2. Voltage, Current and Resistance Voltage is the amount of energy per charge move electrons in one point to another in a circuit. It is measured in volts (v) Current is the rate of charge flow and it is measured in amperes (A). Its unit is (I). Resistance is the opposition of current and it is measured in ohms (W)
3. DC Current vs. AC Current The current from a battery is always in the same direction. One end of the battery is positive and the other end is negative. The direction of current flows from positive to negative. This is called direct current, or DC.
4. DC Current vs. AC Current If voltage alternates, so does current. When the voltage is positive, the current in the circuit is clockwise. When the voltage is negative the current is the opposite direction. This type of current is called alternating current, or AC.
5. Series and Parallel Circuit In a series circuit , the current through each of the components is the same, and the voltage across the components is the sum of the voltages across each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.
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7. Current is the same at all points in a series circuit Parallel Circuits In contrast, in a parallel circuit, there are multiple paths for current flow. Different paths may contain different current flow. This is also based on Ohms Law Total resistance will be less than the smallest resistor
8. In Series circuit; the total current is equal Rtotal = R1 + R2 + R3 Vtotal = V1+ V2+ V3 In Parallel Circuit; the total voltage is equal 1 = 1 +1 + 1 + 1 Rtotal R1 R2 R3Rn Itotal = I1+ I2+ I3
10. Advantages of parallel circuits Parallel circuits have two big advantages over series circuits: 1. Each device in the circuit sees the full battery voltage. 2. Each device in the circuit may be turned off independently without stopping the current flowing to other devices in the circuit.
11. Ohms Law States that the voltage is directly proportional to the current in the circuit. It is discovered by the German Physicist George Simon Ohm. Ohm’s law is: V = IR , I = V/R, R= V/I
12. Unit Modifiers for Reference Smaller Deci = 10-1 Centi = 10-2 Milli = 10-3m Micro = 10-6 Nano = 10-9 Pico = 10-12p Fento = 10-15 Larger Kilo = 103k Mega = 106 Giga = 109 Tera = 1015
13. Series Circuit Analysis Example #1 A 4v battery is placed in a series circuit with a 2 resistor. What is the total current that will flow through the circuit? 2 4v I = ?
17. Parallel Circuit Analysis Example # 3 A 220v is connected in parallel with the load. It has a resistance of 5ohms and 10ohms. Find the Total current and the I1and the I2 10 5 220v
19. Example #4 Find the total voltage and the total resistance of the load if the total current is 15A and it has a R1 of 6 ohms and R2 of 2 ohms. 6 V=? 2 IT =15
20. Solution Given: R1 = 6 R2 = 2 IT =15 A RT = R1 R2 / R1 + R2VT= IT/ RT = (6 x 2) / 8 = 15 A x 1.5 = 1.5 = 22.5 v I1 = V/ R1 I2 = V/ R2 = 22.5 / 6 = 22.5 / 2 = 3.75 A= 11.25 A
21. Electric Power, AC, and DC Electricity The watt (W) is a unit of power. Power is the rate at which energy moves or is used. Since energy is measured in joules, power is measured in joules per second. One joule per second is equal to one watt.
22. Power in electric circuits One watt is a pretty small amount of power. In everyday use, larger units are more convenient to use. A kilowatt (kW) is equal to 1,000 watts. The other common unit of power often seen on electric motors is the horsepower. One horsepower is 746 watts.
23. Formula of Power Power (watts) Voltage (volts) P = VI Current (amps
24. Example # 5 A light bulb with a resistance of 1.5Ω is connected to a 1.5-volt battery in the circuit shown at right. Calculate the power used by the light bulb.
25. Solution Given: V = 1.5v R = 3 I =V / R P = V I = 1.5 / 3 = 1.5 x 0.5 = 0.5 A = 0.75 watts
26. Paying for electricity Electric companies charge for the number of kilowatt-hours used during a set period of time, often a month. One kilowatt-hour (kWh) means that a kilowatt of power has been used for one hour. Since power multiplied by time is energy, a kilowatt-hour is a unit of energy. One kilowatt-hour is 3.6 x 106 joules.
27. Exercises: I. Identification: Direction: Identify the following questions. 1. It is the amount of energy per charge move electrons in one point to another in a circuit. It is measured in volts? 2. It is a type of current which the current from a battery is always in the same direction? 3. It is the opposition of current? 4.It is the rate of charge flow? 5. It is the rate at which energy moves or is used?
28. 6. One horsepower is equivalent to how many watts? 7. In series circuit the total _______ is equal /the same. 8. In parallel circuit the total _______ is equal/the same. 9. A circuit where the voltage across each of the components is the same, and the total current is the sum of the currents through each component. 10. It States that the voltage is directly proportional to the current in the circuit. 11. A circuit where the current through each of the components is the same, and the voltage across the components is the sum of the voltages across each component. 12. Ohm’s Law is discovered by whom a German Physicist.
29. 13. what does kwh means? 14. How many joules are there in a kwh? 15. The other common unit of power often seen on electric motors is? II. Problem Solving: 1. A 115 volt is connected in parallel with the load. It has a I1 3 A and R2 15 ohms. Find the IT and the RT. I1 115v I1 = 3A 15
30. 2. Resistors 1, 2 ,3 has a resistance of 4 , 8 and 12 are connected in series with a total current of 25 A. Find the total voltage and the V1 , V2 , V3. 3. A motor has a voltage of 220 volts and a resistance of 35. Find the current and the power. V1 4 IT = 25A VT = ? 8 12
31. Key Answers I. Identification 1. voltage 2. direct current 3. resistance 4. current 5. power 6. 746watts 7. current 8. voltage
32. 9. Series Circuit 10 Ohm’s Law 11. Parallel Circuit 12. George Simon Ohm 13. kilo watt hour 14. 3.6 x 106joules 15. horsepower II. Problem Solving 1. Given: V = 115 v R2 = 15 I1 = 3A R1 = V / I1 I2 = V/ R2 RT = R1 + R2IT = V/ RT = 115/ 3 A = 115/15 = 15 + 38.33 = 115 / 53.33 = 38.33 = 7.67A = 53.33 = 2.16A
