# Electrical Energy and Power

Surely you have faced a situation where some important appliance stops working because the cells run out. What does that mean? That means the cell is no more able to give current or we can say that it has no more energy stored.Â  This means that the energy that is the chemical energy is consumed in the electric circuits. So in order to find out the amount of energy consumed, we study the electric energy or electric power.

## Electric Energy

To under the concept of electric energy, let us consider a conductor carrying the current I and potential difference V between the two endpoints A and B. Let us denoted the electric potential of A and B as V(A) and V(B). As we know that current is flowing from A to B soÂ  V(A) >V(B) and the potential difference across AB isÂ V = V(A) – V(B) > 0

NOW, in a time intervalÂ Î”t, an amount of chargeÂ Î”Q is equal to IÎ”t moves from point A to B of the circuit and the work was done by the electric field is equal to the product of V andÂ Î”Q.

Here if the charges in the conductor move without collisions, theirÂ kinetic energy would also change. Conservation of total energy isÂ Î”K = I V Î”tÂ > 0. The amount of energy dissipatedÂ as heat in a conductor in a time intervalÂ Î”t is,

Î”W = V Î”Q = VI Î”t

### Electric Power

The rate at which the electric energyÂ enters the portion of the circuit is called the electrical power input. The rate at which work is done in bringing the charged particles from one point to another is known as electric power. It is denoted by P.

The SI unit of power is watt (W). One watt is the power consumed by the device catting 1A of current when operated at a potential difference of 1 V.

P = VI

Applying ohms law we can write

P = IÂ² R = VÂ²/R

The above equation is the power loss in a conductor of resistance R which carries the current I. The application of electrical power is that it is transmitted from the power stations which later on reaches our homes and the industrial factories via transmission cables.

Now we know that the transmission of power is very costly. So how do we minimize the power loss in transmission cables? Let us consider a device R to which a power is to be delivered via the cables having resistance Rc.Â Â So if V is the voltage across R and current I then,

P = V I

The wires which are connected to the device from the power station has finite resistance Rc. So,Â PcÂ = IÂ²Â Rc

âˆ´ PÂ²Â RcÂ / VÂ²

Hence the power wasted in connecting the wires is inversely proportional to VÂ². So the resistanceÂ  RcÂ of the transmission cable is considerable.

## Solved Questions

Q1. The circuit given below is for the operation of an industrial fan. The resistance of the fan isÂ 30 ohm. The regulator provided with the fan is a fixed resistor and a variable resistor in parallel. Under what value of the variable resistance given, power transferred to the fans will be maximum? The power source of the fan is a dc source with an internal resistance ofÂ 60 oh.

1. 3 0HM
2. 0
3. âˆž
4. 6 ohm

Solution: The correct option is “B”. The power which transfers to the fan isÂ  P = VÂ²/R where R is the total resistance of the circuit. As power is inversely proportional to total resistance. So for maximum power, the total resistance should be minimum. Total resistance here isÂ R = 6r/6 +r + 3. r is the variable resistance. R is minimum when r = 0

Â Q2.Â  AnÂ electric heater has a resistance of 150 ohms andÂ canÂ bear a maximum current of 1 ampere. If we use the heaterÂ  on 220-volt mains, theÂ leastÂ resistance required in the circuit will be
1. 70 ohms
2. 5 ohms
3. 2.5 ohms
4. 1.4 ohms

Solution: The correct option is “A”. Given that the heater can bear a maximum current of 1 ampere we need to add a resistance to the circuit in series with the heater so that current is less or equal to 1 ampere. Let that resistance be R, then (150+ R)Â Ã— 1 = 220. R = 70 ohm.

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