This document discusses electrical concepts including voltage, current, resistance, Ohm's Law, and circuit types. It defines voltage as energy per charge, current as rate of charge flow, and resistance as opposition to current. It explains that direct current (DC) flows from positive to negative terminals, while alternating current (AC) changes direction. Series circuits have the same current and summed voltages, while parallel circuits have the same voltage and summed currents. Several examples demonstrate calculating values using Ohm's Law in series and parallel circuits. The document also covers power, paying for electricity based on kilowatt-hours, and provides practice problems to solve.

1. By: Mr. Augosto M. Directo

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.

6. Series Vs. Parallel
 Series Circuits
o A series circuit is a circuit in which the current can
only flow through one path.
o 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

7.  In Series circuit; the total current is equal
Rtotal = R1 + R2 + R3
V total = V1 + V2 + V3
In Parallel Circuit; the total voltage is equal
1 = 1 + 1 + 1 + 1
Rtotal R1 R2 R3 Rn
I total = I1 + I2 + I3

8. By Analogy: Series Vs Parallel
E
R2R1
I
E
R1
R3
R2
I1
I3
I2

9. 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.

10. 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

11. 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

12. Series Circuit Analysis
 Example #1
 A 4v battery is placed in a series circuit with a 2W
resistor.
 What is the total current that will flow through the
circuit?
I = ?
2W
4v

13. Solution
 Given:
 V = 4v
 R = 2 W
 I = ?
 I = V/R
 = 4v/2 W
 I = 2A

14.  Example#2
 A 110v supplies a load with a resistance of 3W,5W, and
7W respectively, find the current in the circuit?
I = ?110v
3W
7W
5W

15. Solution
 Given:
 V= 110 v
 R1 = 3W
 R2 = 5W
 R3 = 7W
 I = ?
 Rtotal = R1 + R2 + R3 I = V/R
= 3 + 5 + 7 = 110/12
= 12 W = 9.17 A

16. 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 I1 and the I2
220v 5W 10W

17. Solution
 Given:
 V=220v
 R1 = 5W
 R2 = 10W
 I1 = V/ R1 I2 = V/ R2
= 220 / 5 = 220 / 10
= 44 A = 22 A
 RT = R1 R2 / R1 + R2 IT = V / RT
 = (5 x 10) / 15 = 220/3.33
 = 3.33 W = 66.06A

18.  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.
V=? 6W 2WIT =15

19. Solution
 Given:
 R1 = 6W
 R2 = 2W
 IT =15 A
 RT = R1 R2 / R1 + R2 VT = IT / RT
 = (6 x 2) / 8 = 15 A x 1.5 W
 = 1.5 W = 22.5 v
 I1 = V/ R1 I2 = V/ R2
= 22.5 / 6 = 22.5 / 2
= 3.75 A = 11.25 A

20. 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.

21. 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.

22. Formula of Power
 Power (watts) P = VI
Voltage (volts)
Current (amps

23.  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.

24. Solution
 Given:
 V = 1.5v
 R = 3 W
 I =V / R P = V I
= 1.5 / 3 = 1.5 x 0.5
= 0.5 A = 0.75 watts

25. 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.

26. 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?

27.  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.

28.  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 W

29.  2. Resistors 1, 2 ,3 has a resistance of 4 W, 8 W and 12 W
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 W. Find the current and the power.
V1VT = ?
4 W
8 W
12 W
IT = 25A

30. Key Answers
 I. Identification
 1. voltage
 2. direct current
 3. resistance
 4. current
 5. power
 6. 746watts
 7. current
 8. voltage

31.  9. Series Circuit
 10 Ohm’s Law
 11. Parallel Circuit
 12. George Simon Ohm
 13. kilo watt hour
 14. 3.6 x 106 joules
 15. horsepower
 II. Problem Solving
 1. Given:
 V = 115 v R2 = 15 W
 I1 = 3A
 R1 = V / I1 I2 = V/ R2 RT = R1 + R2 IT = V/ RT
 = 115/ 3 A = 115/15 = 15 + 38.33 = 115 / 53.33
 = 38.33 W = 7.67A = 53.33 W = 2.16A

32.  2. Given:
 R1 = 4 W
 R2 = 8 W
 R3 = 12 W
 IT = 25A
 V1 = IT R1 V2 = IT R2 V3 = IT R3
 = 25 x 4 = 25 x 8 = 25 x 12
 = 100 v = 200v = 300v
 RT = R1 +R2 +R3 VT = IT RT
 = 4 + 8 + 12 = 25 x 24
= 24 W = 600v

33.  3. Given:
 V= 220v
 R = 35 ohms
 Find: I? P?
 I = V/R P = V I
 = 220 / 35 = 220 x 6.29
 = 6.29A = 1382.86 watts

34. God Bless