Simple Circuits&Kirchoff’s Rules

Parallel Circuit Series Circuit

Simple Series Circuits Each device occurs one after the other

sequentially. The Christmas light dilemma: If one light goes

out all of them go out.

R1

R2

+V

R3

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

to zero.

Vsource + V1 + V2 + V3 = 0

Where we consider the source voltage to be positive and the voltage drops of each device to be negative.

Vsource = V1 + V2 + V3

Since V = IR (from Ohm’s Law):

Vsource = I1R1 + I2R2 + I3R3

V1

V2

+V

V3

R1

R2

+V

R3

Simple Series Circuit - Conservation of Charge

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

IT = I1 = I2 = I3

I

Simple Series Circuit – Determining Requivalent

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

Vsource = V1 + V2 + V3

Vsource = I1R1 + I2R2 + I3R3

Due to conservation of charge, ITotal = I1 = I2 = I3, we can factor out I such that

Vsource = ITotal (R1 + R2 + R3) Since Vsource = ITotalRTotal:

RTotal = REq = R1 + R2 + R3

Simple Parallel Circuit A parallel circuit exists where components are

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

homes.

R1 R2

+V

R3

Simple Parallel Circuits

Since each device is connected across the same voltage source:

Vsource = V1 = V2 = V3

V1 V2

+V

V3

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

Your water comes into the house under pressure. Each faucet is like a resistor 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 current from the voltage source (pressurized water supply) is equal to the sum of the currents (flow of water through faucet and drain) in each of the resistors (faucets)

ITotal = I1 + I2 + I3

+V

ITotal

I1 I3I2

ITotal

Simple Parallel Circuit – Determining Requivalent

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

ITotal = I1 + I2 + I3 or

Since Vsource = V1 = V2 = V3 we can substitute Vsource in (1) as follows

)1(3

3

2

2

1

1

R

V

R

V

R

VI total

)2(321 R

V

R

V

R

VI sourcesourcesourcetotal

Simple Parallel Circuit – Determining Requivalent

What it the total resistance in a parallel circuit (cont.)? However, since ITotal = Vsource/RTotal substitute in (2)

as follows

Since Vsource cancels, the relationship reduces to

)2(321 R

V

R

V

R

V

R

V sourcesourcesource

total

source

)3(1111

321 RRRReq

Note: Rtotal has been replaced by Req.

Kirchoff’s Rules Loop Rule (Conservation of Energy):

The sum of the potential drops (Resistors) equals the sum of the potential rises (Battery or cell) around a closed loop.

Junction Rule (Conservation of Electric Charge): The sum of the magnitudes of the

currents going into a junction equals the sum of the magnitudes of the currents leaving a junction.

Rule #1: Voltage Rule (Conservation of Energy)

0Loop

V

R1

R2

+V

R3

ΣV

Vsource – V1 – V2 – V3 = 0

Rule #2: Current Rule (Conservation of Electric Charge)

I1

I3

I2

I1 + I2 + I3 = 0

0Junction

I

Example Using Kirchoff’s Laws

Create individual loops to analyze by Kirchoff’s Voltage Rule.

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

          

 

+

+

R3 = 5Ω

2 = 5V

I1 I2

I3

1 = 3V

R1 = 5ΩR2 = 10Ω

Ex. (cont.)

Apply Kirchoff’s Current Rule (Iin = Iout):

I1 + I2 = I3 (1)

Apply Kirchoff’s Voltage Rule to the left loop (Σv = 0):

1 – V1 – V2 = 0

1 – I1R1 – I3R2 = 0

Substitute (1) for I3 to obtain:

1 – I1R1 – (I1 + I2)R2 = 0 (2)

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

loop:

2 – V3 – V2 = 0

2 – I2R3 – I3R2 = 0

Substitute (1) for I3 to obtain:

2 – I2R3 – (I1 + I2)R2 = 0 (3)

Ex. (cont.) List formulas to analyze.

I1 + I2 = I3 (1)

1 – I1R1 – (I1 + I2)R2 = 0 (2)

2 – I2R3 – (I1 + I2)R2 = 0 (3) Solve 2 for I1 and substitute into (3)

1 – I1R1 – I1R2 – I2R2 = 0

– I1R1 – I1R2 = I2R2 – 1

I1 (R1 + R2) = 1 - I2R2

1 - I2R2

(R1 + R2)

I1 =

(1 - I2R2) (R1 + R2)

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

Ex. (cont.)

2 – I2R3 – + I2 R2 = 0

2 (R1 + R2) – I2R3 (R1 + R2) – 1R2 + I2R22

– I2R2 (R1 + R2) = 0

[

[

1 - I2R2

(R1 + R2)R2 – I2R2 = 02 – I2R3 – [

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

5V(5Ω+10Ω) – I25Ω (5Ω+10Ω) – 3V(10Ω) + I2(10Ω)2 – I210Ω (5Ω+10Ω) = 0

I2 = 0.36 A

Ex. (cont.)

Plug your answer for I2 into either formula to find I1 1 – I1R1 – (I1 + I2)R2 = 0

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

I1 =1 - I2R2

(R1 + R2)

I1 =3V – (0.36A)(10)

(5 + 10)

I1 = -0.04A

Ex. (cont.)

Use formula (1) to solve for I3

I1 + I2 = I3 -0.04A + 0.36A = 0.32A

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