# ELECTRICITY

**HEATING
EFFECT OF ELECTRIC CURRENT**

A battery or a cell is a source of electrical energy. It generates potential difference that sets the electrons in motion to flow the current through a resistor or a system of resistors.

A part of the source energy to
maintain the current may be consumed into useful work (e.g. rotation of an
electric fan). Rest of the energy is lost as heat. E.g., an electric fan
becomes warm if used for longer time.

If an electric circuit is purely
resistive (i.e. a configuration of resistors only connected to a battery), the
source energy is dissipated entirely as heat. This is called **heating effect
of electric current.**

Consider a current *I *flowing through a resistor of
resistance *R*. Let the potential
difference across it be* V* and *t* is the time during which a charge *Q* flows across.

*A steady current in a purely
resistive electric circuit.*

The
work done in moving the charge *Q *through
a potential difference *V* is *VQ*. Therefore, the source must supply
energy equal to *VQ* in time *t.* Hence the power input to the circuit
by the source is

Or
the energy supplied to the circuit by the source in time *t* is *P *×* t*,
i.e., *VIt. *This energy is dissipated
in the resistor as heat (*H)*.

Therefore,
*H = VIt*

Applying
Ohm’s law (V = IR),

*H = I ^{2}
Rt*

This is known as **Joule’s law
of heating.** It implies that heat produced in a resistor is directly
proportional to

- The
square of current for a given
- Resistance for a given current.
- The time for which current flows through resistor.

In practical situations, when an
electric appliance is connected to a voltage source, Eq. *H = I ^{2}
Rt* is used after calculating the current using the
relation

*I = V/R*.

**Problem: **An
electric iron consumes energy at a rate of 840 W when heating is at the maximum
rate and 360 W when the heating is at the minimum. The voltage is 220 V. What
are the current and the resistance in each case?

**Solution**

Power input, *P = V I*

Thus the current *I = P/V*

(a) When heating is at the maximum
rate:

*I = *840
W/220 V = __3.82 A__

Resistance
of the electric iron is

*R = V/I *=
220 V/3.82 A = __57.60 ____Ω__

(b) When heating is at the minimum
rate:

*I = *360
W/220 V = __1.64 A__

Resistance of the electric iron is

*R = V/I *=
220 V/1.64 A = __134.15 ____Ω____.__

**Problem: **100
J of heat is produced each second in a 4 Ω
resistance. Find the potential difference across resistor.

**Solution**

*H*
= 100 J, *R* = 4 Ω,
*t* = 1 s, *V*= ?

*H= I ^{2}Rt*

So, the current through the
resistor is

I = √(H/Rt) = √[100
J/(4 Ω × 1 s)]

=
5 A

Thus the potential difference
across the resistor is

V = IR = 5 A ×
4 Ω = __20 V__

# Practical Applications of Heating Effect of Electric Current

Due to heating effect of electric
current, electrical energy is lost as heat. Also, it may alter the properties
of components in electric circuits. But heating effect (Joule’s heating) has
many useful applications.

a. To make devices such as electric
laundry iron, electric toaster, electric oven, electric kettle & electric
heater.

b. To produce light in electric
bulb. Here, the filament made of metals with high melting point can retain much
heat. So it gets very hot and emits light.

E.g. **tungsten**
(melting point 3380°C) is used to make filaments. Filament should be thermally
isolated, using insulating support. The bulbs are filled with chemically
inactive nitrogen and argon gases to prolong the life of filament. Most of the
power consumed by the filament appears as heat, but a small part is radiated as
light.

c. To make **fuse*** *used*
*in electric circuits.

It protects circuits and
appliances by stopping the overflow of electric current. The fuse is placed in
series with the device. It consists of a piece of wire made of a metal or an
alloy of suitable melting point (aluminium, copper, iron, lead etc.). During
the overflow of the current, the temperature of the fuse wire increases. It melts
the fuse wire and breaks the circuit.

The fuse wire is encased
in a cartridge of porcelain or similar material with metal ends.

The fuses used for
domestic purposes are rated as 1A, 2A, 3A, 5A, 10A, etc. E.g. when an electric
iron which consumes 1 kW electric power is operated at 220 V, 4.54 A current
(1000/220) flows in the circuit. In this case, a 5 A fuse must be used.

**ELECTRIC
POWER**

**Power**
is the rate of doing work or rate of consumption of energy.

Equation *H = I*^{2}*
**Rt* gives the rate at which electric energy is
dissipated or consumed in an electric circuit. This is also termed as **electric
power.** The power *P* is given by

* P = VI*

Or *P = I ^{2}R = V^{2}/R*

The SI unit of electric power is **watt
(W).** It is the power consumed by a device that carries 1 A of current when
operated at a potential difference of 1 V.

Thus, 1 W = 1 volt × 1 ampere = 1 V A

Watt is a very small unit. So, practically
a much larger unit called **kilowatt **(1000 watts) is used.

Electrical energy is the product
of power and time. Its unit is **watt hour (W h**). One watt hour is the
energy consumed when 1 watt of power is used for 1 hour.

The commercial unit of electric
energy is **kilowatt hour (kW h),** commonly known as **‘unit’.**

1 kW h = 1000 watt × 3600 second

= 3.6 × 10^{6} watt second =
3.6 × 10^{6} joule (J)

In
an electric circuit, electrons are not consumed. We pay for energy to move
electrons through electric gadgets.

**Problem:** An
electric bulb is connected to a 220 V generator. The current is 0.50 A. What is
the power of the bulb?

**Solution**

P
= VI

= 220 V × 0.50 A

= 110 J/s =
__110 W.__

**Problem: **An
electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the
energy to operate it for 30 days at Rs 3.00 per kW h?

**Solution**

Total energy consumed by the
refrigerator in 30 days:

400 W ×
8.0 hr/day × 30 days = 96000 W h

=
96 kW h

∴ cost of energy to operate the
refrigerator for 30 days:

96 kW h ×
Rs 3.00 per kW h = __Rs 288.00__

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