Chapter-12 Electricity

Explanation:

An electric circuit is a path or loop through which an electric current can flow. It is made up of electrical components such as wires, resistors, switches, capacitors, and batteries or other power sources, connected together in a complete and closed loop.

The electric circuit provides a means for the electric charge to flow through a closed loop from a power source, through various components that perform specific functions, and back to the power source. The components in a circuit are connected in a specific order to control the flow of electric current and create the desired result, such as lighting a bulb or powering an electronic device.

Electric circuits can be found in a wide variety of applications, from simple circuits used in household appliances to complex circuits used in computer systems and other electronic devices. Understanding the principles of electric circuits is essential for designing and troubleshooting electrical systems and devices.

Explanation:

The unit of electric current is the ampere (symbol: A). It is defined as the amount of electric charge passing through a given point in a circuit per unit time. One ampere of current is defined as the flow of one coulomb of electric charge per second. In other words, if one coulomb of electric charge flows through a circuit in one second, the current flowing through that circuit is one ampere. The ampere is one of the seven SI base units, which are used to express all physical quantities in the International System of Units (SI).

Explanation:

The elementary charge of an electron is approximately equal to 1.6 x 10^-19 Coulombs (C). This means that each electron carries a charge of 1.6 x 10^-19 C. Therefore, the number of electrons constituting one coulomb of charge can be calculated as follows:

1 Coulomb of charge = 1 / (1.6 x 10^-19) electrons

= 6.25 x 10^18 electrons

Therefore, there are approximately 6.25 x 10^18 electrons constituting one coulomb of charge.

Explanation:

A device that helps to maintain a potential difference across a conductor is called a voltage source. A voltage source is a device that provides a fixed voltage or potential difference between two points in an electrical circuit, and it is able to maintain this potential difference even when a current is flowing through the circuit. Examples of voltage sources include batteries, generators, and power supplies.

Explanation:

When we say that the potential difference between two points is 1 volt (1 V), it means that one joule (1 J) of work is required to move one coulomb (1 C) of electric charge from one point to the other against the electric field.

In other words, if we have a circuit with two points between which there is a potential difference of 1 V, and we want to move one coulomb of charge from the point of lower potential to the point of higher potential, we need to do one joule of work to overcome the electric force between the charges and move them against the electric field.

Thus, the potential difference between two points is a measure of the amount of energy required to move electric charge from one point to another, and it is expressed in volts (V).

Explanation:

The energy given to each coulomb of charge passing through a 6 V battery is equal to the potential difference across the battery, which is 6 volts (6 V).

We know that the potential difference, V, between two points in an electric circuit is defined as the amount of energy, E, required to move a unit of charge, Q, from one point to another:

V = E / Q

Rearranging the equation, we can find the energy given to each coulomb of charge passing through the battery:

E = V x Q

Since the potential difference across the 6 V battery is 6 V, the energy given to each coulomb of charge passing through the battery is:

E = 6 V x 1 C = 6 J/C

Therefore, each coulomb of charge passing through the 6 V battery receives 6 joules (6 J) of energy.

Explanation:

The resistance of a conductor depends on the following factors:

1.Length: The resistance of a conductor is directly proportional to its length. The longer the conductor, the higher the resistance.

2.Cross-sectional area: The resistance of a conductor is inversely proportional to its cross-sectional area. The larger the cross-sectional area, the lower the resistance.

3.Material: The resistance of a conductor depends on the material it is made of. Some materials offer higher resistance than others. For example, copper has lower resistance than iron.

4.Temperature: The resistance of a conductor changes with temperature. In general, the resistance of a conductor increases with an increase in temperature.

5.Presence of impurities: The presence of impurities in a conductor can increase its resistance. This is because impurities disrupt the flow of electrons and cause them to collide more frequently, which increases resistance.

Explanation:

Current will flow more easily through a thick wire than a thin wire of the same material, when connected to the same source. This is because the resistance of a wire is inversely proportional to its cross-sectional area. Therefore, a thicker wire will have a lower resistance than a thinner wire of the same material and length.

According to Ohm's law, the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to its resistance. So, when the same voltage is applied to two wires of different thicknesses, the one with lower resistance will allow more current to flow through it. In this case, the thick wire offers less resistance and hence, more current flows through it..

Explanation:

According to Ohm's Law, the current through a component is directly proportional to the potential difference across it, provided the resistance remains constant.

Therefore, if the potential difference across the component decreases to half of its former value, the current through it will also decrease to half of its former value, as long as the resistance of the component remains constant.

In other words, if the resistance of the component remains constant, reducing the potential difference across it by half will result in a proportional decrease in the current flowing through it, also by half.

Explanation:

Coils of electric toasters and electric irons are made of an alloy rather than a pure metal due to several reasons:

1.Higher resistivity: Alloys used in the coils have higher resistivity than pure metals. This means that they offer more resistance to the flow of electric current, which allows them to heat up more quickly and efficiently.

2.Greater strength: Alloys are often stronger and more durable than pure metals, making them better suited for use in devices like toasters and irons that are subjected to repeated heating and cooling cycles.

3.Resistance to oxidation: Many alloys are more resistant to oxidation than pure metals. This is important in electric toasters and irons, which are often exposed to air and moisture, as oxidation can cause the heating element to degrade and fail prematurely.

Explanation:

a. Iron is a better conductor than mercury. Iron has a higher electrical conductivity than mercury, which means it offers less resistance to the flow of electric current.

b. Silver is the best conductor of electricity among all the known elements. It has the highest electrical conductivity of any element, followed closely by copper and gold. Silver's exceptional conductivity makes it the preferred material for use in high-performance electrical applications such as electronics, power generation, and transmission. However, silver is expensive and not always practical for large-scale applications, so copper is often used as a more cost-effective alternative.

Explanation:

A battery of three cells of 2 V each equals to battery of potential 6 V. The circuit diagram below shows three resistors of resistance 12 Ω, 8 Ω and 5 Ω connected in series along with a battery of potential 6 V.

Explanation:

An ammeter should always be connected in series with resistors while the voltmeter should be connected in parallel to the resistor to measure the potential difference as shown in the figure below.

We may find the ammeter and voltmeter readings by applying Ohm's Law.

The circuit has a total resistance of 5 + 8 + 12 = 25.

We may determine the current flowing through the circuit or the resistors using the following formula since we know the potential difference of the circuit is 6 V:

I = V/R = 6/25 = 0.24A

Let V1 to represent the potential difference across the 12 resistor.

The following calculations can be made using the acquired current V1:

V1 = 0.24A × 12 Ω = 2.88 V

As a result, the ammeter will read 0.24 A and the voltmeter 2.88 V.

Explanation:

(a) When 1 Ω and 106 Ω resistors are connected in parallel, the equivalent resistance can be calculated using the formula:

1/R_eq = 1/R_1 + 1/R_2

where R_1 = 1 Ω and R_2 = 106 Ω

Substituting the values, we get:

1/R_eq = 1/1 + 1/106

1/R_eq = 0.9906

R_eq = 1.0094 Ω (approx.)

Therefore, the equivalent resistance when 1 Ω and 106 Ω resistors are connected in parallel is approximately 1.0094 Ω.

(b) When 1 Ω, 103 Ω, and 106 Ω resistors are connected in parallel, the equivalent resistance can be calculated using the same formula:

1/R_eq = 1/R_1 + 1/R_2 + 1/R_3

where R_1 = 1 Ω, R_2 = 103 Ω, and R_3 = 106 Ω

Substituting the values, we get:

1/R_eq = 1/1 + 1/103 + 1/106

1/R_eq = 0.0102

R_eq = 97.96 Ω (approx.)

Therefore, the equivalent resistance when 1 Ω, 103 Ω, and 106 Ω resistors are connected in parallel is approximately 97.96 Ω.

Explanation:

The electric lamp, the toaster and the water filter connected in parallel to a 220 V source can be shown as using a circuit diagram as follows:

The resistance of the electric iron box is 31.25 Ω.

When the electric lamp, toaster, and water filter are connected in parallel, the equivalent resistance of the combination can be calculated as:

1/R_eq = 1/R_1 + 1/R_2 + 1/R_3

where R_1 = 100 Ω, R_2 = 50 Ω, and R_3 = 500 Ω

Substituting the values, we get:

1/R_eq = 1/100 + 1/50 + 1/500

1/R_eq = 0.03

R_eq = 33.33 Ω (approx.)

Therefore, the equivalent resistance of the appliances when connected in parallel is approximately 33.33 Ω.

Now, the electric iron is connected to the same 220 V source and draws the same current as all three appliances combined. Let the resistance of the electric iron be R_i.

According to Ohm's law, the current through a resistor is given by:

I = V/R

where I is the current, V is the voltage, and R is the resistance.

Since the current drawn by the appliances in parallel is the same as the current drawn by the electric iron, we can calculate the current as:

I = V/R_eq

I = 220/33.33

I = 6.60 A (approx.)

The resistance of the electric iron can be calculated as:

I = V/R_i

R_i = V/I

R_i = 220/6.60

R_i = 33.33 Ω (approx.)

Therefore, the resistance of the electric iron connected to the same source that takes as much current as all three appliances is approximately 33.33 Ω, and the current through it is approximately 6.60 A.

Explanation:

Connecting electrical devices in parallel with a battery has several advantages over connecting them in series:

1.Voltage: When devices are connected in parallel, each device receives the full voltage of the battery. This is because the positive and negative terminals of the battery are connected directly to each device. In contrast, when devices are connected in series, the voltage is divided among the devices. For example, if two devices are connected in series to a 12-volt battery, each device will receive only 6 volts.

2.Current: In a parallel circuit, each device receives the current it needs, independently of the other devices. This means that devices can draw as much current as they need without affecting the other devices. In contrast, in a series circuit, the current is the same for all devices, which can limit the amount of current each device can draw.

3.Redundancy: In a parallel circuit, if one device fails, the other devices will continue to function normally. This is because each device is connected to the battery independently. In contrast, in a series circuit, if one device fails, the entire circuit will fail.

4.Easy to add or remove devices: In a parallel circuit, it is easy to add or remove devices without affecting the other devices. Each device is connected independently to the battery. In contrast, in a series circuit, adding or removing devices can affect the voltage and current of the entire circuit.

Explanation:

(a) The circuit diagram below shows the connection of three resistors

From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by

The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows:

Req= 2 Ω +2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) The circuit diagram below, shows the connection of three resistors.

From the circuit, it is understood that all the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:

The total resistance of the circuit is 1 Ω.

Explanation:

(a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest.

Their equivalent resistance connected in parallel is

Hence, the lowest total resistance is 2 Ω.

Explanation:

The heating element in an electric heater is designed to convert electrical energy into heat energy through the process of resistive heating. It is typically made of a high-resistance material that heats up when an electric current passes through it. The cord of an electric heater, on the other hand, is made of a material that has a low resistance, which means that it does not heat up significantly when electricity flows through it.

When an electric current flows through the heating element, it encounters resistance and produces heat. The heating element is specifically designed to have a high enough resistance to convert a significant portion of the electrical energy into heat, causing it to glow and emit heat. However, the cord of the heater has a much lower resistance than the heating element, so most of the electrical energy flowing through it is not converted into heat, and it does not get hot enough to glow.

In summary, the heating element is made of a high-resistance material that is designed to convert electrical energy into heat, while the cord of the heater is made of a low-resistance material that does not convert much electrical energy into heat, and thus does not get hot enough to glow.

Explanation:

The amount of heat generated when charge Q is transferred through a potential difference V is given by the formula:

H = QV

where H is the heat generated in joules (J), Q is the charge in coulombs (C), and V is the potential difference in volts (V).

In this case, we are transferring 96000 coulombs of charge in one hour through a potential difference of 50 V. One hour is equal to 3600 seconds, so the time t taken to transfer the charge is:

t = 1 hour = 3600 seconds

Using the formula, the heat generated is:

H = QV = 96000 C * 50 V = 4,800,000 J

Therefore, the amount of heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V is 4,800,000 joules (or 4.8 megajoules).

Explanation:

The heat developed in a resistor is given by the formula:

H = I^2 * R * t

where H is the heat developed in joules (J), I is the current in amperes (A), R is the resistance in ohms (Ω), and t is the time in seconds (s).

In this case, the electric iron has a resistance of 20 Ω and takes a current of 5 A for 30 s. Plugging these values into the formula, we get:

H = I^2 * R * t = 5^2 * 20 Ω * 30 s = 7500 J

Therefore, the heat developed in the electric iron in 30 s is 7500 joules.

Explanation:

The rate at which energy is delivered by an electric current is determined by two factors: the amount of current flowing through a device and the potential difference across the device.

The amount of current flowing through a device is measured in amperes (A) and represents the rate at which charge flows through the device. The more current that flows through a device, the more energy is delivered per unit of time, and the greater the rate at which energy is delivered.

The potential difference across a device is measured in volts (V) and represents the amount of work required to move a unit of charge through the device. The greater the potential difference across a device, the more energy is delivered per unit of charge, and the greater the rate at which energy is delivered.

The rate at which energy is delivered by a current is therefore determined by the product of these two factors, known as power, which is measured in watts (W):

Power = Current x Voltage

or

P = I x V

So, the rate at which energy is delivered by a current is directly proportional to the current flowing through the device and the potential difference across the device.

Explanation:

The power of the electric motor can be determined by multiplying the current drawn from the line by the voltage across the line:

Power (P) = Current (I) x Voltage (V)

In this case, the motor takes 5 A from a 220 V line, so the power of the motor is:

P = 5 A x 220 V = 1100 W

Therefore, the power of the electric motor is 1100 watts (or 1.1 kilowatts).

The energy consumed by the motor in 2 hours can be determined by multiplying the power of the motor by the time it runs:

Energy (E) = Power (P) x Time (t)

In this case, the motor runs for 2 hours, so the energy consumed by the motor is:

E = 1100 W x 2 hours x 3600 seconds/hour = 7,920,000 joules

Therefore, the energy consumed by the motor in 2 hours is 7,920,000 joules (or 7.92 megajoules).

Explanation:

 When the five equal parts of the wire are connected in parallel, the equivalent resistance R' can be calculated using the formula:

1/R' = 1/R + 1/R + 1/R + 1/R + 1/R

Simplifying this equation, we get:

1/R' = 5/R

Multiplying both sides by R'R, we get:

R = 5R'

Therefore, the ratio R/R' is:

R/R' = 5R' / R' = 5

Hence, the answer is (c) 5.

Explanation:

Option (b) IR^2 does not represent electrical power in a circuit.

The electrical power in a circuit can be calculated using any of the following formulas:

P = IV (where I is the current and V is the voltage)

P = I^2R (where I is the current and R is the resistance)

P = V^2/R (where V is the voltage and R is the resistance)

In option (b), IR^2 does not represent electrical power. It is the product of the current and the square of the resistance, which has no physical meaning or units associated with it.

Therefore, the answer is (b) IR^2.

Explanation:

The power consumed by an electric bulb is given by the formula:

P = V^2/R

Where P is the power, V is the voltage, and R is the resistance.

We can rearrange the formula to solve for R:

R = V^2/P

The resistance of the bulb is constant, regardless of the voltage applied to it. Therefore, if we know the resistance of the bulb, we can use the formula to calculate the power consumed at a different voltage.

The resistance of the bulb can be found by using the given power and voltage:

P = V^2/R

100 W = (220 V)^2 / R

R = (220 V)^2 / 100 W

R = 484 Ω

Now we can calculate the power consumed when the bulb is operated at 110 V:

P = V^2/R

P = (110 V)^2 / 484 Ω

P = 25 W

Therefore, the answer is (d) 25 W.

Explanation:

The heat produced in a wire is given by the formula:

H = I²Rt

Where H is the heat produced, I is the current, R is the resistance, and t is the time.

Let's assume that the wires have a resistance of R each. When the wires are connected in series, the equivalent resistance of the circuit is:

R_series = 2R

When the wires are connected in parallel, the equivalent resistance of the circuit is:

1/R_parallel = 1/R + 1/R = 2/R

R_parallel = R/2

We can assume that the potential difference across the wires is the same in both cases. Therefore, the current through the wires can be calculated using Ohm's law:

V = IR

I = V/R

When the wires are connected in series, the current through each wire is the same and is given by:

I_series = V/(2R)

When the wires are connected in parallel, the total current through the circuit is the sum of the currents through each wire and is given by:

I_parallel = 2V/R

The time for which the wires are connected is assumed to be the same in both cases. Therefore, the ratio of heat produced in the series and parallel combinations is given by:

H_series/H_parallel = (I_series)^2 R t / (I_parallel)^2 R_parallel t

H_series/H_parallel = ((V/2R)^2 R)/(2V/R)^2 (R/2)

H_series/H_parallel = 1/4

Therefore, the answer is (c) 1:4. The heat produced in the series combination is one-fourth the heat produced in the parallel combination.

Explanation:

A voltmeter is a device that is used to measure the potential difference or voltage between two points in an electric circuit. It is always connected in parallel with the component or part of the circuit whose voltage is being measured.

To connect a voltmeter in the circuit, you need to follow these steps:

1.Identify the two points between which you want to measure the potential difference.

2.Turn off the power source to the circuit to avoid any electrical shock.

3.Connect the voltmeter in parallel with the component or part of the circuit whose voltage you want to measure.

4.To do this, you need to connect the positive (red) lead of the voltmeter to the point with higher potential, and the negative (black) lead to the point with lower potential.

5.If the circuit has a switch, make sure it is closed, and turn on the power source.

6.Read the voltage measurement on the voltmeter display.

7.If you are done with the measurement, turn off the power source, and disconnect the voltmeter.

Explanation:

The resistance of the copper wire of length in meters and area of cross-section m2 is given by the formula

The length of the wire is 122.72 m and the new resistance is 2.5 Ω.

Explanation:

The plot between voltage and current is known as IV characteristic. The current is plotted in the y-axis while the voltage is plotted in the x-axis. The different values of current for different values of voltage are given in the table. The I V characteristics for the given resistor is shown below.

The slope of the line gives the value of resistance.

The slope can be calculated as follows:

Slope = 1/R = BC/AC = 2/6.8

To calculate R,

R = 6.8/2 = 3.4 Ω

The resistance of the resistor is 3.4 Ω.

Explanation:

We can use Ohm's law to find the resistance of the unknown resistor. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. The constant of proportionality is known as the resistance of the conductor. Mathematically, Ohm's law can be expressed as:

V = IR

where V is the voltage across the conductor, I is the current through the conductor, and R is the resistance of the conductor.

In this problem, we are given that a 12 V battery is connected across an unknown resistor, and there is a current of 2.5 mA in the circuit. We can use Ohm's law to find the resistance of the resistor as follows:

R = V/I = 12 V / (2.5 mA) = 4800 Ω

Therefore, the value of the resistance of the unknown resistor is 4800 Ω

Explanation:

To find the current that flows through the 12 Ω resistor, we need to first find the total resistance of the circuit, and then use Ohm's law to calculate the current.

The total resistance of a series circuit is the sum of the resistances of all the components in the circuit. In this case, the total resistance of the circuit is:

R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Now we can use Ohm's law to find the current that flows through the 12 Ω resistor:

I = V / R = 9 V / 13.4 Ω ≈ 0.672 A

Therefore, the current that flows through the 12 Ω resistor is approximately 0.672 A.

Explanation:

We can use Ohm's law and the formula for the total resistance of resistors in parallel to solve this problem.

Ohm's law states that the current through a resistor is equal to the voltage across it divided by its resistance:

I = V / R

In a parallel circuit, the total resistance is given by:

1/R_total = 1/R1 + 1/R2 + 1/R3 + ...

where R1, R2, R3, etc. are the resistances of each resistor in the circuit.

We can rearrange this equation to find the total resistance of n resistors in parallel:

R_total = 1 / (1/R1 + 1/R2 + 1/R3 + ... + 1/Rn)

In this problem, we are given that the voltage is 220 V and the current is 5 A. We can use Ohm's law to find the total resistance of the circuit:

R_total = V / I = 220 V / 5 A = 44 Ω

Now we can use the formula for the total resistance of resistors in parallel to find the equivalent resistance of n 176 Ω resistors in parallel:

1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn

1/44 Ω = 1/176 Ω + 1/176 Ω + 1/176 Ω + ... + 1/176 Ω (n times)

Simplifying the equation, we get:

1/44 Ω = n/176 Ω

n = 4

Therefore, we need 4 176 Ω resistors in parallel to carry 5 A on a 220 V line.

Explanation:

If we connect all the three resistors in series, their equivalent resistor would 6 Ω + 6 Ω + 6 Ω =18 Ω, which is not the desired value. Similarly, if we connect all the three resistors in parallel, their equivalent resistor would be

which is again not the desired value.

We can obtain the desired value by connecting any two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

The third resistor is in series, hence the equivalent resistance is calculated as follows:

R = 6 Ω + 3 Ω = 9 Ω

Case (ii)

When two resistors are connected in series, their equivalent resistance is given by

R = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is calculated as follows:

Explanation:

We can use the formula for electrical power to calculate the current drawn by each 10 W bulb:

P = VI

I = P / V = 10 W / 220 V = 0.045 A

Now we can use the maximum allowable current of 5 A to calculate the maximum number of bulbs that can be connected in parallel:

5 A = n * 0.045 A

n = 5 A / 0.045 A = 111.11

Since we cannot have a fractional number of bulbs, the maximum number of bulbs that can be connected in parallel is 111.

Therefore, we can connect a maximum of 111 10 W bulbs in parallel across a 220 V electric supply line if the maximum allowable current is 5 A.

Explanation:

Case (i) When coils are used separately

Using Ohm’s law, we can find the current flowing through each coil as follows:

9.166 A of current flows through each resistor when they are used separately.

Case (ii) When coils connected in series

The total resistance in the series circuit is 24 Ω + 24 Ω = 48 Ω

The current flowing through the series circuit is calculated as follows:

Therefore, a current of 4.58 A flows through the series circuit.

Case (iii) When coils connected in parallel

When the coils are connected in parallel, the equivalent resistance is calculated as follows:

The current in the parallel circuit is 18.33 A.

Explanation:

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series, hence their equivalent resistance is given by 1 Ω + 2 Ω = 3 Ω. The current in the circuit can be calculated using the Ohm’s law as follows:

Therefore, the power consumed by the 2 Ω is 8 W.

(ii) When 12 Ω and 2 Ω resistors are connected in parallel, the voltage across the resistors remains the same. Knowing that the voltage across 2 Ω resistor is 4 V, we can calculate the power consumed by the resistor as follows:

The power consumed by the 2 Ω resistor is 8 W.

Explanation:

The voltage across each of the two bulbs will be the same because they are both linked in parallel.

It is possible to compute the current drawn by a 100 W bulb as follows:

P = V × I

I = P/V

When we replace the values in the equation, we obtain

I = 100 W/220 V = 100/220 A

Similarly, the following formula can be used to determine the current consumed by the 60 W bulb:

I = 60 W/220 V = 60/220 A

the current drawn from the line is as a result.

Explanation:

Equation H = Pt, where P is the appliance's power and t is the time, provides the amount of energy used by electrical equipment.

This formula can be used to determine how much energy a TV with a 250 W power rating uses:

9 105 J is equal to H = 250 W 3600 seconds.

The energy used by a toaster with a power rating of 1200 W is calculated as H = 1200 W 600 s = 7.2 105 J.

From the calculations, it can be concluded that the TV uses more energy than the toaster.

Explanation:

The rate at which the heat develops in the heater can be calculated using the following formula

P = I2 R

Substituting the values in the equation, we get

P = (15A) 2 × 8 Ω = 1800 watt

The electric heater produces heat at the rate of 1800 watt

Explanation:

a. Tungsten is used almost exclusively for the filament of electric lamps because of its high melting point (3422°C) and its ability to resist deformation at high temperatures. When an electric current passes through the tungsten filament, it heats up and emits light. The high melting point of tungsten ensures that the filament does not melt, even at high temperatures, and its ability to resist deformation ensures that the filament maintains its shape, which is crucial for the filament to emit a stable and consistent light.

b. The conductors of electric heating devices, such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because alloys have a higher resistivity than pure metals. This means that they offer more resistance to the flow of electric current, and as a result, they can heat up more quickly and to a higher temperature than pure metals. Additionally, alloys can be tailored to have specific properties such as resistance and melting point, which makes them suitable for use in heating devices.

c. The series arrangement is not used for domestic circuits because if one device in the circuit fails or is turned off, the entire circuit will be broken and none of the other devices in the circuit will work. In a parallel arrangement, each device has its own path to the power source, which means that if one device fails or is turned off, the other devices in the circuit can still function.

d. The resistance of a wire is inversely proportional to its area of cross-section. This means that as the area of cross-section of a wire increases, its resistance decreases. This is because a larger area of cross-section provides more space for the electric current to flow, which reduces the collisions between the electrons and the atoms of the wire, and hence, reduces the resistance.

e. Copper and aluminium wires are usually employed for electricity transmission because they have low resistivity, which means that they offer low resistance to the flow of electric current. This reduces the amount of energy lost as heat during transmission, making the transmission more efficient. Additionally, copper and aluminium are abundant, relatively inexpensive, and easy to work with, which makes them suitable for use in large-scale electricity transmission systems.