1. Electric Circuits

2. What is an Electric Circuit? Electric Circuit: is a closed loop through which charges can continuously move. Need to have complete conducting loop from the positive terminal to the negative terminal. Ben Franklin - conventional current direction.

3. Practice with Circuits. Use the following material to construct 4 ways to complete the circuit.

4. Electric Circuit cont.

5. Simple Circuit Diagram The following diagrams represent the basic components in a simple circuit.

6. Other components in circuits. The following symbols will be used when drawing your circuits in your notebooks. (see page 701 of your text book for more symbols.)

7. Applying Ohm’s Law Ex. Find the current in the following Circuit. ∆V = 10V R = 5 Ω I = ∆V =IR I = 10V/5 Ω I = 2A

8. Series Circuits In a series circuit, each device is connected in a manner such that there is only one pathway by which charge can flow through the circuit. Since there is only one pathway through the circuit, adding more devices results in more overall resistance.

9. Equivalent resistance in Series Circuits. In a series circuit, each device is connected in a manner such that there is only one pathway by which charge can traverse the circuit. The equivalent resistance of loads in series is the sum of the resistances of the individual loads. R eq = R 1 + R 2 + R 3 +...

10. Determine the equivalent resistance and current for the following circuit. R1 = 3 Ω R2 = 10 Ω R3 = 5 Ω Req = R1 + R2 + R3 Req = 3 Ω +10 Ω +5 Ω ∆V = IReq Req = 18 Ω I=∆V/Req I=9V/18 Ω I = 0.5A

11. Voltage drop for series circuits Voltage drop: is the loss in electric potential energy as the charge moves through the circuit. (At each resistor) For series circuits. I battery = I 1 = I 2 = I 3 =... And V battery = V 1 + V 2 + V 3 +... So, we must calculate the individual voltage drop at each resistor. V 1 = I R 1 V 2 = I R 2 V 3 = I R 3

12. Voltage drop for series circuits cont. From previous example: I = 0.50A R1 = 3 Ω ∆V 1 = I R 1 = (0.50A)(3 Ω ) = 1.5V R2 = 10 Ω ∆V 2 = I R 2 = (0.50A)(10 Ω ) = 5V R3 = 5 Ω ∆ V 3 = I R 3 = (0.50A)(5 Ω ) = 2.5V Note that the voltage drops add up to the total voltage supplied by the battery.

13. Parallel Circuits In a parallel circuit, each device is placed in its own separate branch. The presence of branch lines means that there are multiple pathways by which charge can traverse the circuit. The inverse of the equivalent resistance is the sum of the inverses of the individual resistances. 1 / R eq = 1 / R 1 + 1 / R 2 + 1 / R 3 +...

14. Determine the equivalent resistance and current in the following circuit. R1 = 10 Ω R2 = 2 Ω R3 = 1 Ω 1/Req = 1/R1 + 1/R2 + 1/R3 ∆V = IReq 1/Req = 1/10 Ω + 1/2 Ω + 1/1 Ω I = ∆V /Req 1/Req = (1 + 5 + 10)/10 I = 9V/0.625 Ω Req = (10/16) Ω = 0.625 Ω I = 14.4A

15. Voltage drop for Parallel Circuits. In a parallel circuit, a charge does not pass through every resistor; rather, it passes through a single resistor. (Different paths) V battery = V 1 = V 2 = V 3 =... And I total = I 1 + I 2 + I 3 +... We must calculate the current at each resistor I 1 = V 1 / R 1 I 2 = V 2 / R 2 I 3 = V 3 / R 3

16. Voltage drop for Parallel Circuits cont. From previous example: Itot = 14.4A R1 = 10 Ω I 1 = V 1 / R 1 = (9V/ 10 Ω ) = 0.9A R2 = 2 Ω I 2 = V 2 / R 2 = (9V/ 2 Ω ) = 4.5A R3 = 1 Ω I 3 = V 3 / R 3 = (9V/1 Ω ) = 9A Note the that sum of the 3 currents equals the total current of the circuit.