Potential Divider Circuits

• Sometimes elements on circuit need less than the total voltage potential.

• Circuit that can be adjusted to divide the p.d. (voltage) called potential divide.

• Which type of circuit divides voltage?

• Series circuits divide total input voltage between resistors.

A simple series circuit divides the voltage among resistors. Sketch the circuit below.Derivation of the potential divider equation.

V out

To find partial voltage: V2 = (Vtot) R2

R1 + R2.

Basic pd divide equation.

Around R2 I = V2/R2 Sub in & rearrange

V2 = Vtot R2 (R1 + R2)

• I = Vtot

– Req

• I = Vtot

(R1 + R2)

The equation: V2 = (Vtot) R2

R1 + R2.

Describes the output voltage of a particular resistor as a fraction of the total current based on the fraction of resistance offered by a particular resistor.

Ex 1: 2 resistors are in series to 10V. Use the equation to find V1& V2.around R1 & R2.

V1 = 10 V x (3 ) = 2 V

15

3 12

10 V

V2 = V – V1 = 10 – 2 – 8V.

V2 = (Vtot) R2

R1 + R2.

Ex 2: Two resistors are connected to 8V as below. Find V1& V2.

5 k 12 k8 V

• V1 = 2.4 V

• V2 = 5.6 V

V1 V2

Ex 3: Given 3, 5 k resistors as below, find the Vab, Vbc.

Compose an equivalent circuit

a

b

c

8V

a

b

c

5 k

2.5 k

8V

Apply the equation

• Vab = 8 x (5 x 7.5) = 5.3 V.

• Vbc = 8 - 5.3 = 2.7 V.

Proportionality Equation for series circuit.

• Since I = V1/R1 = V2/R2.

• V1/R1 = V2/R2.

• V1/V2 = R1/R2.

R1 R2

I

V1V2

Ex 4: Use proportionality to find V1.

• 2.5 V

4 8

V1 5 V

Ex 5: Choose an equation to solve for V on the 10 resistor.

• Using potential divide equation

• 50 V

7 10

100 V

3

Potentiometersvariable voltage

Kerr 145 potentiometer

• Demo, explanation, problem.

• Nov 04 pg 15

• Kerr pg 150 #32.

Use your data booklets to sketch the circuit using the IB

symbols.

It’s as if wire is many resistors in series. The voltmeter can read the pd around some of the resistance by touching specific points.

V

6.0 V

Simplest potentiometer has slider contact.

• At top V = 2 V

• Halfway V = 1 V

• At bottom V = 0V.

2V

• This is equivalent.

• Why?

2V

Ex: If there are 30cm above & 70cm below the contact, what is the voltmeter reading?

• The reading around70cm.

• Partial R =• 70 cm

• Total R =• 100 cm

• V = 2V x (70/100)

• 1.4 V

2V30cm

70cm

Ex: If there are 40cm above & 60cm below the contact, what is the voltmeter reading?

• The reading around 60cm.

• Partial R =• 60 cm

• Total R =• 100 cm

• V = 2V x (60/100)

• 1.2 V

2V40cm

60cm

I can add another resistor to the circuit and determine the reading on the voltmeter.

The maximum resistance of this variable resistor is 100 . When the sliding contact, S, is at contact A, what will the voltmeter read?

• It will read 6V (the resistor is bypassed): the maximum reading of voltage in this circuit.

• What is the reading of the voltmeter when the sliding contact is moved to B?

We have, in effect, the following situation.

Therefore, the voltmeter will read 3V.

It is not possible to make the reading of the voltmeter vary from zero up to the full voltage of the supply using this circuit.

When all 3 connections of the 100 variable resistor are used as below, what will the voltmeter reading be? X is a wire, the contact is at B.

• Zero. The contact bypasses the bulb like a short circuit. The wire x is in parallel with the bulb.

The variable resistor is 100 . What resistance does it supply on this circuit?

• 50 , since resistance is directly proportional to length.

Uses of potentiometers

• To control volume.

• To control lights (dimmer switch).

Electrical Sensors

• Devices whose resistance changes with changing physical conditions.

Strain Gauge

When stretched R increases b/c it gets longer &

skinnier.

Find symbol put in notes.

Light SensorLDR – light dependent resistor

Resistance decreases when light shines on it.

See symbol.

Thermistor – resistance varies with temperature. Usually the R decreases w/increasing temperature (counter to our wires).

Symbol.

Using Potential Divide with Sensors

12V

LDR - When light stops shining, its resistance increases and Vout increases. If it Vout gets high enough, the switch will activate perhaps putting on lights. The p.d. divide is needed to create a p.d. to activate switch.

switch

Use of Sensors

Ex: The cell has an emf of 12-V and no internal resistance. The p.d. required to activate the switch is 5-V. Find the

value of R1 that will cause the switch to activate when the resistance of the LDR is 200 k.

switch

Rearrange:

Vout = (Vin) R2

R1 + R2.

solve for R1.

R1 = (R2) Vin - Vout

V out.

Solve for R1.

280 k .

Which would be a good sensor to use with a fire alarm?

What would be a good use for a strain gauge?

Hwk. Read Hamper.115 – 123

do pg 123 #21

Nov 04 pg 15