Simple Circuits&Kirchoff’s Rules

Simple Series Circuits Each device occurs

sequentially. The light dilemma: If light

goes all of them go .

Simple Series Circuit - Conservation of Energy In a series circuit, the of the is

equal to .

Vsource +

Where we consider the source voltage to be and the voltage drops of

each device to be .

Vsource =

Since V = ( ):

Vsource =

R1

R2

+V

R3

Simple Series Circuit - Conservation of Charge

In a series circuit, the same amount of passes through each device.

IT =

Simple Series Circuit – Determining Requivalent

What it the total in a series circuit? Start with of

Vsource =

Vsource = Due to conservation of charge, ITotal = I1 = I2 =

I3, we can factor out I such that

Vsource = Since Vsource = :

RTotal = REq =

Simple Parallel Circuit A parallel circuit exists where components are

connected across the same . Parallel circuits are similar to those used in

.

+V

Simple Parallel Circuits

Since each device is connected across the :

Vsource =

+V

In parallel circuits, the is equal to the of the through each individual leg. Consider your home plumbing:

Your water comes into the house under pressure. Each faucet is like a that occupies a leg

in the circuit. You turn the valve and the water flows.

The drain reconnects all the faucets before they go out to the septic tank or town sewer.

All the water that flows through each of the faucets adds up to the total volume of water coming into the house as well as that going down the drain and into the sewer.

This analogy is similar to current flow through a parallel circuit.

Simple Parallel Circuits AnalogyHow Plumbing relates to current

Simple Parallel Circuits – Conservation of Charge & Current

The total from the voltage source (pressurized water supply) is equal to the sum of the (flow of water through faucet and drain) in each of the (faucets)

ITotal =

+V

Simple Parallel Circuit – Determining Requivalent

What it the total resistance in a parallel circuit? Using conservation of charge

ITotal =

or

Since Vsource = V = V = V we can substitute Vsource in (1) as follows

Simple Parallel Circuit – Determining Requivalent

What it the total resistance in a parallel circuit (cont.)? However, since ITotal = / substitute

in (2) as follows

Since Vsource cancels, the relationship reduces to

Note: Rtotal has been replaced by .

Kirchoff’s Rules Loop Rule (Conservation of ):

The sum of the ( )equals the sum of the

( ) around a closed loop.

Junction Rule (Conservation of Electric ):

The sum of the magnitudes of the going into a junction

equals the sum of the magnitudes of the leaving a junction.

Rule #1: Voltage Rule (Conservation of )

R1

R2

+V

R3

ΣV

Vsource – V1 – V2 – V3 =

Rule #2: Current Rule (Conservation of Electric )

I1

I3

I2

I1 + I2 + I3 =

Example Using Kirchoff’s Laws

Create individual loops to analyze by Kirchoff’s .

Arbitrarily choose a direction for the current to flow in each loop and apply Kirchoff’s

.

          

 

+

+

R3 = 5Ω

2 = 5V

I1 I2

I3

1 = 3V

R1 = 5ΩR2 = 10Ω

Ex. (cont.) Apply Kirchoff’s Current Rule ( = ):

I1 + I2 = (1)

Apply Kirchoff’s Voltage Rule to the left loop (Σv = ):

1 – – =

1 – – = Substitute (1) to obtain:

1 – – ( ) = (2)

Ex. (cont.) Apply Kirchoff’s Voltage Rule to the right

loop:

2 – – =

2 – – = Substitute (1) to obtain:

2 – – ( ) = (3)

Ex. (cont.) List formulas to analyze.

I1 + I2 = (1)

1 – – ( ) = (2)

2 – – ( ) = (3) Solve 2 for I1 and substitute into (3)

1 – – – =

– – = – 1

I1 ( ) = 1 -

1 - ( )

I1 =

( ) ( )

Plug in known values for R1, R2, R3, 1 and 2 and then solve for I2 and then I3.

Ex. (cont.)

2 – – + =

2 ( ) – ( ) – + – ( ) = 0

[

[

( ) – = 2 – – [[

Multiply by (R1 + R2) to remove from denominator.

I2 = A

Ex. (cont.)

Plug your answer for I2 into either formula to find I1 1 – – ( ) =

What does the tell you about the current in loop 1?

I1 =1 - ( )

I1 =3V – ( )( ) ( + )

I1 =

Ex. (cont.)

Use formula (1) to solve for I3

I1 + I2 = + =

How to use Kirchhoff’s LawsA two loop example:

•Analyze the circuit and identify all circuit nodes and use KCL.

(2) 1 I1R1 I2R2 = 0(3) 1 I1R1 2 I3R3 = 0(4) I2R2 2 I3R3 = 0

1

2

R1

R3R2

I1 I2

I3

(1) I1 = I2 + I3

• Identify all independent loops and use KVL.

How to use Kirchoff’s Laws

1

2

R1

R3R2

I1 I2

I3

•Solve the equations for I1, I2, and I3:

First find I2 and I3 in terms of I1 :

1 1 2 1 11 1

2 3 2 3

( )R R

I IR R R R

1 1 2

2 31

1 1

2 3

1

R RI

R RR R

2 1 1 1 2( ) /I I R R

3 1 2 1 1 3( ) /I I R R

From eqn. (2)

From eqn. (3)

Now solve for I1 using eqn. (1):

Let’s plug in some numbers

1

2

R1

R3R2

I1 I2

I3

1 = 24 V 2 = 12 V R1= 5R2=3R3=4

Then, and

I1=2.809 A I2= 3.319 A, I3= -0.511 A