Contents
The study of Physics Topics has helped humanity develop technologies like electricity, computers, and space travel.
What is Electrical Power and its Units?
If an electrical source sends current I for time t, in a circuit under potential difference V, then the electrical work done in the circuit,
w = vIt
We know that, the rate of work done with respect to time is called power.
∴ P = \(\frac{W}{t}\) = \(\frac{V I t}{t}\)
or, P = VI
i.e., electrical power = potential difference × current
Unit of power: The units of current and potential difference are volt (V) and ampere (A) respectively. Again, the unit of work is joule (J) and the unit of power is joule/second (J/s) or watt (W). So, from equation (1) we have,
watt = volt × ampere ……… (2)
Definition: If the potential difference at the two terminals of an electrical appliance is 1 volt and the current passing through it is 1 ampere, then the power of the appliance is 1 watt.
We have seen that the relation between the electrical work done in an electrical circuit or in a portion of the circuit is electrical work = active energy + exothermic energy [section 3.2; equation (4)]
or, VIt = V0It + I2Rt
For t = 1s; VI = V0I + I2R ……… (3)
Equation (3) expresses that,
electrical power applied = active power + power consumed in the production of heat
The active energy/power is of little concern to us. We are only interested in the latter, for which we shall find alternative expression.
Alternative expression of the power of passive resistance: If V’ be the terminal potential difference and I be the current flowing through a passive resistance R, then accord-ing to Ohm’s law,
V’ = IR
The power consumed in this resistance is, P’ = I2R [equation (3)]
= IR ᐧ 1
= V’I, as already shown.
Since, I = \(\frac{V^{\prime}}{R}\)
∴ P’ = \(\left(V^{\prime} \cdot \frac{V^{\prime}}{R}\right)\) = \(\frac{V^{\prime 2}}{R}\)
So, P’ = V’I = I2R = \(\frac{V^{\prime 2}}{R}\) ……… (4)
In most electrical circuits, there is no active device. All the resistances are passive. So, the entire power applied is consumed in the form of heat. In that case, the applied potential difference and the potential difference across the passive resistance (or resistances) become equal to each other.
That is, in this condition, V = V’ and P = P’. So equation (4) can be written as, P = VI = I2R = \(\frac{V^2}{R}\) …… (5)
It is evident from equations (4) and (5) that, the power consumed in the external resistance can be expressed in three different ways. P = VI is the general expression of power and it is applicable in all types of electrical appliances. On the other hand, the relations P = I2R or, P = \(\frac{V^2}{R}\) are used as the expressions for power consumed in exothermic devices.
Wastage of energy ¡n transmission of electric power: Suppose, electrical power is to be transmitted over long distances from a generating station. Usually, a long line wire is used for this purpose. Now, as current flows through this wire, huge amount of heat is generated due to Joule effect, i.e, a large amount of electrical energy is wasted as heat during transmission. So, the most important condition to transmit the electrical power from one place to another ¡s that, the wastage of energy should be reduced as much as possible.
A solution to minimise wastage of energy: From the expression of the power P = VI, it can be easily understood that higher the value of V, lower will be the value of I. Again, from equation (3), it is seen that, if the value of I is small, the value of power consumed in the production of heat (I2R) will also diminish, i.e., wastage of energy will also diminish. So, a high voltage power transmission allows for lesser resistive losses over long distances in wiring.
In actual practice, electrical energy is transmitted along the long line wire at the high voltage, 11000V or more. At the place where the supply is required, the high voltage is stepped down to 220 V or 440 V by a step down transformer.
Horsepower or hp: This is a large unit of power. From the discussion of mechanical energy and power, we know,
1 hp = 746 W
Generally, horsepower unit is used to express the power of those machines (motor, pump etc.), where electrical energy is transformed into mechanical energy.
Power Consumed in an Electrical Circuit
Suppose, in an electrical circuit, the emf of the battery is E. If internal resistance of the battery is r and R is the equivalent resistance of all the resistances used in the external circuit, [Fig.] then,
current in the circuit, I = \(\frac{E}{R+r}\)
∴ Power consumed in the circuit, P0 = I2(R + r) = \(\frac{E^2}{R+r}\)
Maximum power n the external circuit: A part of the total power, in the circuit equivalent to I2r, is consumed due to the internal resistance of the battery and it cannot be utilised in any other way. The rest part of the power is spent in the external circuit and is utilised for running different electrical appliances connected in the external circuit. Since, the equivalent resistance of the external circuit is R, the available power in the external circuit,
P = I2R = \(\frac{E^2 R}{(R+r)^2}\)
In case of constant emf and internal resistance of the battery (E = constant, r = constant), the condition of availability of maximum power in the external circuit is given by,
i.e., maximum power is obtained in the external circuit, if the equivalent resistance of the external circuit is equal to the inter-nal resistance of the battery[Fig.]. The value of maximum power is
Pmax = \(\frac{E^2 R}{(R+R)^2}\) = \(\frac{E^2}{4 R}\) = \(\frac{E^2}{4 r}\)
Numerical Examples
Example 1.
A 20 Ω resistor can dissipate a maximum of 2 kW power as heat without being damaged. Should this resistor be connected directly across a 300V dc source of negligible Internal resistance? [WBJEF ‘01]
Solution:
If the resistor is connected directly with a 300V dc source, its power would be,
P = \(\frac{V^2}{R}\) = \(\frac{(300)^2}{20}\) = 4500W = 4.5kW
Since, this power is greater than 2 kW, so the resistor should not be directly connected to a 300V dc source.
Example 2.
Three resistors of equal resistances when connected in series across a voltage source, dissipate 100 watt of power. What would be the power dissipated, If the resistors are connected in parallel across the same source of emf?
Solution:
If each resistance be r, then the equivalent resistance in series, R1 = 3r; and the equivalent resistance in parallel,
R2 = \(\frac{r}{3}\).
In both the combinations, potential difference across the two ends is equal to V (say).
∴ In case of series combination, power, P1 = \(\frac{V^2}{R_1}\) and in case of parallel combination, power, P2 = \(\frac{V^2}{R_2}\).
∴ \(\frac{P_1}{P_2}\) = \(\frac{R_2}{R_1}\) or, P2 = P1 ᐧ \(\frac{R_1}{R_2}\) = 100 × \(\frac{3 r}{\frac{r}{3}}\) = 900W
Example 3.
The coil of a heater connected to a 200V line, consumes a power of 100 W. The coil is divided into two equal parts. The two parts are combined in parallel and connected to a 200V line. What will be the power consumed by the new combination? [HS 06]
Solution:
If R be the resistance of the coil, then power consumed in the potential difference V, P = \(\frac{V^2}{R}\)
In the first case, 100 = \(\frac{(200)^2}{R}\)
In the second case, resistance of each of the two equal parts = \(\frac{R}{2}\)
So, the equivalent resistance in the parallel combination
= \(\frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2}+\frac{R}{2}}\) = \(\frac{R}{4}\)
So, power consumed,
P2 = \(\frac{(200)^2}{\frac{R}{4}}\) = 4 × \(\frac{(200)^2}{R}\) = 4 × 100 = 400 W
Example 4.
The power consumed in the circuit shown in the Fig. is 150 W . What is the value of R?
[AIEEE ‘02]
Solution:
Equivalent resistance
Example 5.
A cell of emf 1.5 V and of internal resistance 0.1 Ω when connected with a resistor and an ammeter of negligible resistance in series, the ammeter shows a 2.0 A steady current. Find
(i) the rate of energy dissipated within the cell and
(ii) the power consumed in the resistor. [HS ‘01]
Solution:
(i) The rate of energy dissipated within the cell
= EI = 1.5 × 2.0 = 3W
ii) I = \(\frac{E}{R+r}\) ; R = resistance of the resistor
or, R = \(\frac{E}{I}\) – r = \(\frac{1.5}{2}\) – 0.1 = 0.65Ω
∴ Power consumed in the resistor,
I2R = (0.2)2 × 0.65 = 2.6 W
Example 6.
A balanced Wheatstone bridge has resistances 100 Ω, 10Ω, 500 Ω and 50 Ω respectively in its four arms. Determine the ratio of powers consumed in its different arms.
Solution:
In Fig., Wheatstone bridge has been shown.
Example 7.
A factory requires a power of 90 kW. The energy is transmitted to the factory through a 2.5 Ω line wire. If 10% of the power generated is lost in transmission, calculate
(i) the transmission line current,
(ii) the potential difference at the power generating station and
(iii) the potential drop due to line resistance.
Solution:
or, power of the generating station = 90 × \(\frac{100}{90}\) = 100 kW
i) Power lost = 100 – 90 = 10kW = 10000W
This power is lost in transmission line.
So, according to the equation, P = I2R
Current in the transmission line,
I = \(\sqrt{\frac{P}{R}}\) = \(\sqrt{\frac{10000}{2.5}}\) = \(\sqrt{4000}\) = 63.25 A (approx.)
ii) According to the equation, P = VI,
potential difference at the generating station,
V = \(\frac{P}{I}\) = \(\frac{10000}{63.25}\) = 1581 V(approx.)
iii) Potential drop due to line resistance,
V’ = IR = 63.25 × 2.5 = 158.1V (approx,)
Example 8.
Electrical energy is transmitted at the rate of 2.2 MW through the Une wire. The resistance of the Une wire is 25 Ω. Calculate the percentage of heat dissipation of the electrical energy for each line voltage
(i) 22000V and
(ii) 110kV.
Solution:
VI = 2.2 MW = 2.2 × 106W
i) V = 22000 V; I = \(\frac{V I}{V}\) = \(\frac{2.2 \times 10^6}{22000}\) = 100A
∴ Percentage of heat dissipation = \(\frac{I^2 R}{V I}\) × 100%
= \(\frac{(100)^2 \times 25}{2.2 \times 10^6}\) × 100%
= 11.4%
ii) V = 110 kV = 110000V ; I = \(\frac{V I}{V}\) = \(\frac{2.2 \times 10^6}{110000}\) = 20A
∴ Percentage of heat dissipation = \(\frac{I^2 R}{V I}\) × 100%
= \(\frac{(20)^2 \times 25}{2.2 \times 10^6}\) × 100%
= 0.45%
[Obviously, it is seen that dissipation of energy becomes less if the line voltage is high.]