Basic laws of electric circuits
2.1 Power and energy .1 Work
You may have learned in physics that work is the result when a force acts on an object and causes it to move a certain distance. Work (W ) is the product of the force (F ) and the displacement (S) in the direction of the motion.
Work W¼ F6S
where W is, if using for example a force of 1 N to lift an object to 1 m, the 1 J of work done in overcoming the downward force of gravity as shown in Figure 2.1.
Quantity Quantity Symbol Unit Unit symbol
Work W Joule J
Force F Newton N
Displacement S Meter m
Note: When the force (F) and displacement (S) do not point in the same direction, the formula to calculate work will be:
W¼ (F cos y)S
● where the angley is the angle between force (F) and displacement (S),
● wheny is 0 degree, cos08 ¼ 1, W ¼ (Fcosy)S ¼ FS.
It is the same in an electric circuit. Work is done after the electrons or charges are moved to a certain distance in a circuit as a result of applying an electric field force from the power supply.
2.1.2 Energy
Energy is the ability to do work; it is not work itself, but a transfer of energy.
Even though you can’t ever really see it, you use energy to do work every day.
For example, after you eat and sleep, your body converts the stored energy to keep you doing daily work, such as walking, running, reading, writing, etc.
The law of conservation of energy is one of most important rules in natural science. It states that energy can neither be created nor destroyed, but can only be converted from one form to another. ‘Converted’ means ‘never disappeared’
in physics terms. For example:
Electrical generator: mechanical energy! electrical energy.
Lamp: electrical energy! light energy.
Battery: chemical energy! electrical energy.
2.1.3 Power
Power refers to the speed of energy conversion or consumption; it is a measure of how fast energy is transforming or being used. For example in Figure 2.1, 1 N object lifted to 1 m may have different time rates depending on the amount of power applied. If a higher power is applied to the object (an adult is lifting it), it will take a shorter period of time to lift it; and if a lower power is applied to the object (a kid is lifting it), it will take a longer period of time to lift it. So power is defined as the rate of doing work, or the amount of work done per unit of time.
Our daily consumption of electricity is electrical energy, and not electrical power. The hydro bill that you receive is for electrical power – the amount of electrical energy consumed in 1 or 2 months.
1 N
1 m F
1 N
Figure 2.1 Work
Energy and power
● Energy is the ability to do work.
● Power is the speed of energy conversion, or work done per unit of time: Power¼ Work=time or P ¼ W=t
Electrical power is the speed of electrical energy conversion or consumption in an electric circuit, and it is a measure of how fast electrons or charges are moving in a circuit.
Since current is the amount of charge (Q) that flows past a given point at the certain time: I¼ Q=t and voltage is the amount of work that is required to move electrons between two points: V¼ W=Q or Wc ¼ QV:
Substituting work W into the power equation gives P¼ W=t ¼ QV=t ¼ IV:
It can also be expressed as the form of a derivative:
p¼ ðdw=dtÞ ¼ ðdw=dqÞðdq=dtÞ ¼ vi
Substituting Ohm’s law into the power equation P¼ IV obtains the other two different power equations:
P¼ VI ¼ ðIRÞI ¼ I2R ðOhm’s law : V ¼ IRÞ P¼ VI ¼ VV
R¼V2
R Ohm’s law : I¼V R
(Electrical) PowerP
P¼ IV ¼ I2R¼ V2=R ðor P ¼ IE ¼ E2=RÞ
Quantity Quantity symbol Unit Unit symbol
Electrical Power:
Work or Energy W Joule J
Time t Second s
Power P Watt W
Or: Kilowatt-hour kWh
Hour h
Watt W
Quantity Quantity symbol Unit Unit symbol
Power P Watt W
The above three power equations can be illustrated in Figure 2.2 as the mem-ory aid for power equations. By covering power in any diagram, the correct equation will be obtained to calculate the unknown power.
Example 2.1: In a circuit, voltage V¼ 10 V, current I ¼ 1 A and resistance R¼ 10 O, calculate the power in this circuit by using three power equations, respectively.
Solution:
P¼ IV ¼ ð1 AÞð10 VÞ ¼ 10 W P¼ I2R¼ ð1 AÞ2ð10 OÞ ¼ 10 W P¼ V2=R ¼ ð10 VÞ2=10 O ¼ 10 W
Example 2.1 proved that the three power equations are equivalent since each equation leads to the same value of power at 10 W.
If power is given in a circuit, using mathematical skill to manipulate the power equations and solving for current I and voltage V, respectively, we can express current I and voltage V as follows:
since P¼ I2Ror I2¼ P/R, so I ¼ ffiffiffiffiffiffiffiffiffiffi P=R p since P¼ V2/R or V2=PR, so V¼ ffiffiffiffiffiffiffi
pPR
Example 2.2: If power consumed on a 2.5O resistor is 10 W in a circuit, cal-culate the current flowing through this resistor.
Solution:
2.1.4 The reference direction of power
When a component in a circuit has mutually related reference polarity of cur-rent and voltage (refer to chapter 1, section 1.6.3), power is positive, i.e. P4 0, meaning the component absorption (or consumption) of energy. When a component in a circuit has non-mutually related reference polarity of current and voltage, power is negative, i.e. P5 0, meaning the component releasing (or providing) of energy. The concept of the reference direction of power can be illustrated in Figure 2.3.
V I
P P
I2 R P R
V2
Figure 2.2 Memory aid for power equations
The reference direction of power
● If a circuit has mutually related reference polarity of current and voltage: P4 0 (absorption energy).
● If a circuit has non-mutually related reference polarity of current and voltage: P5 0 (releasing energy).
Example 2.3: Determine the reference direction of power in Figure 2.4(a and b).
Solution:
(a) P¼ IV ¼ (2 A)(3 V) ¼ 6 W (P 4 0, the resistor absorbs energy).
(b) P¼ I (7V) ¼ (2 A)(73 V) ¼ 76 W (P 5 0, the resistor releases energy).
Example 2.4: I¼ 2 A, V1¼ 6 V, V2¼ 14 V and E ¼ 20 V in a circuit as shown in Figure 2.5. Determine the powers dissipated on the resistors R1, R2, and R1
and R2in series in this figure.
Solution:
Power for R1(a to b): P1¼ V1I¼ (6 V)(2 A) ¼ 12 W (absorption).
Power for R2(b to c): P2¼ V2I¼ (14 V)(2 A) ¼ 28 W (absorption).
Power for R1 and R2(a to d): P3¼ (7E)I ¼ (720 V)(2 A) ¼ 740 W (releasing).
P1þ P2þ P3¼ 12 W þ 28 W 7 40 W ¼ 0 (energy conservation).
I ⫹
⫺ V
⫺
⫹ V I
(a) P > 0 (b) P < 0
Figure 2.3 The reference direction of power
I 2 A
V 3 V
I 2 A
V 3 V
(a) (b)
Figure 2.4 Illustrations for Example 2.3