1. Ohm My Goodness!! Ohm’s Law! Regan Via Chemphys 1-2

2. Ohm’s Law Ohm’ law describes the relationship between three important components of electrical circuits: – Voltage – Current – Resistance

3. 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

4. Current Measured in amperes (Coulombs/second) Describes the rate at which charge flows past a point per unit time I=dq/dt

5. 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

6. 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

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

8. Examples 6 Ω 8 Ω 9 V 18V 12 3 4 9 V 0.6 amps 3 V 6 Ω 8 Ω 3 V 1.8 amps 9 V

9. Example 1 We can find the total current of the circuit by dividing the voltage by the total resistance R total = 6+8=14 Ω I=V/R=(9 V)/(14 Ω)=0.64 amps 6 Ω 9 V 1 8 Ω

10. 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 2 9 V 0.6 amps 3 V R total =(9 V)/(0.6 amps)=15 Ω R bulb =(3 V)/(0.6 amps)=5 Ω R resistor =15 Ω-5 Ω=10 Ω

11. 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 18V 3 6 Ω 8 Ω R total =6+8=14 Ω I=(18 V)/(14 Ω)=1.29 amps

12. 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 4 3 V 1.8 amps 9 V R total =(9 V)/(1.8 amps)=5 Ω R bulb =(3 V)/(1.8 amps)=1.67 Ω R resistor =5 Ω-1.67 Ω=3.33 Ω