We also will classified sources as Independent and Dependent sources Independent source establishes a voltage or a current in a circuit without relying on a voltage or current elsewhere in the circuit Dependent sources establishes a voltage or a current in a circuit whose value depends on the value of a voltage or a current elsewhere in the circuit We will use circle to represent Independent source and diamond shape to represent Dependent sources Dependent sources Independent source

Independent voltage source Independent current source Independent and dependent voltage and current sources can be represented as 3 A 5 V Independent voltage source Independent current source 4 ix V 4 vx A were ix is some current through an element were vx is some voltage across an element Dedependent current source Current depend on voltage Dedependent voltage source Voltage depend on current

The dependent sources can be also as 4 vx V 7 ix A were vx is some current through an element were ix is some voltage across an element Dedependent voltage source Voltage depend on voltage DE dependent current source Current depend on current

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

Example 2.7 for the circuit shown apply KVL to each designated path in the circuit

We will redraw the circuit and assign currents and voltages as follows Example 2.8 for the circuit shown use Kirchcoff’s laws and Ohm’s law to find io ? Solution We will redraw the circuit and assign currents and voltages as follows Since io is the current in the 120 V source , therefore there is only two unknown currents in the circuit namely: io and i1

Therefore we need two equations relating io and i1 Applying KCL to the circuit nodes namely a,b and c will give us the following Node a Nothing new ! Node b Node c The same as node b Therefore KCL provide us with only one equation relating io and i1 namely We need another equation to be able to solve for io That equation will be provided by KVL as shown next

We have three closed loops However only one loop that you can apply KVL to it namely abca Since the other two loops contain a current source namely 6 A and since we can not relate the voltage across it to the current through it , therefore we can not apply KVL to that loop Applying KVL around loop abca clockwise direction assigning a positive sign to voltage drops ( + to - ) and negative sign to voltage rise , we have Combinning this with the KCL equation we have two equations and unknowns which can be solved simultaneously to get io= -3 A i1= 3 A

2.5 Analysis of a circuit containing dependent sources For the circuit shown we want apply Kirchhof’s and Ohm law to find vo ? Solution We are going to device a strategy for solving the circuit Let io be the current flowing on the 20 W resistor → therefore we seek io Since vo=20 io

KCL will provide us with one equation relating io with iΔ namely Node b We need two equations to be able to solve for io KVL and Ohm’s law will provide us with the additional equation relating io with iΔ as will be shown next

The circuit has three closed loops However only the loop abca is the one that you can apply KVL to it Since the other two loops contain a current source that, you will not be able to write the voltage across it in terms of the current (i.e, you will not be able to write KVL around the loop the one that you can apply KVL to it ) Therefore

Therefore we have two equations relating io with iΔ namely Two equations and two unknowns io and iΔ , we can solve simultaneously and get io = 24 A and iΔ = 4 A Therefore vo=20 io = (20)(24) = 480 V

DEPENDENT VOLTAGE AND CURRENT SOURCES A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage. Example: You watch a certain voltmeter V1 and manually adjust a voltage source Vs to be 2 times this value. This constitutes a voltage-dependent voltage source. V1 + - Circuit A 2V1 + - Circuit B This is just a manual example, but we can create such dependent sources electronically. We will create a new symbol for dependent sources.

+ V Element X DEPENDENT VOLTAGE AND CURRENT SOURCES We can have voltage or current sources depending on voltages or currents elsewhere in the circuit. Here, the voltage V provided by the dependent source (right) is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way. + - V X Element X A diamond-shaped symbol is used for dependent sources, just as a reminder that it’s a dependent source. Circuit analysis is performed just as with independent sources.

THE 4 BASIC LINEAR DEPENDENT SOURCES Constant of proportionality Parameter being sensed Output Voltage-controlled voltage source … V = Av Vcd Current-controlled voltage source … V = Rm Ic Current-controlled current source … I = Ai Ic Gm Vcd Voltage-controlled current source … I = Gm Vcd Gm Vcd I = Gm Vcd Ai Ic + _ + _ Av Vcd Rm Ic

NODAL ANALYSIS WITH DEPENDENT SOURCES 4 VSS +  ISS 3 1 V a b A v c 6 2

NODAL ANALYSIS WITH DEPENDENT SOURCES 4 I 2 ISS 3 1 V b + - m NODAL ANALYSIS WITH DEPENDENT SOURCES a

INTUITION A dependent source is like a “krazy resistor”, except its voltage (current) depends on a different current (voltage). If the controlling voltage or current is zero, the dependent voltage source will have zero voltage (and the dependent current source will have zero current). There must be independent sources in the circuit for nonzero voltages or currents to happen! Math explanation: Node voltage analysis leads to The A matrix is nonsingular for real circuits, and is made of resistor values and dependent source parameters. The b vector is made of independent voltage & current source values. If b is zero and A is nonsingular, the solution must be zero! Ax = b