Fundamentals of Electric Circuits

Lecture 4

Kirchoff’s Laws

2

3

Node – any point where 2 or more circuit elements are connected togetherWires usually have negligible resistanceEach node has one voltage (w.r.t. ground)

Branch – a circuit element between two nodesLoop – a collection of branches that form a

closed path returning to the same node without going through any other nodes or branches twice

Circuit Definitions

4

Example

How many nodes, branches & loops?

+ -

Vs Is

R1

R2 R3

+

Vo

-

5

Example

Three nodes

+ -

Vs Is

R1

R2 R3

+

Vo

-

6

Example

5 Branches

+ -

Vs Is

R1

R2 R3

+

Vo

-

7

Example

Three Loops, if starting at node A

+ -

Vs Is

R1

R2 R3

+

Vo

-

A B

C

8

THE ALGEBRAIC SUM OF THE CURRENTS ENTERING A NODE IS ZERO

Kirchhoff’s Current Law

I = 0

I1

I2

I3

I4

I5

I1 + I2 = I3 + I4 + I5

9

EQUALS CURRENTCURRENT OUTOUT

CURRENT CURRENT ININ

Example 0.5 A

0.3 A

? A

0.5 A - 0.3 A = 0.2 A

10

Kirchoff’s Current Law at B

Assign current variables and directions

Add currents IN, subtract currents OUT: I1 – I2 – I3 + Is = 0

+ -

Vs Is

R1

R2 R3

+

Vo

-

AB

C

I2

I1

I3

The sum of currents flowing into a node must be equal to sum of currents flowing out of the node.

Kirchoff’s Current Law

ii11 flows into the node

ii22 flows out of the node

ii33 flows out of the node

ii11 = = ii22 + + ii33

0i 0ii1 – i2 – i3 = 0

ii22 ii33

nodei1

Kirchoff’s Current Law

Example

Q: How much are the currents i1 and i2 ?

A: i2 = 10 mA – 3 mA = 7 mAi1 = 10 mA + 4 mA = 14 mA

+_

i1 4 mA i2

10 mA

3 mA

4 mA + 3 mA + 7 mA = 14 mA

node

13

Example 1Determine I, the current flowing out of the voltage

source.Use KCL

1.9 mA + 0.5 mA + I areentering the node.

3 mA is leaving the node.

V1 is generating power.

mAI

mAmAmAI

mAImAmA

6.0

)5.09.1(3

35.09.1

Example: Calculate the unknown currents in the following circuits.

Kirchhoff’s Current Law

15

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 areentering the node.

1.9 mA is leaving the node.

V1 is dissipating power.

mAI

mAmAmAI

mAImAmA

6.1

)5.03(9.1

9.15.03

Example 2

16

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

––vv11 + + vv22 + + vv3 3 = 0= 0

++__

+ + vv22 – – + +

vv33

– –

+ +

vv44

– –

+ +

vv11

– –

The sum of voltages around a closed loop is zero.

Example : • Calculate the unknown voltages in the given

circuit.

Kirchhoff’s Voltage Law

19

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.

20

Example 3 (con’t)

There are three possible loops in this circuit – only two include R1. Either loop may be used to determine VR1.

21

Example 3 (con’t)

If the outer loop is used: Follow the loop clockwise.

22

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.

23

Example 3 (con’t)

By convention, voltage drops are added and voltage rises are subtracted in KVL.

VV

VVV

R

R

2

035

1

1

Kirchhoff’s Voltage Law