1. What we will do today Introduce potential divider circuits. Devise a formula comparing resistances and voltages in potential divider circuit.

2. Potential Dividers (Discussion) A potential divider circuit is a series circuit where the voltage (potential) is split between two resistors. What will our circuit diagram look like? Draw the circuit diagram in your class jotter.

3. Potential Dividers A potential divider circuit is a series circuit where the voltage (potential) is split between two resistors. It is also known as a voltage divider. It is often shown on its side:

4. Potential Dividers From the equation for V s in a series circuit we know that V s = V 1 + V 2 Therefore there is a voltage drop across each component in a series circuit. We can calculate these potential differences using a voltmeter. However we can also calculate these values when their resistance is known.

5. Experiment Connect two resistors in series to a battery. Record the resistance of each resistor. Measure the voltage of the battery. Measure the voltage of each resistor. Fill in the table on the next slide.

6. Results R 1 (  )R 2 (  )R1R2R1R2 V 1 (V)V 2 (V)V1V2V1V2

7. Conclusion We can see that the ratio R 1 / R 2 is the same as V 1 / V 2. R1= V1R1= V1 R2 V2R2 V2 A potential divider circuit consists of a supply voltage and two resistors in series which divide the supply voltage between them.

8. What we have learned For two resistors in series, we have the ratio R 1 = V 1 R2 V2R2 V2 Using this ratio we can now work out the p.d. across components when knowing their resistances.

9. Example 1 R 1 = ? R 2 = 200 Ω V 1 = 1.5 V V 2 = 3 V R 1 = V 1 R2 V2R2 V2 R 1 = 1.5 200 3 R 1 = 1.5x 200 3 R 1 = 100 Ω

10. What if you’re asked about R 2 or V 2 ? If asked about these values (the denominators) then simply flip both sides of the equation. So: R 1 = V 1 R2 V2R2 V2 Becomes: R2=V2R2=V2 R1 V1R1 V1 Then input values as in example 1

11. Example 2 R 1 = 100 Ω R 2 = 200 Ω V 1 = 2 V V 2 = ? R 1 = V 1 R2 V2R2 V2 R2=V2R2=V2 R1 V1R1 V1 200=V 2 100 2 200 x 2=V 2 100 V 2 = 4V

12. 2009 Qu: 10 E