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Heating Effect of Electric Current

  • Last Updated : 19 Apr, 2021

Electrons are very minute small particles that exist inside the molecular structure of a substance. Sometimes very commonly, these electrons are tightly held, and some other times they are very loosely held. When electrons are very loosely held by the nucleus, they are able to travel freely within the limits of the appliance or source. Electrons are negatively charged particles therefore they tend to move a number of electrical charges, we say this movement of electrons is the electric current. It has to be noted and observed that the number of electrons that tend to move governs the ability of a particular substance to conduct electricity. Some materials allow the electrical current to move better than others. Based on the ability of the material to conduct electricity, materials are classified into conductors which allow current flow and insulators which do not allow current flow. This flow of current in any appliance apart from the useful work liberates or exerts the residual energy in form of heat or magnetism. In this article, we will explain the heating effect of the electrical current. Also, the force that is required to drive the current flow through the conductor is known as voltage.

Heating Effect Of Electric Current

When we use some electrical appliances, the chemical reactions which occur in their cells on which they run generate some potential difference between its terminals which sets the electrons in motion. To maintain the flow of current the source needs to expend some energy. A part of the energy is used in doing some sort of useful work like moving the fan blades in case of electricity generated fans, etc. The remaining energy is exploited or expended in the form of heat which raises the temperature of the appliance. If we are using a circuit in an appliance that is purely resistive then a lot of energy is dissipated entirely in the form of what we call heat. This is called or known as the heating effect of electric current.

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More simply, when an electrical current is passed through a conductor, it generates excess heat due to the resistance caused by the electrons in conductor to the flowing current. The work done in overcoming this resistance to the current generates what we call heat in that conductor.The electrical heating effect of the electrical current is most commonly and widely applied and used in our daily life. For example, the electrical irons, kettles, toasters, electrical heaters, etc. are used widely as alternatives as the conventional methods of cooking and also laundry. This same effect is used widely in electrical bulbs which are the alternatives of conventional incandescent lamps. These devices have modernized and revolutionized the new sustainable world over the years.

Circuit Diagram

Let us assume a current I that is flowing through a resistor that has a resistance of R as shown in the circuit. Let the potential difference across ends of the terminals of the battery be V. Let us assume to be the time during which a charge of Q amount flows across the circuit. The work which is done in moving that charge Q through a potential difference V is V × I. Hence, the source has to supply energy equal to V × I in time t. Therefore, the power input to the electrical circuit by the source is

= V\times\frac{Q} {t}=V\times I

Or the energy that is being supplied to the circuit by the source in time t is P × t, that is, V × I × t. This extra energy generated gets dissipated in the resistor in the form of heat. Therefore, for a steady and fixed current I, the amount of heat denoted by H that is produced in time t is

H = V × I × t

Joules Law Of Heating

The very famous physicist James Prescott found that the amount of heat generated per second that develops in a conductor having current is directly proportional to the electrical resistance of the wire and also with the square of the current given. This heat which is liberated or generated because of the electrical current that flows in an electrical wire is expressed in Joules.

Joule’s First Law

By applying ohms law to the equation H = V × I × t. We can deduce the Joules law or joules first law which gives the relationship between the heat that is produced by flowing charges of electric current through a conductor. It is directly proportional to the square of the supplied current, the electrical resistance exerted by the appliance, and the time for which we used it. This is known or called as joule’s law of heating. The following is its expression:

H = I2 × R × t

Where,

  • H gives or indicates the amount of heat.
  • I shows the amount ofthe electrical current supplied.
  • R is the value or amount of electric resistance exerted in the conductor.
  • t denotes the time for which the appliance is operated.

Factors On Which Heat Depends

  • The amount of liberated or generated heat is directly proportional to the given wire’s electrical resistance when the electrical current in the given circuit and the flow of supplied current is not altered or changed.
  • The amount of liberated or generated heat in the conductor carrying current is directly proportional to the square of the electrical current that flows through the given circuit when the electrical resistance and current supply is kept constant.
  • The amount of the heat generated or produced because of the electrical current flow is directly proportional to the time of usage of flow when the electrical resistance and the current flow is constant.

Applications Of Heating Effect of Electric Current

Following are some of the widely used common devices in which the heating effect of current is used and harnessed for other purposes:

Electric Iron: Between the metal part and the electrical coil in an iron, Mica is placed which is by nature is an insulator. The coil of the iron becomes warm or heated with the continuous passage of current which is then passed on or transferred to the metallic part through the mica used. Finally, after a while, the metallic part becomes very heated or whatever temperature we have set, which is then used for ironing different material clothes according to our wish.

Electric Bulb: Electrical bulb contains a very thick metallic wire which is in turn made up of highly resistive tungsten metal. This metal is always kept in an inert environment so that it doesn’t react and with a neutral gas or vacuum. When the electrical current flows through the used tungsten wire, it becomes warm or heated, and then it emits light. Most of the electrical power which is drawn in the electrical circuit from the electrical source is liberated or dissipated in the form of heat and the rest is given or emitted in the form of light energy. The tungsten filament used also has a high resistivity and has a very high melting point so that it doesn’t get heated easily when used.

Electric Heater: In an electrical heater, a very high resistance nichrome wire is mostly and commonly used as a coil. The coil is rotated or wound on grooves which are made up of the ceramic material of the iron plate or china clay plate. Whenever the electrical current flows in the coil, it quickly becomes warm or heated, which is then widely used to heat our cooking vessels. In mountain areas, electrical room heaters are used to keep their rooms warm and heated to save themselves from the exhaustive cold outside.

Electric Fuse: In any electrical instrument which we sometimes use due to a sudden rise in the amount of current, the instrument or appliance gets overheated or burnt down which sometimes may result in a severe fire. A conducting wire with a very low melting point is joined or connected in series connection with the device or appliance circuit to avoid any mishap or this type of accident. Whenever the current value somehow accidentally rises, the wire inside the fuse melts due to the excessive heating and thus results in breaking the electrical circuit saving the device as well as our lives. We choose or select the fuse according to the appliance used. A device or an appliance that works on a higher current needs a greater value of fuse and vice versa.

For using and harnessing the heating effect of electric current, the element of appliances needs or is required to have a high melting point to retain more heat.

Electric Power

Electrical power is defined or stated as the rate of doing work. Or in other words, the rate of consumption of the energy. The power has an S.I. unit of Watt. One watt is also defined as the energy consumed by a device or appliance which carries 1 A of current ad operated at 1 V. Different appliances have a different amount of power consumption according to their needs. Some have a high power consumption for example the A.C, microwave, washing machine, T.V, refrigerator, etc. Whereas some like electrical toaster, iron, charger, hairdryer, etc. have a lesser power consumption. The electrical power of a given device or appliance is seen before we install them in our home because we have to use a suitable electrical fuse for them in case they consume high power, as in that case, the extent of short circuit is very high, so we see the rated power of the devices before using them to avoid any kind of mishap or misfortune to the appliance and to us. The expression or equation of power is given in many forms:

P = V × I

P = \frac{V^{2}}{R}

P = I2R

Energy Consumption in Daily Life

Therefore, we in actual practice use a very much larger unit called ‘kilowatt’ instead of a watt. It is merely equal to 1000 watts. Since the electrical energy is given as the product of power and the time, the deduced unit of electrical energy is watt-hour (W h). One watt-hour (W h) is the energy consumed or used when only 1 watt of power is used for a time of 1 hour. The commercially used unit of electrical energy is a kilowatt-hour (kW h), which is widely or commonly known as a ‘unit’.

1 kW h = 1000watt × 3600second

1 kW h = 3.6 × 106 watt second

1 kW h = 3.6 × 106 joule (J)

Sample Problem

Question 1: 100 J of heat energy is produced by an electrical appliance which is used for 1 sec and is having a resistance of 4 ohms. Find and calculate the potential difference of the appliance.

Solution:

As we know,

H = I2×R×t

Given, H = 100 J, R = 4 ohm, t = 1 sec

Therefore, 100J=I2×4×1

\therefore I^{2} = \frac{100}{4}

\therefore I = \sqrt{\frac{100}{4}} A

\therefore I =\frac{10} {2} A

I = 5 A

From ohm’s law we know

V=I×R

Therefore, V = 5A × 4Ω

V = 20 V

Therefore, the potential difference generated is 20 V.

Question 2: Find out the heat produced by the electric toaster when it is used for 5 minutes. The current given was 2 A and its resistance is 3 ohm.

Solution:

As we know,

H = I2×R×t

Given, I = 2 A, R = 3 ohm, t = 5 min = 300 sec

Therefore, H=22×3×300J

H = 22×900 J

Therefore, H=4×900 J

H = 3600 J

Hence, the heat that is produced or liberated by the electrical toaster is 3600 J.

Question 3: Find out the resistance of electrical iron if it has produced a heat of 100 J when operated at 2 A current and for 50 sec.

Solution:

As we know,

H = I2×R×t

Given, H = 100 J, I = 2 A, t = 50 sec

Therefore, 100J=22×R×50

\therefore R = \frac{100}{4 \times 50} \Omega

\therefore R = \frac{1}{2} \Omega

R = 0.5Ω

Therefore, the resistance of the electrical iron is found to be 0.5Ω

Question 4: The heat generated is 100 J from an electrical fan which has a potential difference of 10 V and the time for which it is used is 10 sec. Find what the amount of electrical current that is used is?

Solution:

As we know,

H = V × I × t

Given, H = 100 J, V = 10 V, t = 10 sec

Therefore, 100J=10V×I×10

∴100J = 100 × I

\therefore I = \frac{100}{100} A

I = 1 A.

Therefore, the current used is 1 A.

Question 5: The heat generated is 100 J from an electrical fan which has a current of 10 A and the time for which it is used is 10 sec. find what the amount of electrical potential difference that is generated?

Solution:

As we know,

H = V × I × t

Given, H = 100 J, I = 10 A, t = 10 sec

∴100J=10A×V×10

∴100J=100×V

\therefore V = \frac{100}{100} A

V = 1 V.

Therefore, the potential difference used is 1 V.

Question 6: Determine the heat liberated or generated by a toaster when it is operated at a potential difference of 6 V having a current supply of 6 A and is used for 50 sec.

Solution:

As we know,

H = V × I × t

Given, V = 6 V, I = 6 A, t = 50 sec

∴H = 6 × 6 × 50J

H = 36 × 50 J

H = 1800 J

Hence, heat or energy liberated is 1800 J.

Question 7: What is the power consumed if a device or appliance is operated at 1 V potential difference and 6 A current?

Solution:

As we know,

P = V × I

Given, V = 1V AND I = 6 A

P = 1 V × 6 A

P = 6 W

Hence, the power that has been consumed is 6 W.

Question 8: What is the current consumed if a device or appliance is operated at 10 V potential difference and power of 2 W?

Solution:

As we know,

P = V × I

Given, V = 10 V and P = 2W

2 = 10 V × I

\therefore I = \frac{2}{10} A

I = 0.2 A

Hence, the current that has been consumed is 0.2 A.

Question 9: What is the potential difference used if a device or appliance is operated at 10 A current supply and power of 2 W?

Solution:

As we know,

P = V × I

Given, I = 10 A and P = 2W

∴2 = V × 10A

V = \frac{2}{10} V

V = 0.2 V

Hence, the Potential difference that has been used is 0.2 A.

Question 10: What is the power consumed if a toaster having 3 ohm resistance is operated at 1 A current?

Solution:

As we know,

P = I2 × R

Given, I = 1 A and R = 3 ohm

∴P = 12 × 3Ω

P = 3 W.

Therefore, the power consumed has been calculated to be 3 W.

Question 11: What is the power consumed if an electrical heater having 13 ohm resistance is operated at 1 A current?

Solution:

As we know,

P = I2 × R

Given, I = 1 A and R = 13 ohm

∴P = 12 × 13Ω

P = 13 W.

Therefore, the power consumed has been calculated to be 13 W.

Question 12: What is the resistance offered, if a toaster having 10W power is operated at 1 A current?

Solution:

As we know,

P = I2 × R

Given, I = 1 A and P = 10 W

∴P = 12 × RΩ

R = 10 Ω

Therefore, the resistance offered is R = 10 Ω

Question 13: A bulb rated 10W, 220V is used for 6 days for 2 hour daily. Find the resistance of the bulb and energy that is consumed in units.

Solution:

As we know,

P = \frac{V^{2}}{R}

Given, P = 10 W and V = 220 V

\therefore 10 = \frac{220^{2}}{R}

\therefore R = \frac{220^{2}}{10}

\therefore R = \frac{48400}{10} \Omega

∴R = 4840Ω

Therefore, the resistance of bulb is R = 4840Ω

We also know that,

H = V × I × t

P = V × I

Therefore, we can write H = P × t

Given, P = 10 W and t = 2 hr. and it is used for 6 days, therefore the energy consumed in 6 days is

H = 10 × 2 × 6 W hr.

H = 120 W hr.

H = \frac{120}{1000}  KW hr.

\therefore H = \frac{120}{1000}  KW hr.

H = 0.12 KW hr. or units

Therefore, the bulb consumed 0.12 units.

Question 14: A lamp bulb has a resistance of 44 ohm, and it is operated at 220 V supply voltage. Find the power consumed by it.

Solution:

As we know,

P = \frac{V^{2}}{R}

Given, V = 220 V and R = 44 Ohm

\therefore P = \frac{220^{2}}{44}

\therefore P = \frac{48400}{44}

P = 1100 W

Therefore, the power consumed by the lamp bulb is 1100 W.

Question 15: A bulb rated 100W, 220V is used for 6 days for 2 hours daily. Find the offered resistance of the bulb.

Solution:

As we know,

P = \frac{V^{2}}{R}

Given, P = 100 W and V = 220 V

\therefore 100 = \frac{220^{2}}{R}

\therefore R = \frac{220^{2}}{100}

\therefore R = \frac{48400}{100} \Omega

∴R = 484Ω

Therefore, the resistance of bulb is R = 484Ω

Question 16: What is the power consumption if an electric iron is operated at 220 V supply and the amount of charge that flows for 10 sec is 10 coulomb.

Solution:

As we know,

P = V\frac{Q} {t}

Given, V = 220 V, Q = 10 Coulomb and t = 10 sec

\therefore P = 220 \frac{10}{10}

P = 220 W

Hence, the power that is consumed by the electric iron is 220 W.

Question 17: What is the power consumption if an electric iron is operated at 220 V supply and the amount of charge that flows for 10 sec is 20 coulomb.

Solution:

As we know,

P = V\frac{Q}{t}

Given, V = 220 V, Q = 20 Coulomb and t = 10 sec

\therefore P = 220 \frac{20}{10}

\therefore P = 220\times 2 W

P = 440 W

Hence, the power that is consumed by the electric iron is 440 W.

Question 18: What is the time for which the bulb was operated if the power used by it was 10 W, and it was given a 220 V supply and 10-coulomb charges have flown through it.

Solution:

As we know,

P = V\frac{Q}{t}

Given, V = 220 V, Q = 10 Coulomb and P = 10 W

\therefore 10 = 220 \frac{10}{t}

\therefore t = 220 \frac{10}{10} sec

t = 220 sec = 3 min 40 sec

Hence the time for which is was used for 220 sec or 3 min 40 sec.




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