OHMS LAW Ohms law states that the current

OHM’S LAW Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship: where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current.

1 OHM Resistance of conductor is said to be 1 ohm when it allows 1 ampere current to flow through it on applications of 1 volt across its terminal.

CIRCUIT ANALYSIS In circuit analysis, three expressions of Ohm's law interchangeably: equivalent are used I=V/R or V=IR or R=V/I. The interchangeability of the equation may be represented by a triangle, where V (voltage) is placed on the top section, the I (current) is placed to the left section, and the R (resistance) is placed to the right. The line that divides the left and right sections indicate multiplication, and the divider between the top and bottom sections indicates division (hence the division bar).

RESISTIVE CIRCUITS Resistors are circuit elements that impede the passage of electric charge in agreement with Ohm's law, and are designed to have a specific resistance value R. In a schematic diagram the resistor is shown as a zig-zag symbol. An element (resistor or conductor) that behaves according to Ohm's law. Resistors which are in series or in parallel may be grouped together into a single "equivalent resistance" in order to apply Ohm's law in analyzing the circuit.

CIRCUIT DEFINITIONS Node – any point where 2 or more circuit elements are connected together Wires usually have negligible resistance Each node has one voltage (w. r. t. ground) Branch – a circuit element between two nodes Loop – a collection of branches that form a closed path returning to the same node without going through any other nodes or branches twice

EXAMPLE How many nodes, branches & loops? R 1 + - Vs + R 2 R 3 Is Vo -

EXAMPLE Three nodes R 1 + - Vs + R 2 R 3 Is Vo -

EXAMPLE 5 Branches R 1 + - Vs + R 2 R 3 Is Vo -

KIRCHOFF’S LAW There are two types of law: 1. KIRCHOFF’S CURRENT LAW(KCL). 2. KIRCHOFF’S VOLTAGE LAW(KVL).

CURRENT LAW This law is also called Kirchhoff's first law, Kirchhoff's point rule, or Kirchhoff's junction rule (or nodal rule). The principle of conservation of electric charge implies that: At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node, or: The algebraic sum of currents in a network of conductors meeting at a point is zero.

VOLTAGE LAW This law is also called Kirchhoff's second law, Kirchhoff's loop (or mesh) rule, and Kirchhoff's second rule.

EXAMPLE CIRCUIT Solve for the currents through each resistor And the voltages across each resistor

EXAMPLE CIRCUIT + I 1∙ 10Ω + I 2∙ 8Ω - + I 3∙ 6Ω + I 3∙ 4Ω - Using Ohm’s law, add polarities and expressions for each resistor voltage

EXAMPLE CIRCUIT + I 1∙ 10Ω + I 2∙ 8Ω - + I 3∙ 6Ω + I 3∙ 4Ω - Write 1 st Kirchoff’s voltage law equation -50 v + I 1∙ 10Ω + I 2∙ 8Ω = 0

EXAMPLE CIRCUIT + I 1∙ 10Ω + I 2∙ 8Ω - + I 3∙ 6Ω + I 3∙ 4Ω - Write 2 nd Kirchoff’s voltage law equation -I 2∙ 8Ω + I 3∙ 6Ω + I 3∙ 4Ω = 0 or I 2 = I 3 ∙(6+4)/8 = 1. 25 ∙ I 3

EXAMPLE CIRCUIT A Write Kirchoff’s current law equation at A +I 1 – I 2 - I 3 = 0

EXAMPLE CIRCUIT We now have 3 equations in 3 unknowns, so we can solve for the currents through each resistor, that are used to find the voltage across each resistor Since I 1 - I 2 - I 3 = 0, I 1 = I 2 + I 3 Substituting into the 1 st KVL equation -50 v + (I 2 + I 3)∙ 10Ω + I 2∙ 8Ω = 0 or I 2∙ 18 Ω + I 3∙ 10 Ω = 50 volts

EXAMPLE CIRCUIT But from the 2 nd KVL equation, I 2 = 1. 25∙I 3 Substituting into 1 st KVL equation: (1. 25 ∙ I 3)∙ 18 Ω + I 3 ∙ 10 Ω = 50 volts Or: I 3 ∙ 22. 5 Ω + I 3 ∙ 10 Ω = 50 volts Or: I 3∙ 32. 5 Ω = 50 volts Or: I 3 = 50 volts/32. 5 Ω Or: I 3 = 1. 538 amps