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Electricity Class 10 Physics Notes

ELECTRIC CURRENT AND CIRCUIT

Just as flowing water creates a water current in a river, the flow of electric charge through a conductor (like a metal wire) creates an electric current.

An electric circuit is a continuous and closed path through which an electric current can flow. If this path is broken at any point, the current stops. A switch is a simple device that opens or closes a circuit, either stopping or allowing the current to flow.

Defining and Measuring Electric Current

Electric current is defined as the rate of flow of electric charge. It measures the amount of charge passing through a specific area in a given amount of time.

The formula for electric current is: $I = \frac{Q}{t}$ Where:

I is the electric current
Q is the net charge flowing
t is the time

Units of Charge and Current

The SI unit of electric charge is the coulomb (C). One coulomb is the charge contained in approximately $6 \times 10^{18}$ electrons. An electron has a negative charge of $1.6 \times 10^{-19} \text{ C}$ .
The SI unit of electric current is the ampere (A), named after Andre-Marie Ampere. One ampere is defined as the flow of one coulomb of charge per second ( $1 \text{ A} = 1 \text{ C/s}$ ).
Smaller currents are often measured in milliampere (mA), where $1 \text{ mA} = 10^{-3} \text{ A}$ , or microampere ( $\mu$ A), where $1 \mu\text{A} = 10^{-6} \text{ A}$ .

Direction of Current In metallic wires, the charges that flow are negatively charged electrons. However, when electricity was first discovered, electrons were unknown. By convention, electric current was considered the flow of positive charges. Therefore, the conventional direction of electric current is taken as opposite to the direction of the flow of electrons. Current flows from the positive terminal to the negative terminal of a cell in the external circuit.

Measuring Current An ammeter is an instrument used to measure electric current in a circuit. It must always be connected in series in the circuit, so that all the current being measured flows through it.

Example

Example A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.

Given

Current, $I = 0.5 \text{ A}$
Time, $t = 10 \text{ min} = 10 \times 60 \text{ s} = 600 \text{ s}$

To Find

The amount of electric charge, $Q$

Formula

$Q = I \times t$

Solution

Substitute the given values into the formula:

$Q = 0.5 \text{ A} \times 600 \text{ s}$ $Q = 300 \text{ C}$

Final Answer The amount of electric charge that flows through the circuit is $300 \text{ C}$ .

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

What causes electric charges to flow? Charges in a conductor, like a copper wire, don't move on their own. They need a "push," similar to how water in a horizontal pipe needs a pressure difference to flow.

This "electric pressure" is called potential difference. It is the driving force that sets charges in motion. A battery or an electric cell creates this potential difference across its terminals through chemical reactions. When a conductor is connected to a battery, the potential difference makes the charges flow, creating an electric current. To keep the current flowing, the cell must continuously expend its stored chemical energy.

Potential difference (V) between two points is defined as the work done (W) to move a unit charge (Q) from one point to the other.

The formula for potential difference is: $V = \frac{W}{Q}$

Unit of Potential Difference The SI unit of electric potential difference is the volt (V), named after Alessandro Volta. One volt is the potential difference between two points when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. $1 \text{ volt} = \frac{1 \text{ joule}}{1 \text{ coulomb}}$ $1 \text{ V} = 1 \text{ J C}^{-1}$

Measuring Potential Difference A voltmeter is an instrument used to measure potential difference. It is always connected in parallel across the two points where the potential difference is to be measured.

Example

Example How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?

Given

Charge, $Q = 2 \text{ C}$
Potential difference, $V = 12 \text{ V}$

To Find

The work done, $W$

Formula

$W = V \times Q$

Solution

Substitute the given values into the formula:

$W = 12 \text{ V} \times 2 \text{ C}$ $W = 24 \text{ J}$

Final Answer The work done is $24 \text{ J}$ .

CIRCUIT DIAGRAM

It is often convenient to represent an electric circuit using a schematic diagram. In these diagrams, different electrical components like cells, batteries, bulbs, and switches are represented by standard symbols. This makes it easier to draw and understand circuit connections.

OHM'S LAW

Is there a relationship between the potential difference across a conductor and the current flowing through it? In 1827, German physicist Georg Simon Ohm discovered this relationship.

Ohm's law states that the potential difference ( $V$ ) across the ends of a metallic wire in an electric circuit is directly proportional to the current ( $I$ ) flowing through it, provided its temperature remains the same.

In other words: $V \propto I$ This can be written as an equation: $V = IR$ Here, R is a constant for a given conductor at a given temperature and is called its resistance.

Resistance

Resistance is the property of a conductor to resist the flow of charges through it. It controls the amount of current in a circuit for a given voltage.

From Ohm's law, we can define resistance as: $R = \frac{V}{I}$

The SI unit of resistance is the ohm, represented by the Greek letter $\Omega$ . One ohm is the resistance of a conductor if a potential difference of 1 volt across its ends causes a current of 1 ampere to flow through it. $1 \text{ ohm} = \frac{1 \text{ volt}}{1 \text{ ampere}}$

From the formula $I = V/R$ , we can see that the current through a resistor is inversely proportional to its resistance. If you double the resistance, the current gets halved (for the same voltage).

A rheostat is a device used in a circuit to change the resistance, thereby regulating the current without changing the voltage source.

FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS

The resistance of a conductor is not the same for all materials or all shapes. Experiments show that the resistance of a uniform metallic conductor depends on three main factors:

Length of the conductor (l): Resistance is directly proportional to the length. A longer wire has more resistance. $R \propto l$
Area of cross-section (A): Resistance is inversely proportional to the area of cross-section. A thicker wire (larger area) has less resistance. $R \propto \frac{1}{A}$
Nature of the material: Different materials have different abilities to resist the flow of charge.

Combining these factors, we get the formula for resistance: $R = \rho \frac{l}{A}$ Here, $\rho$ (rho) is a constant of proportionality called the electrical resistivity of the material.

Electrical Resistivity

Resistivity ( $\rho$ ) is a fundamental property of a material that measures how strongly it resists electric current. Its SI unit is the ohm-meter ( $\Omega$ m).

Good conductors, like metals and alloys, have very low resistivity (e.g., $10^{-8} \Omega \text{ m}$ to $10^{-6} \Omega \text{ m}$ ).
Insulators, like rubber and glass, have very high resistivity (e.g., $10^{12} \Omega \text{ m}$ to $10^{17} \Omega \text{ m}$ ).

Alloys generally have higher resistivity than their constituent pure metals. They also do not oxidize (burn) easily at high temperatures. This makes them ideal for use in heating devices like electric irons and toasters. Tungsten is used for filaments in light bulbs due to its extremely high melting point, while copper and aluminium are used for transmission lines because of their very low resistivity.

Example

Example (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is

1200 \Omega

? (b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is

100 \Omega

?

Given

Voltage, $V = 220 \text{ V}$
(a) Bulb resistance, $R_{bulb} = 1200 \Omega$
(b) Heater resistance, $R_{heater} = 100 \Omega$

To Find

(a) Current drawn by the bulb, $I_{bulb}$ (b) Current drawn by the heater, $I_{heater}$

Formula

$I = \frac{V}{R}$

Solution

(a) Current drawn by the electric bulb

$I_{bulb} = \frac{220 \text{ V}}{1200 \Omega} = 0.18 \text{ A}$

Answer for part (a) = $0.18 \text{ A}$

(b) Current drawn by the electric heater

$I_{heater} = \frac{220 \text{ V}}{100 \Omega} = 2.2 \text{ A}$

Answer for part (b) = $2.2 \text{ A}$

Note

Notice the large difference in current drawn by the bulb and the heater from the same 220 V source. The heater has a much lower resistance, so it draws significantly more current.

Example

Example The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Given

Initial potential difference, $V_1 = 60 \text{ V}$
Initial current, $I_1 = 4 \text{ A}$
New potential difference, $V_2 = 120 \text{ V}$

To Find

The new current, $I_2$

Formula

$R = \frac{V}{I}$ $I = \frac{V}{R}$

Solution

First, we calculate the resistance of the heater, which remains constant.

$R = \frac{V_1}{I_1} = \frac{60 \text{ V}}{4 \text{ A}} = 15 \Omega$

Now, we use this resistance to find the new current when the potential difference is 120 V.

$I_2 = \frac{V_2}{R} = \frac{120 \text{ V}}{15 \Omega} = 8 \text{ A}$

Final Answer The heater will draw a current of $8 \text{ A}$ .

Example

Example Resistance of a metal wire of length 1 m is

26 \Omega

at

20^{\circ}\mathrm{C}

. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.

Given

Resistance, $R = 26 \Omega$
Length, $l = 1 \text{ m}$
Diameter, $d = 0.3 \text{ mm} = 3 \times 10^{-4} \text{ m}$

To Find

The resistivity of the metal, $\rho$

Formula

The area of cross-section $A = \pi (\frac{d}{2})^2 = \frac{\pi d^2}{4}$ . The formula for resistivity is derived from $R = \rho \frac{l}{A}$ : $\rho = \frac{RA}{l} = \frac{R(\pi d^2 / 4)}{l} = \frac{R\pi d^2}{4l}$

Solution

Substitute the given values into the formula:

$\rho = \frac{26 \Omega \times \pi \times (3 \times 10^{-4} \text{ m})^2}{4 \times 1 \text{ m}}$ $\rho = \frac{26 \times 3.14 \times 9 \times 10^{-8}}{4} \Omega \text{ m}$ $\rho = 1.84 \times 10^{-6} \Omega \text{ m}$

Final Answer The resistivity of the metal is $1.84 \times 10^{-6} \Omega \text{ m}$ . Looking at Table 11.2, this value corresponds to the resistivity of manganese.

Example

Example A wire of given material having length

l

and area of cross-section

A

has a resistance of

4 \Omega

. What would be the resistance of another wire of the same material having length

l/2

and area of cross-section

2A

?

Given

For the first wire: Resistance $R_1 = 4 \Omega$ , length $l_1 = l$ , area $A_1 = A$ .
For the second wire: Length $l_2 = l/2$ , area $A_2 = 2A$ .
The material is the same, so resistivity $\rho$ is constant.

To Find

The resistance of the second wire, $R_2$

Formula

$R = \rho \frac{l}{A}$

Solution

For the first wire, we have: $R_1 = \rho \frac{l_1}{A_1} = \rho \frac{l}{A} = 4 \Omega$

Now, for the second wire: $R_2 = \rho \frac{l_2}{A_2} = \rho \frac{l/2}{2A}$

Let's simplify this expression: $R_2 = \rho \frac{l}{4A} = \frac{1}{4} \left( \rho \frac{l}{A} \right)$

Since we know that $\rho \frac{l}{A} = R_1 = 4 \Omega$ , we can substitute this value: $R_2 = \frac{1}{4} R_1 = \frac{1}{4} \times 4 \Omega = 1 \Omega$

Final Answer The resistance of the new wire is $1 \Omega$ .

RESISTANCE OF A SYSTEM OF RESISTORS

In electrical circuits, resistors are often connected together in different combinations. There are two basic ways to join resistors: in series and in parallel.

Resistors in Series

When resistors are connected end-to-end, they are said to be in series.

Key characteristics of a series circuit:

Current is the same: The same amount of current flows through every resistor in the series combination.
Potential difference is divided: The total potential difference (voltage) across the combination is equal to the sum of the potential differences across the individual resistors. $V = V_1 + V_2 + V_3$

To find the equivalent resistance ( $R_s$ ) of a series combination, which is the single resistance that could replace the entire combination without changing the total current and voltage, we use the following formula: $R_s = R_1 + R_2 + R_3 + \dots$ The equivalent resistance in a series circuit is always greater than any of the individual resistances.

Example

Example An electric lamp, whose resistance is

20 \Omega

, and a conductor of

4 \Omega

resistance are connected to a 6 V battery. Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Given

Lamp resistance, $R_1 = 20 \Omega$
Conductor resistance, $R_2 = 4 \Omega$
Battery voltage, $V = 6 \text{ V}$

To Find

(a) Total resistance, $R_s$ (b) Total current, $I$ (c) Potential difference across the lamp ( $V_1$ ) and conductor ( $V_2$ )

Formula

$R_s = R_1 + R_2$ $I = \frac{V}{R_s}$ $V_1 = I R_1 \quad \text{and} \quad V_2 = I R_2$

Solution

(a) Calculate the total resistance

The components are connected in series, so we add their resistances: $R_s = 20 \Omega + 4 \Omega = 24 \Omega$ Answer for part (a) = $24 \Omega$

(b) Calculate the current through the circuit

Using Ohm's law with the total resistance: $I = \frac{V}{R_s} = \frac{6 \text{ V}}{24 \Omega} = 0.25 \text{ A}$ Answer for part (b) = $0.25 \text{ A}$

(c) Calculate the potential difference across each component

The current (0.25 A) is the same through both the lamp and the conductor. Potential difference across the electric lamp: $V_1 = I R_1 = 0.25 \text{ A} \times 20 \Omega = 5 \text{ V}$ Potential difference across the conductor: $V_2 = I R_2 = 0.25 \text{ A} \times 4 \Omega = 1 \text{ V}$ Answer for part (c) = $V_{lamp} = 5 \text{ V}$ , $V_{conductor} = 1 \text{ V}$

Note

Notice that the sum of the individual potential differences (

5 \text{ V} + 1 \text{ V}

) equals the total battery voltage (

6 \text{ V}

), as expected for a series circuit.

Resistors in Parallel

When resistors are connected between the same two points, they are said to be in parallel.

Key characteristics of a parallel circuit:

Potential difference is the same: The potential difference (voltage) is the same across every resistor in the parallel combination.
Current is divided: The total current flowing from the source is divided among the branches, and the total current is the sum of the currents in the individual branches. $I = I_1 + I_2 + I_3$

To find the equivalent resistance ( $R_p$ ) of a parallel combination, we use the following formula: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$ The equivalent resistance in a parallel circuit is always less than the smallest individual resistance in the combination.

Advantages of Parallel Circuits in Homes

Independent Operation: If one appliance stops working, the others are unaffected because the circuit for each branch remains complete. In a series circuit, if one bulb fuses, the entire circuit breaks.
Proper Current: Different appliances require different amounts of current to operate correctly. A parallel circuit divides the current, allowing each appliance to draw the specific current it needs.
Constant Voltage: All appliances are connected to the same voltage source (e.g., 220 V in many countries), ensuring they operate at their correct power rating.

Example

Example In the circuit diagram, suppose the resistors

R_1

,

R_2

and

R_3

have the values

5 \Omega, 10 \Omega, 30 \Omega

, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

Given

Resistances: $R_1 = 5 \Omega$ , $R_2 = 10 \Omega$ , $R_3 = 30 \Omega$
Battery voltage, $V = 12 \text{ V}$

To Find

(a) Current through each resistor ( $I_1, I_2, I_3$ ) (b) Total current, $I$ (c) Total circuit resistance, $R_p$

Formula

$I = \frac{V}{R}$ $I = I_1 + I_2 + I_3$ $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Solution

(a) Calculate the current through each resistor

In a parallel circuit, the voltage across each resistor is the same as the battery voltage (12 V). Current through $R_1$ : $I_1 = \frac{V}{R_1} = \frac{12 \text{ V}}{5 \Omega} = 2.4 \text{ A}$ Current through $R_2$ : $I_2 = \frac{V}{R_2} = \frac{12 \text{ V}}{10 \Omega} = 1.2 \text{ A}$ Current through $R_3$ : $I_3 = \frac{V}{R_3} = \frac{12 \text{ V}}{30 \Omega} = 0.4 \text{ A}$ Answer for part (a) = $I_1 = 2.4 \text{ A}$ , $I_2 = 1.2 \text{ A}$ , $I_3 = 0.4 \text{ A}$

(b) Calculate the total current in the circuit

The total current is the sum of the currents in each branch: $I = I_1 + I_2 + I_3 = (2.4 + 1.2 + 0.4) \text{ A} = 4 \text{ A}$ Answer for part (b) = $4 \text{ A}$

(c) Calculate the total circuit resistance

We can find the total resistance in two ways. Method 1: Using the formula for parallel resistors. $\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{6+3+1}{30} = \frac{10}{30} = \frac{1}{3}$ Therefore, $R_p = 3 \Omega$ .

Method 2: Using Ohm's law with total voltage and total current. $R_p = \frac{V}{I} = \frac{12 \text{ V}}{4 \text{ A}} = 3 \Omega$ Answer for part (c) = $3 \Omega$

Example

Example If in Fig. 11.12,

R_1 = 10 \Omega, R_2 = 40 \Omega, R_3 = 30 \Omega, R_4 = 20 \Omega, R_5 = 60 \Omega

, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Given

Resistors: $R_1=10 \Omega, R_2=40 \Omega, R_3=30 \Omega, R_4=20 \Omega, R_5=60 \Omega$
Voltage, $V = 12 \text{ V}$
Arrangement: ( $R_1$ || $R_2$ ) in series with ( $R_3$ || $R_4$ || $R_5$ )

To Find

(a) Total resistance, $R_{total}$ (b) Total current, $I_{total}$

Formula

For parallel resistors: $\frac{1}{R_p} = \frac{1}{R_a} + \frac{1}{R_b} + \dots$ For series resistors: $R_s = R_a + R_b + \dots$ Ohm's Law: $I = \frac{V}{R}$

Solution

(a) Calculate the total resistance

First, find the equivalent resistance ( $R'$ ) of the parallel combination of $R_1$ and $R_2$ . $\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10} + \frac{1}{40} = \frac{4+1}{40} = \frac{5}{40} = \frac{1}{8}$ So, $R' = 8 \Omega$ .

Next, find the equivalent resistance ( $R''$ ) of the parallel combination of $R_3, R_4$ , and $R_5$ . $\frac{1}{R''} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2+3+1}{60} = \frac{6}{60} = \frac{1}{10}$ So, $R'' = 10 \Omega$ .

The total resistance of the circuit is the series combination of $R'$ and $R''$ . $R_{total} = R' + R'' = 8 \Omega + 10 \Omega = 18 \Omega$ Answer for part (a) = $18 \Omega$

(b) Calculate the total current

Using Ohm's law with the total resistance and total voltage: $I_{total} = \frac{V}{R_{total}} = \frac{12 \text{ V}}{18 \Omega} \approx 0.67 \text{ A}$ Answer for part (b) = $0.67 \text{ A}$

HEATING EFFECT OF ELECTRIC CURRENT

When a battery supplies energy to a circuit, some of that energy is used for useful work (like spinning a fan), but some is converted into heat, raising the temperature of the components. If a circuit is purely resistive (containing only resistors), all the energy supplied by the source is dissipated entirely as heat. This is known as the heating effect of electric current.

Consider a current $I$ flowing through a resistor $R$ with a potential difference $V$ across it for a time $t$ . The work done to move a charge $Q$ is $W = VQ$ . The power input to the circuit is the rate at which work is done: $P = \frac{W}{t} = \frac{VQ}{t}$ Since current $I = Q/t$ , the power is: $P = VI$ The energy supplied to the circuit in time $t$ is $Energy = P \times t = VIt$ . This energy is dissipated as heat ( $H$ ). Therefore, the heat produced is: $H = VIt$

By applying Ohm's Law ( $V = IR$ ), we can express this in another form, known as Joule's law of heating: $H = (IR)It = I^2Rt$

Joule's law of heating states that the heat produced in a resistor is:

Directly proportional to the square of the current ( $I^2$ ) for a given resistance.
Directly proportional to the resistance ( $R$ ) for a given current.
Directly proportional to the time ( $t$ ) for which the current flows.

Practical Applications of Heating Effect

The heating effect of current is used in many devices:

Heating Appliances: Electric irons, toasters, ovens, and heaters are designed to produce heat. They use heating elements made of alloys like nichrome, which have high resistivity and don't oxidize easily.
Electric Bulb: The filament of a bulb (made of tungsten, which has a very high melting point of $3380^{\circ}\mathrm{C}$ ) gets extremely hot and glows, producing light. The bulbs are filled with inactive gases like nitrogen and argon to prevent the filament from burning out.
Electric Fuse: A fuse is a safety device used to protect circuits and appliances from excessively high currents. It consists of a short wire with a low melting point. If the current exceeds a safe value, the fuse wire heats up, melts, and breaks the circuit, stopping the current flow.

Example

Example 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?

Given

Voltage, $V = 220 \text{ V}$
(a) Maximum power, $P_{max} = 840 \text{ W}$
(b) Minimum power, $P_{min} = 360 \text{ W}$

To Find

The current ( $I$ ) and resistance ( $R$ ) for both cases.

Formula

$P = VI \implies I = \frac{P}{V}$ $R = \frac{V}{I}$

Solution

(a) When heating is at the maximum rate

Current: $I = \frac{P_{max}}{V} = \frac{840 \text{ W}}{220 \text{ V}} \approx 3.82 \text{ A}$ Resistance: $R = \frac{V}{I} = \frac{220 \text{ V}}{3.82 \text{ A}} \approx 57.60 \Omega$ Answer for part (a) = Current is $3.82 \text{ A}$ and resistance is $57.60 \Omega$ .

(b) When heating is at the minimum rate

Current: $I = \frac{P_{min}}{V} = \frac{360 \text{ W}}{220 \text{ V}} \approx 1.64 \text{ A}$ Resistance: $R = \frac{V}{I} = \frac{220 \text{ V}}{1.64 \text{ A}} \approx 134.15 \Omega$ Answer for part (b) = Current is $1.64 \text{ A}$ and resistance is $134.15 \Omega$ .

Example

Example 100 J of heat is produced each second in a

4 \Omega

resistance. Find the potential difference across the resistor.

Given

Heat, $H = 100 \text{ J}$
Time, $t = 1 \text{ s}$
Resistance, $R = 4 \Omega$

To Find

The potential difference across the resistor, $V$

Formula

$H = I^2Rt \implies I = \sqrt{\frac{H}{Rt}}$ $V = IR$

Solution

First, calculate the current flowing through the resistor. $I = \sqrt{\frac{100 \text{ J}}{4 \Omega \times 1 \text{ s}}} = \sqrt{25} \text{ A} = 5 \text{ A}$

Now, use Ohm's law to find the potential difference. $V = I \times R = 5 \text{ A} \times 4 \Omega = 20 \text{ V}$

Final Answer The potential difference across the resistor is $20 \text{ V}$ .

ELECTRIC POWER

Electric power is the rate at which electric energy is dissipated or consumed in an electric circuit.

The formula for power ( $P$ ) can be expressed in three ways:

$P = VI$
$P = I^2R$ (substituting $V=IR$ )
$P = V^2/R$ (substituting $I=V/R$ )

The SI unit of electric power is the watt (W). One watt is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V. $1 \text{ W} = 1 \text{ volt} \times 1 \text{ ampere}$

Since the watt is a small unit, a larger unit called the kilowatt (kW) is often used, where $1 \text{ kW} = 1000 \text{ W}$ .

Commercial Unit of Electrical Energy

Electrical energy is the product of power and time ( $E = P \times t$ ). The commercial unit used to measure electrical energy is the kilowatt-hour (kWh), often called a "unit."

One kilowatt-hour is the energy consumed when 1 kilowatt of power is used for 1 hour. $1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s}$ $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$

Note

When you pay your electricity bill, you are not paying for electrons, but for the energy used to move those electrons through your appliances.

Example

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

Given

Voltage, $V = 220 \text{ V}$
Current, $I = 0.50 \text{ A}$

To Find

The power of the bulb, $P$

Formula

$P = V \times I$

Solution

Substitute the given values into the formula: $P = 220 \text{ V} \times 0.50 \text{ A}$ $P = 110 \text{ W}$

Final Answer The power of the bulb is $110 \text{ W}$ .

Example

Example 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 kWh?

Given

Power, $P = 400 \text{ W} = 0.4 \text{ kW}$
Operating time per day = 8 hours
Total duration = 30 days
Cost per kWh = Rs 3.00

To Find

The total cost of energy for 30 days.

Formula

Total Energy (kWh) = Power (kW) $\times$ Time (hours)

Solution

First, calculate the total operating time in 30 days. Total time = 8 hours/day $\times$ 30 days = 240 hours

Next, calculate the total energy consumed in kWh. Total Energy = $0.4 \text{ kW} \times 240 \text{ h} = 96 \text{ kWh}$

Finally, calculate the total cost. Total Cost = Total Energy $\times$ Cost per kWh Total Cost = $96 \text{ kWh} \times \text{Rs } 3.00/\text{kWh} = \text{Rs } 288.00$

Final Answer The cost to operate the refrigerator for 30 days is Rs 288.00.

ElectricityClass 10 Physics Notes

ELECTRIC CURRENT AND CIRCUIT

Defining and Measuring Electric Current

Given

To Find

Formula

Solution

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

Given

To Find

Formula

Solution

CIRCUIT DIAGRAM

OHM'S LAW

Resistance

FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS

Electrical Resistivity

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

RESISTANCE OF A SYSTEM OF RESISTORS

Resistors in Series

Given

To Find

Formula

Solution

Resistors in Parallel

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

HEATING EFFECT OF ELECTRIC CURRENT

Practical Applications of Heating Effect

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

ELECTRIC POWER

Commercial Unit of Electrical Energy

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution