Chapter 2 – Circuit Elements Voltage and Current Sources Electrical Resistance Construction of a circuit model Kirchhoff’s Laws Circuit Analysis of dependent sources

Circuit Elements 5 Basic Elements: Voltage Sources Current Sources Resisters Inductors Capacitors

Voltage and Current Sources Electrical Source – A Device that is capable of converting nonelectric energy to electric energy and vice verse Ideal Voltage Source – a circuit element that maintains a prescribed voltage across its terminals regardless of the current flowing in those terminals. Ideal Current Source – a circuit element that maintains a prescribed current through its terminals regardless of the voltage across those terminals.

Voltage and Current Sources Independent Sources – Establish a voltage or a current in a circuit without relying on voltages or currents elsewhere in the circuit. Dependent Sources – Establish a voltage or a current in a circuit whose value depends on the value of the voltage or current elsewhere in the circuit. (also known as controlled sources)

Voltage and Current Sources Circuit symbols for Ideal Independent Voltage and Current sources

Voltage and Current Sources Circuit symbols for Ideal dependent Voltage and Current sources The subscript x denotes the controlling factor: 𝑣 𝑥 =𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑙𝑖𝑛𝑔 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖 𝑥 =𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑙𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 The subscript s denotes the supplied factor: 𝑣 𝑠 =𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖 𝑠 =𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝜇 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑟 𝛽 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑟 𝜌 ℎ𝑎𝑠 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑉𝑜𝑙𝑡 𝐴𝑚𝑝𝑒𝑟𝑒 𝛼 ℎ𝑎𝑠 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐴𝑚𝑝𝑒𝑟𝑒 𝑉𝑜𝑙𝑡

Voltage and Current Sources Active Element – a device capable of generating electric energy (Batteries, Generators, vacuum tubes, transistors, silicon-controlled rectifiers (SCRs),etc…) Passive Element – a device that cannot generate electric energy (resisters, capacitors, inductors, transformers, diodes, etc.)

Voltage and Current Sources Example: Using the definitions of the ideal independent voltage and current sources, state which interconnection are permissible and which violate the constraints imposed by the ideal sources. (a) Is valid, same voltage, same polarity across the same terminal (b) Is valid, same current, same direction across the same terminal (c) Is NOT valid, different voltages across the same terminal (d) Is NOT valid, different currents across the same terminal (e) Is valid, same voltage and same current is throughout the entire circuit

Voltage and Current Sources Example: Using the definitions of the ideal independent and dependent sources, state which interconnection are permissible and which violate the constraints imposed by the ideal sources. (a) Is NOT valid, they do not supply the same voltage across the same terminals (b) Is valid, same current and same voltage throughout the entire circuit (c) Is valid, same current and same voltage throughout the entire circuit (d) Is NOT valid, they do not supply the same current across the same terminals

Voltage and Current Sources Example: What value of 𝑣 𝑔 is required in order for the interconnection to be valid? For this value of 𝑣 𝑔 , find the power associated with the 8A source. (a) 𝑣 𝑔 = 𝑖 𝑏 4 = 8 4 =2𝑉 using passive sign convention (pos to neg flow of 𝑖 𝑏 ) so 𝑣 𝑔 =−2𝑉 (b) 𝑝=𝑣𝑖= −2 8 =−16𝑊 Since it is negative it is delivering 16W to the System

Voltage and Current Sources Example: What value of 𝛼 (alpha) is required in order for the interconnection to be valid? For this value of 𝛼, find the power associated with the 25V source. (a) 𝛼 𝑉 𝑥 =15𝐴 therefore 𝛼= 15 𝑉 𝑥 = 15 25 =0.6 𝐴 𝑉 has the units of A/V because it is a current controlled by a voltage (b) 𝑝=𝑣𝑖= 25 15 =375𝑊

Resistance and Conductance Resistance - The capacity of materials to impede the flow of current, the unit for resistance is Ohms (Ω) Conductance – The reciprocal of resistance, it how easily current is able to flow. The unit of conductance is a Siemens (S) 𝐺= 1 𝑅

Ohm’s Law The relationship between Voltage, Current and Resistance V= Voltage I = Current R = Resistance 𝑉=𝐼𝑅 𝐼= 𝑉 𝑅 𝑅= 𝑉 𝐼

Power in terms Voltage, Current, & Resistance 𝑝=𝑣𝑖 𝑝= 𝑖 2 𝑅 𝑝= 𝑉 2 𝑅 Power in terms of Conductance: 𝑃= 𝑣 2 𝐺 or 𝑃= 𝑖 2 𝐺

Example: For the following circuits: Calculate the values of v and i Determine the power dissipated in each resister 𝑝 8Ω = 8 2 8 =8𝑊 𝑣 𝑎 = 1 8 =8𝑉 𝑝 0.2𝑆 = 50 2 0.2 =500𝑊 𝑖 𝑏 = 50 0.2 =10𝐴 𝑣 𝑐 =− 1 20 =−20𝑉 𝑝 20Ω = (−20) 2 20 =20𝑊 𝑖 𝑑 =− 50 25 =−2𝐴 𝑝 25Ω = −2 2 25 =100𝑊

Open and Closed Circuits Short Circuit – Occurs when an event makes 𝑅=0 Open Circuit – Occurs when an event makes 𝑅=∞ = Short = Open = Switch

Open and Closed Circuits Closed circuit – when the circuit is operational and current flows correctly through the circuit.

Constructing a basic circuit:

Node – A Point where two of more circuit elements meet. They are represented with a dot

Kirchhoff’s Current Law (KCL) The Algebraic sum of all the currents at any node in a circuit equals zero Node a 𝑖 𝑠 − 𝑖 1 =0 Node b 𝑖 1 + 𝑖 𝑐 =0 Node c − 𝑖 𝑐 − 𝑖 𝑙 =0 Node d 𝑖 𝑙 − 𝑖 𝑠 =0 An algebraic sign corresponding to a reference direction must be assigned to every current at the node If you assign a positive sign to a current leaving a node, then you must assign a negative sign to a current entering a node

Kirchhoff’s Current Law (KCL) Example: Sum the currents at each node. Node a 𝑖 1 + 𝑖 4 − 𝑖 2 − 𝑖 5 =0 Node b 𝑖 2 + 𝑖 3 − 𝑖 1 − 𝑖 𝑏 − 𝑖 𝑎 =0 Node c 𝑖 𝑏 − 𝑖 3 − 𝑖 4 − 𝑖 𝑐 =0 Node d 𝑖 5 + 𝑖 𝑎 + 𝑖 𝑐 =0

Kirchhoff’s Voltage Law (KVL) The Algebraic sum of all the voltages around any closed path in a circuit equals zero 𝑣 𝑙 − 𝑣 𝑐 + 𝑣 1 − 𝑣 𝑠 =0 An algebraic sign must be assigned to each voltage in the loop

Kirchhoff’s Voltage Law (KVL) Example: Sum the voltages around each designated path. Path a: −𝑣 1 + 𝑣 2 + 𝑣 4 − 𝑣 𝑏 − 𝑣 3 =0 Path b: −𝑣 𝑎 + 𝑣 3 + 𝑣 5 =0 Path c: 𝑣 𝑏 − 𝑣 4 − 𝑣 𝑐 − 𝑣 6 − 𝑣 5 =0 Path d: −𝑣 𝑎 − 𝑣 1 + 𝑣 2 − 𝑣 𝑐 + 𝑣 7 − 𝑣 𝑑 =0

Applying KCL and KVL Example: Use Kirchhoff’s laws and Ohm’s law to find 𝑖 0 . Then verify power generated equals power dissipated.

Applying KCL and KVL Solution: At Node b using KCL 𝑖 1 − 𝑖 0 −6=0 Using KVL −120+ 𝑣 0 + 𝑣 1 =0 Rewrite in terms of current −120+10 𝑖 0 +50 𝑖 1 =0 Solve two equations with two unknowns using a matrix (make sure 𝑖 0 and 𝑖 1 align) or algebra

Applying KCL and KVL Solution: Algebraic Method 𝑖 1 =6+ 𝑖 0 Matrix Method Substitute 𝑖 1 into Rewrite Equations 𝑖 1 − 𝑖 0 =6 50 𝑖 1 +10 𝑖 0 =120 50 𝑖 1 +10 𝑖 0 =120 50 6+ 𝑖 0 +10 𝑖 0 =120 1 −1 50 10 𝑖 1 𝑖 0 = 6 120 Distribute the 50 300+50 𝑖 0 +10 𝑖 0 =120 1 −1 50 10 −1 ∗ 6 120 = 𝑖 1 𝑖 0 Solve for 𝑖 0 𝑖 1 =3𝐴 𝑖 0 =−3𝐴 60 𝑖 0 =−180 𝑖 0 = −180 60 =−3𝐴 Substitute -3 into 𝑖 0 in the first equation to find 𝑖 1 =3𝐴

Applying KCL and KVL Last Part: Verify power generated equals power dissipated. 𝑖 0 =−3𝐴 𝑎𝑛𝑑 𝑖 1 =3𝐴 Power Delivered: Power Dissipated: 𝑣 1 = 𝑖 1 𝑅 50Ω =3 50 =150𝑉 𝑃 50Ω = 3 2 50 =450𝑊 𝑝 6𝐴 = 𝑣 1 6𝐴 =−150∗6=−900𝑊 𝑆𝑖𝑔𝑛 𝑐ℎ𝑎𝑛𝑔𝑒 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑃 10Ω = −3 2 10 =90𝑊 𝑃 120𝑉 =−120 𝑖 0 =−120 −3 =360𝑊 Power Delivered = -900W Power absorbed = 450W+90W+360W=900W

Building a Circuit Model Construct a circuit model of the device inside this box. Using the designed circuit model, predict the power this device will deliver to a 10Ω resister. Part a: Use basic algebra to develop an equation for the device 𝑠𝑙𝑜𝑝𝑒= Δ𝑌 Δ𝑋 = Δ𝑣 Δ𝑖 = 30−0 0−6 =−5Ω Slope intercept is y=mx+b Therefore 𝑣 𝑡 =𝑅 𝑖 𝑡 +𝑉 𝑣 𝑡 =−5 𝑖 𝑡 +30, 𝑛𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑣𝑜𝑙𝑡𝑎𝑔𝑒

Building a Circuit Model Construct a circuit model of the device inside this box. Using the designed circuit model, predict the power this device will deliver to a 10Ω resister. Part b: Attach a 10Ω resister to the device: Knowing the current is the same throughout this circuit, we can apply KVL and write the equation in voltage form (in terms of currents) −30+5𝑖+10𝑖=0 𝑃 10Ω = 𝑖 2 𝑅= 2 2 10 =40𝑊 Solve for i 15𝑖=30 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑖=2𝐴

Section 2.5 later