Ohm My Goodness!! Ohm’s Law!

Regan ViaChemphys 1-2

Ohm’s Law

• Ohm’ law describes the relationship between three important components of electrical circuits:– Voltage– Current– Resistance

Voltage

• Measured in volts• Measures the potential difference across two

points• A potential difference is required for a

current to flow• Voltage is equal to the energy required to

move a charge from one point to another divided by the magnitude of the charge

• V=-∫E dr

Current

• Measured in amperes (Coulombs/second)• Describes the rate at which charge flows past

a point per unit time• I=dq/dt

Resistance

• Measured in ohms (Ω)• Describes a components opposition to the

flow of an electric current• Can be thought of the friction of electrical

circuits• Resistivity of an object is proportional to

length over cross sectional area

How do they all relate???

• Ohm’s law says the voltage across two points is the product of the current and the resistance across the two points– V=IR

• Can be written as– I=V/R– R=V/I

Applications

• Using Ohm’s law we can determine the values of voltage, current and resistance across electrical components in a circuit

Examples6 Ω

8 Ω9 V

18V

1 2

3 4

9 V

0.6 amps

3 V

6 Ω

8 Ω 3 V

1.8 amps

9 V

Example 1

• We can find the total current of the circuit by dividing the voltage by the total resistance

Rtotal= 6+8=14 ΩI=V/R=(9 V)/(14 Ω)=0.64 amps

6 Ω

9 V1 8 Ω

Example 2• We can find the total

resistance by dividing the voltage by the total current

• Resistance across the bulb equals the voltage difference divided by the current

• Resistance across the resistor equals the total resistance minus the resistance of the bulb

29 V

0.6 amps

3 V

Rtotal=(9 V)/(0.6 amps)=15 ΩRbulb=(3 V)/(0.6 amps)=5 ΩRresistor=15 Ω-5 Ω=10 Ω

Example 3

• The circuit is identical to the circuit in example 1 but the voltage is doubled. Let’s see what happens…

• We can find the total current by dividing the voltage by the total resistance

• With double the voltage, the current is doubled

18V3

6 Ω

8 Ω

Rtotal=6+8=14 ΩI=(18 V)/(14 Ω)=1.29 amps

Example 4• The circuit is identical to the

circuit in example 2 but the current is tripled. Let’s see what happens…

• Resistance across the bulb equals the voltage difference divided by the current

• Resistance across the resistor equals the total resistance minus the resistance of the bulb

• The total resistance of the circuit is a third of the resistance in example 2. The lower the resistance, the higher the current

43 V

1.8 amps

9 V

Rtotal=(9 V)/(1.8 amps)=5 ΩRbulb=(3 V)/(1.8 amps)=1.67 ΩRresistor=5 Ω-1.67 Ω=3.33 Ω