment 2.9 + - 5v 54kΩ 6kΩ i1 v 30i1 1v 1.8kΩ 8v a b c d e f i2 Writing KVL around loop e-d-c-f-e 4/30/2019 Analysis with Dependent Sources

Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed. Assessment 2.9 + - 5v 54kΩ 6kΩ i1 v 30i1 1v 1.8kΩ 8v a b c d e f i2 (c) The dependent source is not generating Source Other Power 5v 25µA 125µW 1v 25µW v 750µA -1500µW 8v 6000µW total generated = 6150µW 4/30/2019 Analysis with Dependent Sources

Analysis with Dependent Sources Find (a) i1 in microamps, (b) v in volts, (c) the total power generated and (d) the total power absorbed. Assessment 2.9 + - 5v 54kΩ 6kΩ i1 v 30i1 1v 1.8kΩ 8v a b c d e f i2 (d) 4/30/2019 Analysis with Dependent Sources

Assessment 2.10 2iφ i1 a The current iφ is 2A. Calculate vs. b 10Ω iφ 30Ω 10Ω vs iφ 2iφ i1 The current iφ is 2A. Calculate vs. a b Label the nodes Define a current, i1, through the 10Ω resistor c KCL at node a: 1 2 Solving for i1: KVL around loop b-a-c-b: 3 4 Solving for vs: 4/30/2019 Analysis with Dependent Sources

Assessment 2.10 iφ=2A 2iφ i1=1A vs=70v 30Ω 10Ω vs iφ 2iφ (b) Calculate the power absorbed by the independent voltage source. i1 a b is To calculate power absorbed let’s define a downhill current, is c Then power absorbed by the independent voltage source is So power absorbed by the independent voltage source: Writing KCL at b: Solving for is: 4/30/2019 Analysis with Dependent Sources

Assessment 2.10 iφ=2A 2iφ i1=1A vs=70v is=3A 30Ω 10Ω vs iφ 2iφ (c) Calculate the power delivered by the independent current source. i1 a b is vi + - To calculate power delivered let’s define a voltage, vi, so the current is uphill c For an uphill current, power delivered is: To find vi, write KVL around the lefthand loop c-a-c: So & 4/30/2019 Analysis with Dependent Sources

Assessment 2.10 vφ + - 5A 30Ω 10Ω vs iφ 2iφ (d) Calculate the power delivered by the controlled current source. i1 a b is vi + - Again define voltage, vφ so the current is uphill c For an uphill current, power delivered is: iφ=2A i1=1A vs=70v is=3A vi=60v Note that vφ also appears across the 10Ω resistor (KVL) So, by Ohm’s Law: Giving power delivered by the controlled current source: 4/30/2019 Analysis with Dependent Sources

Analysis with Dependent Sources Assessment 2.10 vφ + - 5A 30Ω 10Ω vs iφ 2iφ (e) Calculate the total power absorbed by the two resistors. i1 a b is vi + - Since we know the currents we’ll just use i2R c iφ=2A i1=1A vs=70v is=3A vi=60v Power check 4/30/2019 Analysis with Dependent Sources