Electric Circuits

• Direction of current flow in circuits for AP is from + to -

Circuit Diagrams

• Standard symbols:

Wires connected

cell

+ - switch

Wires not connected

Voltmeter

Ammeter

Fuse

Resistor

Bulb

• A resistor is any device that converts electrical potential

energy into other forms of energy, such as heat or light

Circuit Diagram

SERIES CIRCUITS

• Components connected one

after the other

•There is only one path

for the current to travel

PARALLEL CIRCUITS

• There is more than one path for the current to flow

• No wire is a perfect

conductor. There is

always some friction

applied to the current

flow.

• Unit is the Ohm ()

ELECTRICAL RESISTANCE

Factors that affect

resistance:

1)Type of material

• Metals have low

resistance

• some materials

have virtually no

resistance

2) Length of the conductor: the

longer the conductor, the greater

the resistance

3) cross-sectional area of the

conductor: the greater the area,

the lower the resistance

4) Temperature:

resistance increases

as temperature

increases.

• The resistance of a material at a certain

temperature is 𝑅 =𝜌l𝐴

• is the resistivity of the material (m)

• l is the length

• A is the cross sectional area

Ohm’s Law

• the potential difference between any two points in a circuit depends on the current flow and the resistance between the points

• V = IR

• V = potential difference in volts across 2 points in a circuit

• I = current in amperes

• R = resistance in ohms

• The slope of a V-I graph is the resistance

• Total resistance = VT/IT

• R1 = V1/I1

• R2 = V2/I2

Example

• A circuit has 36.0 mA flowing in it when it is connected to a 13.5 V power supply. Determine the resistance of the circuit.

A .

V .R

I

VR

310036

513

Watt's Law

• the power in an electric circuit is measured in watts and depends on the voltage and the current

• P = VI (Watt’s Law)

• since V = IR, P = I2R

• the power dissipated as heat in a resistor depends on the square of the current and the resistance

Example

• Which light bulb has the greatest resistance: a 100 W bulb or a 60.0 W bulb? Both are connected to 120 V

R100 W = 144

R60.0 W = 240

Resistors in Series and Parallel • resistors connected

end to end are in series

• the resistance of each add to the total

Rt = R1+ R2 + R3 ...

• each resistor in series has the same current flowing in it

Parallel Resistors • when resistors are in

parallel, some of the current will flow in each branch

• the total resistance in the circuit can be found by

3

R1

2R1

1R1

TR1

the total resistance will be less than the smallest individual resistor

Example

• If R1 = 10 , R2 = 10 , and R1 = 20 , what is the total resistance of the circuit?

• No calculator!

• 4

Example A) Determine the total current in the circuit below.

Solution • Resistance of parallel section: 6.00

• Resistance of total circuit: 27.0

• I = 0.852 A

Example B) What happens to the brightness of the bulb when a wire is connected between points A and B?

Solution The wire creates a short circuit, all the current flows through the wire, so the total resistance of the circuit is 21.0 Total I = 1.10 A, so the bulb gets brighter.

Kirchhoff's Laws

• batteries and generators increase potential differences in circuits

• electrons lose potential energy as they pass through resistors and other components

Kirchhoff’s Voltage Law

• another way of stating the Law of Conservation of Energy

• the total increase in potential difference (from batteries, etc.) = the total decrease in potential difference across resistors

• VT = V1 + V2 + V3

Example

• Determine the potential difference across each resistor.

• Hint: find total resistance first.

Solution

• Total Resistance = 10

• Total current = 4.0 A (through each resistor)

• V1 = I1R1 = 24 V (voltage drop across R1)

• V2 = I2R2 = 16 V (voltage drop across R2)

• VT = V1 + V2 = 40 V

• Potential difference across parallel resistors is the same.

• V = V1 = V2

Equivalent Circuit

V = V1 = V2

Example

• Determine I1 if R1 = 100 and V = 45.0 V.

• I1 = 0.450 A

Kirchhoff’s Current Law (Junction Rule)

• based on the Law of Conservation of Charge

• the current flowing into any point of a circuit must equal the current flowing out of the point

• I0 = I1 + I2

Example

• V = 45.0 V, R1 = 100 , R2 = 300 , calculate I0 and I2

Solution

• V = 45.0 V, R1 = 100 , R2 = 300 , I1 = 0.450 A

• RT = 75.0

• I0 = 0.600 A

• I2 = V/R2 = 0.150 A or:

• I2 = I0 – I1 = 0.150 A

Example

• Find the current through each resistor and the value of R2.

Solution

• Voltage drop across both resistors= 21V

11.5

1

21

14 A

VI

R

VI

• I1 = 1.5 A

• I2 = IT - I1

• I2 = 5.0 A - 1.5 A = 3.5 A

2

2

21 6.0

3.5

VR

I

VR

A

Example

• Find the current in each resistor and the voltage across R3.

Solution

• Step 1: find the total R of circuit.

• RT = Rparallel + R3 = 3.0 + 3.0 = 6.0

Solution

• Find the total current in circuit.

• IT = 2.0 A

• IT = current in R3 since all current must pass through R3.

Solution

• IT = 2.0 A = current in R3

• Since R1 = R2, ½ of the current will go through each resistor; I1 = I2 = 1.0 A.

Solution

• V3 = I3R3

• V3 = 2.0 A(3.0 ) = 6.0 V