Power and Energy Consider the product of voltage and current. V  I (J/C)  (C/s) = J/s (1 J/s = 1 W: power!) Now multiply by time. V  I  t (J/C)  (C/s)  s = J (energy!)

Power and Energy P = V  I Recall that V = I R. So P = V  I = I R  I = I2 R Also, I = V/R. So P = V  I = V  (V/R) = V2/R

Mathematical Conventions If positive current leaves the positive voltage terminal, the element is delivering or furnishing power (active). If positive current enters the positive voltage terminal, the element is absorbing or dissipating power (passive).

Independent Sources Independent Voltage Source The voltage across the source is independent of the current drawn from the source. Independent Current Source The current delivered by the source is independent of the voltage across the source. Strong, fresh battery Lightning

Dependent Sources Voltage-Controlled Voltage Source Current-Controlled Current Source Current-Controlled Current Source

Resistors in Series - Review

Resistors in Parallel - Review

Resistors Combination - Practice

RECALL: Types of connections Resistors connected in series… When JUST two elements connect at a SINGLE node, they are said to be in series and series-connected elements carry the SAME CURRENT (why? KCL)

Resistors connected in parallel… RECALL: Types of connections Resistors connected in parallel… When two elements connect at a single node pair, they are said to be in parallel and they have the SAME VOLTAGE across their terminals ! (why? KVL)

Step by step process to find equivalent resistance

Resistors Combination - Practice

Voltage and Current Division The current and voltage in a resistive circuit get divided between different resistors based on their resistance values. How does that happen? KVL and KCL are universal rules and should be satisfied in every circuit – including (obviously) resistive circuits.

Voltage Division From Ohm’s Law I = VT/Req Req = R1 + R2 + R3 Voltage applied to series circuit applies a fraction of the voltage across each element

Voltage Division Section 2.3 Voltage Divider (Series Circuits) Voltage Divider (Series Circuits) Voltage Division Principle: the fraction of the total voltage across a single resistor in a series circuit is the ratio of the given resistor to the total series resistance So if V1 = VT R1 / (R1 + R2 + R3) , then what is V2 and V3

VOLTAGE DIVISION V1 = VT R1/ (R1 + R2) …… …. V6 = VT R6/ (Req) Voltage division is a simple method/procedure that allows one to determine the voltage across a resistor in a series combination, if the total voltage across all series-connected resistors is known. The relationships are shown below: V1 = VT R1/ (R1 + R2) …… …. V6 = VT R6/ (Req)

Current Division Total current in a parallel circuit is divided among resistances. What is i2?

Current Division Current Divider (Parallel Circuits) Used to determine the current through one of several parallel resistors, if the total current entering the parallel combination is known

Once again… WHEN YOU HAVE RESISTORS CONNECTED IN SERIES WHEN YOU HAVE RESISTORS CONNECTED IN PARALLEL YOU CAN USE VOLTAGE DIVISION TO FIND INDIVIDUAL VOLTAGE VALUES ACROSS THE RESISTORS YOU CAN USE CURRENT DIVISION TO FIND THE INDIVIDUAL CURRENT VALUES THROUGH THE RESISTORS

Let’s solve some examples…

And an example on current division…

How can we combine these techniques to solve circuits How can we combine these techniques to solve circuits? Let’s try to find io and vo !

First thing we can do is to assign variables to each element – hopefully we’ll be able to solve for those !

First thing we can do is to assign variables to each element – hopefully we’ll be able to solve for those !

Node-Voltage Analysis In addition to analyzing circuits by combining series and parallel resistors and applying the voltage and current – division principles, there is the NODE-VOLTAGE Analysis. Recall a node is a point at which two or more circuit elements are joined together

Node-Voltage Analysis CONVENTION: usually write expressions for I, current, leaving the node under consideration and set sum to zero Step 1. Select or find reference node Step 2. Label the node voltages - reference and other nodes

Node-Voltage Analysis CONVENTION: usually write expressions for I, current, leaving the node under consideration and set sum to zero Step 3. Observe node voltage relationship to element voltage (e.g. KVL says -v2 + vx + v3 = 0, then vx = v2 – v3) Step 4. Write current equations at each of the nodes for ALL currents leaving the node

Node-Voltage Analysis Step 5. Node-voltage equations can be written for each node in the form of current leaving one node and entering another node e.g. i4 through R4 leaving Node 2 and entering ground gives i4 = V2/R4

Node-Voltage Analysis Node Voltage Equation for Node 3 KCL iR1 + iR5 + iR3 = 0 where iR1 = (v3 – v1)/R1 iR5 = v3/R5 iR3 =( v3 – v2)/R3 Then the Node Voltage Equation for Node 3 is (v3 – v1)/R1 + v3/R5 + (v3 – v2)/R3 = 0

Node-Voltage Analysis What is the Node Voltage Equation for Node 1

Node-Voltage Analysis What is the Node Voltage Equation for Node 2

Node-Voltage Analysis RECALL: Node-voltage equations can be written for each node in the form of current leaving one node and entering another node Another Example

Node-Voltage Analysis Another Example

Section 2.4 Node Voltage Analysis Another Example CONVENTION: usually write expressions for I, current, leaving the node under consideration and set sum to zero

Mesh-Current Analysis Then there is MESH- CURRENT Analysis. Use KVL to write a vrises = vdrops equation for each mesh mesh – a closed path that contains no other closed paths

Mesh-Current Analysis 1. Label the meshes. # of independent KVL equations for planar network are equal to # of open areas defined by the network layout (2 OPEN areas in this circuit, hence 2 mesh currents, i1, and i2)

Section 2.5 Mesh Current Analysis 2. Follow KVL around each mesh Mesh 1 -Va + i1R1 + V3 = 0 where v3 = R3(i1 - i2) Thus i1R1 + R3(i1 - i2) = Va The same can be found for Mesh 2

Section 2.5 Mesh Current Analysis

Mesh-Current Analysis

Section 2.5 Mesh Current Analysis

Section 2.5 Mesh Current Analysis Exercise. Find ia

Section 2.5 Mesh Current Analysis Exercise. Find ib

Section 2.5 Mesh Current Analysis Mesh with Controlled Sources Combine MESH 1 and 2 – Supermesh Voltage Controlled Current Source referenced as i2 - i1 = vx/4 where vx = 2i2