Ch 23: Series and Parallel Circuits O H M S L A W A N D R E S O L V I N G C I R C U I T S

Transcription

1 Ch 23: Series and Parallel Circuits O H M S L A W A N D R E S O L V I N G C I R C U I T S

2 Series Circuits When charge has only one complete path to follow, the current, I, is the same everywhere. This is a series circuit. A break anywhere stops all current from flowing. From Ohm s law: I = V / R we can calculate the current, I, in the circuit. The equivalent Resistance, R eq, in a circuit is the sum of the individual resistors. R eq = R 1 + R 2 + R Regardless of the resistor value, each gets the same current. How does this effect brightness?

3 Series Circuits

4 More on resistors in Series If the battery voltage is constant, the more objects added to the circuit, the the overall resistance. As resistance increases, what happens to current? Which circuit would be brighter? 6v, 3Ω; or 6v, 6Ω?

5 Practice Problem 1 Three 20Ω resistors are connected in series across a 120 V generator. What is the equivalent resistance? What is the current in the circuit? R eq = sum R eq = 20Ω + 20Ω + 20Ω = 60Ω I = V source /R eq = 120V / 60Ω = 2A

6 Voltage drops in Series circuits Voltage or potential, in a circuit comes from either a battery (DC) or a generator (AC). The net change in potential as current moves through the circuit is zero. In a series circuit, voltage is dropped off at each resistor to supply the needed current. Voltages drops add back up to equal the voltage source: V source = V 1 + V 2 + V 3 + Voltage drop is calculated by I total * R obj

7 Practice Problem 2 A 10Ω resistor, a 15Ω resistor, and a 5Ω resistor are connected in series to a 90 V battery. What is the equivalent resistance of the circuit? What is the current in the circuit? Calculate the voltage drops across each resistor.

8 Practice Problem 2 Solution R eq = 10Ω + 15Ω + 5Ω = 30Ω I = V / R eq = 90V / 30Ω = 3A, (everywhere) V drop for 10Ω = I*R = 3A * 10Ω = 30 V V drop for 15Ω = I * R = 3A * 15Ω = 45 V V drop for 5Ω = I * R = 3A * 5Ω = 15 V V batt = V 1 + V 2 + V 3 90 V = 30 V + 45 V + 15 V Do CD 35-1 ws (only parts about Series circuits)

9 Parallel Circuits A circuit where there are several paths for current to take, is called a parallel circuit. Each branch of the circuit has its own path to the battery. A break in any path, stops current in that path only! (Other paths still carry current)

10 Parallel Circuits

11 Voltage in Parallel Whatever voltage exists at the source, the same voltage is dropped across EACH branch of a parallel circuit. V batt = V 1 = V 2 = V 3 =

12 Equivalent Resistors in Parallel The more paths a circuit has, the harder the battery or voltage source must work. However, the more paths, the LOWER the overall resistance! If resistance is low, current is higher that s why the potential source works so hard! We solve for R eq by summing the reciprocal of each resistor. 1 R eq 1 R 1 1 R 2 1 R 3

13 Practice Problem Three resistors, 60Ω, 30Ω, 20Ω, are connected in parallel across a 90 V battery. Find the current through each branch of the circuit. Find the equivalent resistance of the circuit. Calculate the current through the battery.

14 Practice Problem solution Sketch the diagram Recall what is constant in a parallel circuit: voltage across each branch. Use Ohm s law to find the current through each R branch. I=V/R I=90V / 60Ω = 1.5A, 90V / 30Ω = 3A, 90V / 20Ω = 4.5A. I thru battery = sum of I s I battery = 1.5A + 3A + 4.5A 1 R eq I battery = 9A R eq 10

15 Your turn to Practice a little Please do the Right side of CD 35-1 worksheet about Parallel Circuits now. Remember the Rules for Series and Parallel circuits are DIFFERENT.

16 Power in Circuits Electric power is measured in watts, W. Since 1W = 1J/s like all other power types, and 1Amp is 1C/s, then when current, I is multiplied by voltage, V (1J/C) we get power. Ex: 3A * 120 V = 3 C/s * 120 J/C = 360 W Power = Current x Voltage P = I * V REMEMBER PIVVIR P=I*V, V=I*R

17 Energy Transfer in Circuits Recall Electric energy in a circuit can take on many forms: light, heat, mechanical energy, sound, etc The rate of changing energy is called Power. P = E / t. Remember current is charge/sec (q/t) and power is I x V, so in any resistor V = I x R and by substitution, Power P = I 2 R. The total energy that can be converted in a resistor is E = P * t = I 2 R*t and most energy is measured in kilowatt*hours (or watt*sec.) A kw-h is 1000watts delivered for 1 continuous hour.

18 Your turn to Practice Please do Ch 23 Practice problems page 534 #3,#4 Practice Problems pg 540 #11 Ch 23 Review pg #s 4, 5, 12, 15, 19, 21, 22 Ch 23 Review pg 552 #s 24, 28, 35