Module 4 Electrons, Waves & Photons

Unit 3 Electrical Circuits

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4.3 Electrical Circuits

• 4.3.1 Series & Parallel Circuits

• 4.3.2 Internal Resistance

• 4.3.3 Potential Dividers

4.3.1 Series & Parallel Circuits

What are Kirchhoff’s

Laws?

Series Circuits

• Some things to remember first...– Kirchhoff’s first law:

• The sum of the currents entering an electrical junction will always be the same as the sum of the currents leaving the junction.

– Definition of Resistance:• R=V/I• Units are volts per Ampere• A resistance of 4.8Ω requires a voltage of 4.8V to produce a

current of 1A.

– Definition of Voltage:• V=W/Q• Units are Joules per Coulomb (Volts)• Energy transferred per unit charge

Some calculations in a series circuit

• What is the resistance of the whole circuit?

• What is the current flowing though each resistor?

• What is the pd across each resistor?

• What is the power to each resistor?

Resistance in a series circuit...

• Resistance in a series circuit is simply the sum total of all resistances.

Rtotal = R1 + R2 + R3 + ...

• Resistance of the circuit is therefore 30+10+60Ω = 100 Ω

Current in a series circuit...

• Current flowing through a series circuit is the same for all parts of the circuit.

• If total resistance is 100 Ω and emf is 12V,

• Current = 12V/ 100 Ω = 0.12AtotalR

emfCurrent =

PD in a series circuit...

• The pd across each component in a series circuit is calculated using V=IR

PD across R1 is 0.12A x 30Ω = 3.6V

PD across R2 is 0.12A x 10Ω = 1.2V

PD across R3 is 0.12A x 60Ω = 7.2V

If we now add these up their total is equal to the emf of the cell.

Power in a series circuit...

• The power to each component in a series circuit is calculated using P=IV

Power to R1 is 0.12A x 3.6V = 0.432W

Power to R2 is 0.12A x 1.2V = 0.144W

Power to R3 is 0.12A x 7.2V = 0.864W

The total power used by the components (0.432 + 0.144 + 0.864) equals the total power supplied to the circuit (12V x 0.12A = 1.44W).

Kirchhoff’s First Law

• The law of conservation of charge:

• The total current flowing into a junction equals the total current flowing out of the junction.

Σ𝐼𝑖𝑛= Σ𝐼𝑜𝑢𝑡

Kirchhoff’s Second Law

• The law of conservation of Energy.

• In any closed loop in a circuit, the sum of the electromotive forces is equal to the sum of the potential differences.

• Or, the energy put into a circuit must exactly match that which leaves the circuit.

Σ𝜀𝑖𝑛= Σ𝑉𝑜𝑢𝑡

Using Kirchhoff’s laws in series & parallel circuits.

How do we calculate

resistance in parallel circuits?

Remember, resistance in a series

circuit...

• Resistance in a series circuit is simply the sum total of all resistances.

Rtotal = R1 + R2 + R3 + ...

• Resistance of the circuit is therefore 30+10+60Ω = 100 Ω

Resistance in Parallel Circuits

• What happens to the supply current and total resistance as each switch is closed?

• With just the lamp on:– Supply current is 230V/1000Ω = 0.23A

• Include the computer:– 230V/250Ω = 0.92A is added

• And the telly:– 230V/210Ω = 1.10A is added

• Then turn on the heater:– 230V/50Ω = 4.60A is added

• Supply current is now 6.85A.• As more items are switched on, current increases which

would imply total resistance decreases.– Total resistance is now 230V/6.85A = 33.6 Ω

So,

• The total resistance of resistors in parallel must always be lower than the smallest individual resistor.

• Because adding current reduces total resistance.

What is the total resistance of this circuit?

• Assume negligible internal resistance (more on this later)

• Using Kirchhoff’s 2nd Law for the R1 resistor:– The PD across R1 equals the EMF of the battery, E.

• Similarly for R2 and R3.

• Using Kirchhoff’s 1st Law for current:

• And so:

E

321 IIII ++=

321

321R

E

R

E

R

EIIII ++=++=

• Since total current, I, equals E/R where R is the total resistance, we get:

• Cancelling E throughout gives:

321 R

E

R

E

R

E

R

E++=

321

1111

RRRR++=

What is internal

resistance?

Internal Resistance

• Resistor r is inside the cell. (it represents the resistance of the cell’s materials).

– Internal resistance is shown by the two dots at either side.

– Some diagrams show a dotted box around the components.

Given that Rtotal = R1 + R2 + R3 + ...

Rtotal = R + r

So, E = I(R+r) (Where E=emf)

And, E = IR +Ir

Therefore, E = V + Ir

Internal Resistance

So, E = I(R+r) (Where E=emf, and I=current)

And, E = IR +Ir

Therefore, E = V + Ir (where V is the pd across R)

This equation shows the difference between the emf of the cell (E) and the pd across its terminals (V). From this we can see the efficiency of the cell.

Given that Rtotal = R1 + R2 + ...

Then, Rtotal = R + r

Circuits with 2 sources of EMF

• The circuit here is similar to that in a car.

• There are 2 sources of EMF:– The battery

• 12V negligible internal resistance.

– The charging unit• 15V, 0.5Ω internal

resistance.

Why is the current supplied by the charging unit 6A?

Circuits with 2 sources of EMF

• Due to its negligible internal resistance, the battery maintains 12V across it.

• Therefore the terminal PD of the charging units must also be 12V.

• Therefore 3V is lost to the internal resistance of the charging unit.

• Since this resistance is 0.5Ω, the current through it must be 6.0A (3/0.5).

Why is the current supplied to the car’s overall resistance 10.0A?Why is the current supplied by the battery 4.0A?

Circuits with 2 sources of EMF

• If the total resistance of the car’s equipment is 1.20Ω and there is a PD of 12V across it, the current supplied must be 10.0A (12/1.20).

• If the charging unit supplies 6.0A, then 4.0A must be supplied by the battery.

What will happen to the current through the battery if the car’s headlamps are switched off and the overall resistance increases to 5.0 ohms?

Circuits with 2 sources of EMF

• Current to the Car’s remaining equipment will drop to 2.4A (12/5.0).

• The charging unit will still supply 6.0A.

• A current of 3.6A (6.0-2.4) will therefore flow in the opposite direction through the battery.– This current recharges the

battery.

4.3.1 Series & Parallel Circuits (review)

4.3.2 Internal Resistance

How can we calculate internal

resistance graphically?

All emf sources have an internal resistance

Interpreting the data

Variation of Power with Resistance

• With this circuit, if external resistance, R is varied the voltage across R and r changes.– Voltage across R and r

must add to 9V.

R /Ω r /Ω PD /V E /V

0 2 0.0 9.0

2 2 4.5 4.5

8 2 7.2 1.8

Infinite 2 9.0 0

Since,PD + E = 9 andRαV and soR/PD = r/E

r

R

V

9V2Ω

v

Lost volts

Variation of Power with Resistance

• With this circuit, Current will also vary with R:

I = E/RT

R /Ω r /Ω RT /Ω I /A

0 2 2 4.5

2 2 4 2.3

8 2 10 0.9

r

R

V

9V2Ω

v

Maximum current occurs when circuit is shorted (R=0).

Variation of Power with Resistance• Combining these 2

graphs we can see how output power varies with R:(Power = IV)

r

R

V

v

Maximum power output occurs when R = r.

More Circuit Calculations

• Copy & complete the table

6.0Ω

30Ω

20Ω

9.0V 2.0Ω

Component Resistance /Ω Current /A PD /V Power /W % of total Power

Resistor A

Resistor B

Resistor C

Resistor D (int)

Totals

4.3.2 Internal Resistance (review)

4.3.3 Potential Dividers

What are potential divider circuits and how are they used?

Potential Divider Circuits

• A potential divider circuit is essentially a simple series circuit with 2 resistors.

• With a fixed input voltage, the same current will flow through both resistors.

• Therefore the PD across each resistor is proportional to its resistance.

• The output voltage (PD) has been divided.

PD α

PD α

Try this...

• What is the PD across R1 (A to B) and across R2 (B to C)

2 Methods to use...

1. Find the current (I = V/RT) and then multiply this by each resistance.

2. Since the currents are the same through each resistor, and the total resistance is 10Ω, VABmust be 3/10 of 20V and VBC must be 7/10 of 20V.

• Either way VAB is 6V and VBC is 14V.

I like the “finding the current” method but I also quite enjoy the “proportion”

method, which is better?

• There’s only one way to find out...

• Actually, the proportion method is often the quickest and easiest since you don’t need to know the current – just that the current will be the same through each resistor.

In equation form...

• Where:– Vin is the input EMF– V1 is the PD across R1– V2 is the PD across R2

inVRR

RV

+=

21

22

inVRR

RV

+=

21

11

or...

• Where:– Vin is the input EMF– V1 is the PD across R1– V2 is the PD across R2

2

211

R

VRV =

2

1

2

1

R

R

V

V=

Uses of the Potential Divider Circuit

• Producing a variable PD.

• Controlling the heat output from a heater.

• Controlling streetlights when it gets dark/light.

Producing a variable potential difference

• Here we have a 9V supply connected to a 2kΩresistor and a 2kΩ variable resistor.

• What is the maximum output across the variable resistor?– 4.5V

• What is the minimum output across the variable resistor?– 0V

Controlling heat output from a heater

• This circuit uses a thermistor (12kΩ at 0oC, 0.25kΩ at 50oC).

• What is the voltage across the thermistor at these temperatures?– At 0oC, 9 x 12/(12+4)=6.75V

– At 50oC, 9 x 0.25/(0.25+4)=0.53V

• If we place a heater into the output voltage we can control the temperature of a room.

• Describe what happens as the room temperature drops.

H

Street lamp circuit

• Design a street lamp circuit and describe how it works.

The streetlamp circuit

• An LDR has a large resistance in the dark and low resistance in light.

• This circuit produces a high output voltage in the dark and a low voltage in light.

• The lamp is not connected as shown but via a switch called a relay.

Practical

• Draw a simple potential divider & calculate the PD voltages across each resistor, both resistors and the EMF. Use a multimeter to check these values after building the circuit.

• Build this circuit and describe how it could be used as a temperature probe.

• Replace the thermistor for an LDR and suggest some uses of this new circuit.

V

Potentiometers in Potential Divider Circuits

Produce a potential divider circuit using a potentiometer which acts as a dimmer switch for a lamp.

4.3.3 Potential Dividers (review)

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