Fundamentals of Electric Circuits Lecture 4 Kirchoff’s Laws

RESIRESISTORS IN SERIES RESIRESISTORS IN PARALLEL Voltage Dividers Kirchoff’s Laws Circuit Definitions? Kirchoff’s Current Law (KCL) Kirchoff’s Voltage Law (KCL) RESIRESISTORS IN SERIES RESIRESISTORS IN PARALLEL Voltage Dividers Current Dividers

Branch – a circuit element between two nodes 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

How many nodes, branches & loops? Example How many nodes, branches & loops? + - Vs Is R1 R2 R3 Vo

Example Three nodes + - Vs Is R1 R2 R3 Vo

Example 5 Branches + - Vs Is R1 R2 R3 Vo

Three Loops, if starting at node A Example Three Loops, if starting at node A + - Vs Is R1 R2 R3 Vo A B C

Kirchhoff’s Current Law THE ALGEBRAIC SUM OF THE CURRENTS ENTERING A NODE IS ZERO I = 0 I3 I4 I1 I2 I5 I1 + I2 = I3 + I4 + I5

EQUALS CURRENT IN CURRENT OUT ? A 0.5 A - 0.3 A = 0.2 A 0.5 A Example

Kirchoff’s Current Law at B + - Vs Is R1 R2 R3 Vo A B C I2 I1 I3 Assign current variables and directions Add currents IN, subtract currents OUT: I1 – I2 – I3 + Is = 0

Kirchoff’s Current Law The sum of currents flowing into a node must be equal to sum of currents flowing out of the node. i2 i3 node i1 i1 flows into the node i2 flows out of the node i3 flows out of the node i1 = i2 + i3 i1 – i2 – i3 = 0

Kirchoff’s Current Law Example Q: How much are the currents i1 and i2 ? A: i2 = 10 mA – 3 mA = 7 mA i1 = 10 mA + 4 mA = 14 mA + _ i1 4 mA i2 10 mA 3 mA node 4 mA + 3 mA + 7 mA = 14 mA

Example 1 Determine I, the current flowing out of the voltage source. Use KCL 1.9 mA + 0.5 mA + I are entering the node. 3 mA is leaving the node. V1 is generating power.

Kirchhoff’s Current Law Example: Calculate the unknown currents in the following circuits.

Example 2 Suppose the current through R2 was entering the node and the current through R3 was leaving the node. Use KCL 3 mA + 0.5 mA + I are entering the node. 1.9 mA is leaving the node. V1 is dissipating power.

Kirchoff’s Voltage Law (KVL) THE ALGEBRAIC SUM OF VOLTAGES AROUND EACH LOOP IS ZERO V = 0 E-V1-V2-V3-V4=0

Kirchoff’s Voltage Law v1 = v2 + v3 This equation can also be written in the following form –v1 + v2 + v3 = 0 + _ + v2 – v3 – v4 v1 The sum of voltages around a closed loop is zero.

Kirchhoff’s Voltage Law Example : Calculate the unknown voltages in the given circuit.

Example 3 Find the voltage across R1. Note that the polarity of the voltage has been assigned in the circuit schematic. First, define a loop that include R1.

There are three possible loops in this circuit – only two include R1. Example 3 (con’t) There are three possible loops in this circuit – only two include R1. Either loop may be used to determine VR1.

If the outer loop is used: Example 3 (con’t) If the outer loop is used: Follow the loop clockwise.

Example 3 (con’t) Follow the loop in a clockwise direction. The 5V drop across V1 is a voltage rise. VR1 should be treated as a voltage rise. The loop enters R2 on the positive side of the voltage drop and exits out the negative side. This is a voltage drop as the voltage becomes less positive as you move through the component.

Example 3 (con’t) By convention, voltage drops are added and voltage rises are subtracted in KVL.

Kirchhoff’s Voltage Law