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Chapter 12: Electricity
Electricity is a fundamental force that powers much of modern society. It's a versatile and easily controlled form of energy used in countless applications, from lighting homes to running industries. Understanding what electricity is, how it flows, what regulates its flow, and its effects is essential.
Electric Current And Circuit
Just as flowing water constitutes water current, the flow of electric charge through a conductor constitutes an electric current.
In metallic wires, the electric charge is carried by the movement of electrons. Conventionally, the direction of electric current is defined as the direction of flow of positive charges, which is opposite to the direction of electron flow (since electrons are negative). This convention was established before electrons were discovered.
Electric current is quantified as the rate of flow of electric charge through a particular area of a conductor.
If a net charge $Q$ flows through a cross-section of a conductor in time $t$, the electric current $I$ is given by:
$$ I = \frac{Q}{t} $$The SI unit of electric charge is the coulomb (C). One coulomb is the charge of approximately $6 \times 10^{18}$ electrons (the charge of a single electron is about $-1.6 \times 10^{-19}$ 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 units of current are milliampere (mA, $1 \text{ mA} = 10^{-3} \text{ A}$) and microampere ($\mu\text{A}$, $1 \mu\text{A} = 10^{-6} \text{ A}$).
An ammeter is an instrument used to measure electric current in a circuit. It is always connected in series with the component through which the current is to be measured.
An electric circuit is a continuous and closed path along which an electric current flows. A switch acts as a key to close or open the circuit, allowing or stopping the flow of current.
Example 12.1. 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.
Answer:
Given: Current, $I = 0.5$ A. Time, $t = 10$ minutes.
Convert time to seconds: $t = 10 \text{ min} \times 60 \text{ s/min} = 600 \text{ s}$.
Using the formula $I = Q/t$, we can find the charge $Q = I \times t$:
$Q = 0.5 \text{ A} \times 600 \text{ s} = 300 \text{ C}$.
The amount of electric charge that flows through the circuit is 300 coulombs.
Electric Potential And Potential Difference
For electric charge to flow in a conductor, there must be an 'electric pressure' difference along the conductor, similar to how a pressure difference causes water to flow in a pipe. This electric pressure difference is called potential difference.
A battery or an electric cell is a device that creates and maintains a potential difference across its terminals. The chemical reactions within a cell generate this potential difference.
When a cell is connected to a conducting circuit, the potential difference causes the charges (electrons in a wire) to move, producing an electric current. The cell expends its stored chemical energy to maintain this potential difference and current flow.
The electric potential difference ($V$) between two points in an electric circuit is defined as the work done ($W$) to move a unit charge ($Q$) from one point to the other.
$$ V = \frac{W}{Q} $$The SI unit of electric potential difference is the volt (V), named after Alessandro Volta. One volt is defined as the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move 1 coulomb of charge from one point to the other:
$1 \text{ V} = 1 \text{ J/C}^{-1}$.
A voltmeter is an instrument used to measure potential difference. It is always connected in parallel across the points between which the potential difference is to be measured.
Example 12.2. How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
Answer:
Given: Potential difference, $V = 12$ V. Charge, $Q = 2$ C.
Using the formula $V = W/Q$, we can find the work done $W = V \times Q$:
$W = 12 \text{ V} \times 2 \text{ C} = 24 \text{ J}$.
The work done in moving the charge is 24 joules.
Circuit Diagram
An electric circuit can be represented using a schematic diagram where standard symbols are used for various electrical components. This simplifies drawing and understanding circuits.
Table 12.1 lists some commonly used symbols:
| Sl. No. | Components | Symbols |
|---|---|---|
| 1 | An electric cell | |
| 2 | A battery or a combination of cells | |
| 3 | Plug key or switch (open) | |
| 4 | Plug key or switch (closed) | |
| 5 | A wire joint | |
| 6 | Wires crossing without joining | |
| 7 | Electric bulb | |
| 8 | A resistor of resistance R | |
| 9 | Variable resistance or rheostat | |
| 10 | Ammeter | |
| 11 | Voltmeter |
Ohm’s Law
There is a fundamental relationship between the potential difference across a conductor and the electric current flowing through it.
In 1827, Georg Simon Ohm discovered this relationship, known as Ohm's Law.
Ohm's Law: The potential difference ($V$) across the ends of a given metallic wire in an electric circuit is directly proportional to the current ($I$) flowing through it, provided its temperature remains the same.
$V \propto I$
This proportionality can be written as:
$V = R \times I$
where $R$ is a constant called the resistance of the conductor for a given temperature. Rearranging, we get:
$$ \frac{V}{I} = R $$Resistance ($R$) is a property of a conductor that opposes or resists the flow of electric charges (electrons) through it. The SI unit of resistance is the ohm ($\Omega$).
One ohm ($\Omega$) is the resistance of a conductor if a potential difference of 1 volt (V) across its ends causes a current of 1 ampere (A) to flow through it:
$1 \Omega = 1 \text{ V/A}$.
From Ohm's Law, the current through a resistor is inversely proportional to its resistance ($I = V/R$). Doubling the resistance will halve the current for the same potential difference.
A component used to vary the resistance in a circuit without changing the voltage source is called a variable resistance or rheostat.
Different materials have different resistances. Good conductors (like metals) offer low resistance, while poor conductors (resistors) offer appreciable resistance. Insulators offer very high resistance.
An experiment to verify Ohm's Law involves setting up a circuit with a variable voltage source (multiple cells), a resistor (e.g., nichrome wire), an ammeter, and a voltmeter. By changing the number of cells (varying V) and measuring the corresponding current (I), the ratio V/I is found to be constant, and the V-I graph is a straight line through the origin.
Factors On Which The Resistance Of A Conductor Depends
The resistance of a conductor depends on several factors related to its dimensions and the material it is made of.
Precise measurements show that the resistance ($R$) of a uniform metallic conductor is:
- Directly proportional to its length ($l$): Doubling the length doubles the resistance ($R \propto l$).
- Inversely proportional to its area of cross-section ($A$): Doubling the area of a wire halves its resistance ($R \propto 1/A$). Thicker wires have lower resistance than thinner wires of the same material and length.
- Depends on the nature of the material: Different materials have different resistances.
- Varies with temperature: Resistance of a material changes with temperature.
Combining the proportionalities with length and area:
$R \propto \frac{l}{A}$
Introducing a constant of proportionality, $\rho$ (rho), called the electrical resistivity of the material:
$$ R = \rho \frac{l}{A} $$Resistivity is a characteristic property of the material of the conductor. Its SI unit is ohm-metre ($\Omega$ m). Resistivity does not depend on the length or area of the conductor, only on the material and temperature.
Resistivity Values:
- Metals and alloys: Very low resistivity (e.g., $10^{-8}$ to $10^{-6} \Omega$ m). Good conductors. Alloys generally have higher resistivity than their constituent metals.
- Insulators (e.g., rubber, glass): Very high resistivity (e.g., $10^{12}$ to $10^{17} \Omega$ m). Poor conductors.
Alloys are often used in electrical heating devices because they have high resistivity and do not oxidise readily at high temperatures (e.g., nichrome in heaters, tungsten in bulb filaments).
| Material | Resistivity ($\Omega$ m) |
|---|---|
| Conductors | |
| Silver | $1.60 \times 10^{-8}$ |
| Copper | $1.62 \times 10^{-8}$ |
| Aluminium | $2.63 \times 10^{-8}$ |
| Tungsten | $5.20 \times 10^{-8}$ |
| Nickel | $6.84 \times 10^{-8}$ |
| Iron | $10.0 \times 10^{-8}$ |
| Chromium | $12.9 \times 10^{-8}$ |
| Mercury | $94.0 \times 10^{-8}$ |
| Manganese | $1.84 \times 10^{-6}$ |
| Alloys | |
| Constantan (Cu-Ni) | $49 \times 10^{-6}$ |
| Manganin (Cu-Mn-Ni) | $44 \times 10^{-6}$ |
| Nichrome (Ni-Cr-Mn-Fe) | $100 \times 10^{-6}$ |
| Insulators | |
| Glass | $10^{10} - 10^{14}$ |
| Hard rubber | $10^{13} - 10^{16}$ |
| Ebonite | $10^{15} - 10^{17}$ |
| Diamond | $10^{12} - 10^{13}$ |
| Paper (dry) | $10^{12}$ |
Example 12.3. (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 $\Omega$?
Answer:
Given: Voltage source, $V = 220$ V.
(a) For the electric bulb: Resistance, $R = 1200 \Omega$.
Using Ohm's Law, $I = V/R$:
$I_{\text{bulb}} = \frac{220 \text{ V}}{1200 \Omega} \approx 0.183 \text{ A}$.
The electric bulb will draw approximately 0.183 A of current.
(b) For the electric heater coil: Resistance, $R = 100 \Omega$.
Using Ohm's Law, $I = V/R$:
$I_{\text{heater}} = \frac{220 \text{ V}}{100 \Omega} = 2.2 \text{ A}$.
The electric heater coil will draw 2.2 A of current. Note the significant difference in current drawn despite being connected to the same voltage source, due to their different resistances.
Example 12.4. 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?
Answer:
First, find the resistance of the heater using the initial conditions.
Given: $V_1 = 60$ V, $I_1 = 4$ A.
Using Ohm's Law, $R = V_1/I_1$:
$R = \frac{60 \text{ V}}{4 \text{ A}} = 15 \Omega$.
The resistance of the electric heater is 15 $\Omega$. Assuming the temperature (and thus resistance) remains constant, we can use this resistance for the second case.
Second case: Potential difference, $V_2 = 120$ V. Resistance, $R = 15 \Omega$.
Using Ohm's Law, $I_2 = V_2/R$:
$I_2 = \frac{120 \text{ V}}{15 \Omega} = 8 \text{ A}$.
If the potential difference is increased to 120 V, the heater will draw a current of 8 A.
Example 12.5. Resistance of a metal wire of length 1 m is 26 $\Omega$ at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 12.2, predict the material of the wire.
Answer:
Given: Resistance, $R = 26 \Omega$. Length, $l = 1$ m. Diameter, $d = 0.3$ mm.
Convert diameter to metres: $d = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m} = 3 \times 10^{-4} \text{ m}$.
The area of cross-section of the wire is $A = \pi (d/2)^2 = \pi (r)^2$, where $r$ is the radius.
$r = d/2 = (3 \times 10^{-4} \text{ m}) / 2 = 1.5 \times 10^{-4} \text{ m}$.
$A = \pi (1.5 \times 10^{-4} \text{ m})^2 = \pi \times (1.5)^2 \times (10^{-4})^2 \text{ m}^2 = \pi \times 2.25 \times 10^{-8} \text{ m}^2$.
Using the formula $R = \rho \frac{l}{A}$, we can find resistivity $\rho = \frac{R \times A}{l}$:
$\rho = \frac{26 \Omega \times (\pi \times 2.25 \times 10^{-8} \text{ m}^2)}{1 \text{ m}}$
$\rho \approx 26 \times 3.14 \times 2.25 \times 10^{-8} \Omega \text{ m}$
$\rho \approx 183.69 \times 10^{-8} \Omega \text{ m} \approx 1.84 \times 10^{-6} \Omega \text{ m}$.
The resistivity of the metal at 20°C is approximately $1.84 \times 10^{-6} \Omega$ m. Looking at Table 12.2, this value corresponds to the resistivity of Manganese.
Example 12.6. 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?
Answer:
Let the resistance of the first wire be $R_1$ and the second wire be $R_2$.
For the first wire: Length = $l$, Area = $A$, Resistance $R_1 = 4 \Omega$.
Using the formula $R = \rho \frac{l}{A}$, we have $R_1 = \rho \frac{l}{A} = 4 \Omega$.
For the second wire: Length = $l/2$, Area = $2A$. The material is the same, so resistivity $\rho$ is the same.
$R_2 = \rho \frac{l/2}{2A} = \rho \frac{l}{4A}$.
We can rewrite $R_2$ in terms of $R_1$:
$R_2 = \frac{1}{4} \times \rho \frac{l}{A}$.
Since $R_1 = \rho \frac{l}{A} = 4 \Omega$, substitute this into the equation for $R_2$:
$R_2 = \frac{1}{4} \times R_1 = \frac{1}{4} \times 4 \Omega = 1 \Omega$.
The resistance of the second wire would be 1 $\Omega$.
Resistance Of A System Of Resistors
In electric circuits, multiple resistors are often used and connected in various combinations. There are two primary ways to connect resistors: in series and in parallel.
Resistors In Series
When resistors are connected end-to-end, they form a series combination. In a series circuit, the current is the same through every resistor, and the total potential difference across the combination is the sum of the potential differences across each individual resistor.
For resistors $R_1, R_2, R_3$ connected in series (Fig. 12.6), the total potential difference across the combination $V$ is the sum of the potential differences across each resistor ($V_1, V_2, V_3$).
$V = V_1 + V_2 + V_3$
The current $I$ is the same through all resistors: $I_1 = I_2 = I_3 = I$.
According to Ohm's Law, $V_1 = IR_1$, $V_2 = IR_2$, $V_3 = IR_3$.
If $R_s$ is the equivalent resistance of the series combination (a single resistor that would replace the combination and have the same total potential difference $V$ across it for the same total current $I$), then $V = IR_s$.
Substituting these into $V = V_1 + V_2 + V_3$:
$IR_s = IR_1 + IR_2 + IR_3$
Dividing by $I$ (since $I \neq 0$), we get the formula for equivalent resistance in series:
$$ R_s = R_1 + R_2 + R_3 $$For $n$ resistors in series, $R_s = R_1 + R_2 + ... + R_n$.
The equivalent resistance of resistors in series is equal to the sum of their individual resistances and is always greater than the largest individual resistance. This increases the total resistance in the circuit, thereby reducing the current.
Example 12.7. An electric lamp, whose resistance is 20 $\Omega$, and a conductor of 4 $\Omega$ resistance are connected to a 6 V battery (Fig. 12.9). 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.
Answer:
Given: Resistance of lamp, $R_1 = 20 \Omega$. Resistance of conductor, $R_2 = 4 \Omega$. Battery voltage, $V = 6$ V. The components are connected in series.
(a) Total resistance of the circuit ($R_s$): For resistors in series, $R_s = R_1 + R_2$.
$R_s = 20 \Omega + 4 \Omega = 24 \Omega$.
The total resistance of the circuit is 24 $\Omega$.
(b) Current through the circuit ($I$): Using Ohm's Law for the entire circuit, $I = V/R_s$.
$I = \frac{6 \text{ V}}{24 \Omega} = 0.25 \text{ A}$.
The current through the circuit is 0.25 A.
(c) Potential difference across the electric lamp ($V_1$) and conductor ($V_2$): Using Ohm's Law for each component ($V=IR$). Note that the current $I$ is the same (0.25 A) for both as they are in series.
$V_1 = I \times R_1 = 0.25 \text{ A} \times 20 \Omega = 5 \text{ V}$.
$V_2 = I \times R_2 = 0.25 \text{ A} \times 4 \Omega = 1 \text{ V}$.
The potential difference across the lamp is 5 V, and across the conductor is 1 V. Note that $V_1 + V_2 = 5 \text{ V} + 1 \text{ V} = 6 \text{ V}$, which is equal to the total battery voltage $V$.
Resistors In Parallel
When resistors are connected across the same two points, they form a parallel combination. In a parallel circuit, the potential difference is the same across each resistor, and the total current flowing into the combination is divided among the individual resistors.
For resistors $R_1, R_2, R_3$ connected in parallel (Fig. 12.7), the potential difference $V$ is the same across each resistor: $V_1 = V_2 = V_3 = V$.
The total current $I$ flowing into the combination is the sum of the currents through each branch ($I_1, I_2, I_3$).
$I = I_1 + I_2 + I_3$
According to Ohm's Law, $I_1 = V/R_1$, $I_2 = V/R_2$, $I_3 = V/R_3$.
If $R_p$ is the equivalent resistance of the parallel combination, then $I = V/R_p$.
Substituting these into $I = I_1 + I_2 + I_3$:
$V/R_p = V/R_1 + V/R_2 + V/R_3$
Dividing by $V$ (assuming $V \neq 0$), we get the formula for equivalent resistance in parallel:
$$ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$For $n$ resistors in parallel, $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$.
The reciprocal of the equivalent resistance of resistors in parallel is equal to the sum of the reciprocals of their individual resistances. This means the equivalent resistance of resistors in parallel is always less than the smallest individual resistance. Connecting resistors in parallel decreases the total resistance in the circuit, thereby increasing the total current.
Advantages of Parallel Connection:
- The voltage is the same across each component, allowing devices requiring different currents to operate correctly when connected to the same source voltage.
- If one component fails (circuit breaks in one branch), the current can still flow through other parallel branches, so the other components continue to work. (Unlike a series circuit, where one failure breaks the entire circuit).
- The total resistance is reduced, which can be beneficial for drawing required total current from the source.
For these reasons, electrical devices in domestic circuits are connected in parallel.
Example 12.8. In the circuit diagram given in Fig. 12.10, suppose the resistors R₁, R₂ and R₃ 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.
Answer:
Given: $R_1 = 5 \Omega$, $R_2 = 10 \Omega$, $R_3 = 30 \Omega$. Battery voltage, $V = 12$ V. The resistors are connected in parallel.
In parallel connection, the potential difference across each resistor is the same as the battery voltage, $V_1 = V_2 = V_3 = V = 12$ V.
(a) Current through each resistor: Using Ohm's Law, $I = V/R$ for each resistor.
$I_1 = V/R_1 = \frac{12 \text{ V}}{5 \Omega} = 2.4 \text{ A}$.
$I_2 = V/R_2 = \frac{12 \text{ V}}{10 \Omega} = 1.2 \text{ A}$.
$I_3 = V/R_3 = \frac{12 \text{ V}}{30 \Omega} = 0.4 \text{ A}$.
The currents through the resistors are 2.4 A, 1.2 A, and 0.4 A, respectively.
(b) Total current in the circuit ($I$): The total current is the sum of the currents in each parallel branch.
$I = I_1 + I_2 + I_3 = 2.4 \text{ A} + 1.2 \text{ A} + 0.4 \text{ A} = 4.0 \text{ A}$.
The total current in the circuit is 4.0 A.
(c) Total circuit resistance ($R_p$): Using the formula for parallel resistance, $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
$\frac{1}{R_p} = \frac{1}{5 \Omega} + \frac{1}{10 \Omega} + \frac{1}{30 \Omega}$
Find a common denominator (e.g., 30):
$\frac{1}{R_p} = \frac{6}{30 \Omega} + \frac{3}{30 \Omega} + \frac{1}{30 \Omega} = \frac{6+3+1}{30 \Omega} = \frac{10}{30 \Omega} = \frac{1}{3 \Omega}$.
$R_p = 3 \Omega$.
The total circuit resistance is 3 $\Omega$. Note that this is less than the smallest individual resistance (5 $\Omega$).
Example 12.9. If in Fig. 12.12, R₁ = 10 $\Omega$, R₂ = 40 $\Omega$, R₃ = 30 $\Omega$, R₄ = 20 $\Omega$, R₅ = 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.
Answer:
The circuit has a combination of series and parallel resistors.
Resistors $R_1$ (10 $\Omega$) and $R_2$ (40 $\Omega$) are in parallel. Let their equivalent resistance be $R_{p1}$.
$\frac{1}{R_{p1}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10 \Omega} + \frac{1}{40 \Omega} = \frac{4}{40 \Omega} + \frac{1}{40 \Omega} = \frac{5}{40 \Omega} = \frac{1}{8 \Omega}$.
$R_{p1} = 8 \Omega$.
Resistors $R_3$ (30 $\Omega$), $R_4$ (20 $\Omega$), and $R_5$ (60 $\Omega$) are also in parallel. Let their equivalent resistance be $R_{p2}$.
$\frac{1}{R_{p2}} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{30 \Omega} + \frac{1}{20 \Omega} + \frac{1}{60 \Omega}$.
Find a common denominator (e.g., 60):
$\frac{1}{R_{p2}} = \frac{2}{60 \Omega} + \frac{3}{60 \Omega} + \frac{1}{60 \Omega} = \frac{2+3+1}{60 \Omega} = \frac{6}{60 \Omega} = \frac{1}{10 \Omega}$.
$R_{p2} = 10 \Omega$.
Now, the equivalent resistance $R_{p1}$ is connected in series with the equivalent resistance $R_{p2}$.
(a) The total resistance in the circuit ($R_{\text{total}}$) is the sum of these series resistances:
$R_{\text{total}} = R_{p1} + R_{p2} = 8 \Omega + 10 \Omega = 18 \Omega$.
The total resistance in the circuit is 18 $\Omega$.
(b) The total current flowing in the circuit ($I$) is given by Ohm's Law for the entire circuit: $I = V/R_{\text{total}}$.
Given battery voltage, $V = 12$ V.
$I = \frac{12 \text{ V}}{18 \Omega} = \frac{2}{3} \text{ A} \approx 0.67 \text{ A}$.
The total current flowing in the circuit is approximately 0.67 A.
Heating Effect Of Electric Current
When electric current flows through a resistor, some electrical energy is converted into heat. This is an inevitable consequence of current flow and is known as the heating effect of electric current or Joule's heating.
The electrical energy supplied by the source to maintain the current is dissipated as heat in the resistor. This effect is utilised in various heating devices like electric irons, toasters, heaters, kettles, etc.
Consider a resistor of resistance $R$ with a potential difference $V$ across it, through which a current $I$ flows for time $t$. The work done in moving charge $Q$ through potential difference $V$ is $W = VQ$. Since $Q = It$, the work done is $W = VIt$. This work done is the energy supplied by the source, which is dissipated as heat in the resistor.
So, the amount of heat $H$ produced in time $t$ is:
$H = VIt$
Using Ohm's Law ($V = IR$), we can express heat in terms of current and resistance:
$H = (IR)It$
$$ H = I^2Rt $$This is Joule's Law of Heating. It states that the heat produced in a resistor is directly proportional to:
- The square of the current ($I^2$).
- The resistance ($R$).
- The time ($t$) for which the current flows.
Joule's heating is undesirable in components like connecting wires (where it can cause temperature rise and alter properties) but is the basis of operation for heating appliances.
Example 12.10. 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?
Answer:
Given: Voltage, $V = 220$ V. Power input $P = VI$.
From this, we can find current $I = P/V$ and then resistance $R = V/I = V/(P/V) = V^2/P$.
(a) When heating is at the maximum rate: Power, $P_{\text{max}} = 840$ W.
Current, $I_{\text{max}} = P_{\text{max}}/V = \frac{840 \text{ W}}{220 \text{ V}} \approx 3.82 \text{ A}$.
Resistance, $R_{\text{max}} = V/I_{\text{max}} = \frac{220 \text{ V}}{3.82 \text{ A}} \approx 57.60 \Omega$. (Or $R_{\text{max}} = V^2/P_{\text{max}} = (220 \text{ V})^2 / 840 \text{ W} = 48400 / 840 \Omega \approx 57.62 \Omega$)
At maximum heating, the current is approximately 3.82 A and the resistance is approximately 57.6 $\Omega$.
(b) When heating is at the minimum rate: Power, $P_{\text{min}} = 360$ W.
Current, $I_{\text{min}} = P_{\text{min}}/V = \frac{360 \text{ W}}{220 \text{ V}} \approx 1.64 \text{ A}$.
Resistance, $R_{\text{min}} = V/I_{\text{min}} = \frac{220 \text{ V}}{1.64 \text{ A}} \approx 134.15 \Omega$. (Or $R_{\text{min}} = V^2/P_{\text{min}} = (220 \text{ V})^2 / 360 \text{ W} = 48400 / 360 \Omega \approx 134.44 \Omega$)
At minimum heating, the current is approximately 1.64 A and the resistance is approximately 134.15 $\Omega$. (Note: The change in heating rate is achieved by changing the resistance, likely using a switch to include more or less of the heating coil in the circuit).
Example 12.11. 100 J of heat is produced each second in a 4 $\Omega$ resistance. Find the potential difference across the resistor.
Answer:
Given: Heat produced, $H = 100$ J. Time, $t = 1$ s. Resistance, $R = 4 \Omega$.
Using Joule's Law of Heating, $H = I^2Rt$. We can find the current $I$:
$100 \text{ J} = I^2 \times 4 \Omega \times 1 \text{ s}$.
$100 = 4 I^2$
$I^2 = \frac{100}{4} = 25$.
$I = \sqrt{25} = 5 \text{ A}$ (taking the positive value for current magnitude).
Now find the potential difference $V$ across the resistor using Ohm's Law, $V = IR$:
$V = 5 \text{ A} \times 4 \Omega = 20 \text{ V}$.
The potential difference across the resistor is 20 V.
Practical Applications Of Heating Effect Of Electric Current
Joule's heating has numerous useful applications in everyday life and technology.
Examples:
- Heating Appliances: Electric iron, toaster, oven, kettle, heater. These devices use heating elements (usually made of alloys like nichrome) with high resistance, converting electrical energy into heat effectively.
- Electric Bulb: The filament of an electric bulb gets heated to a very high temperature ($> 2000^\circ$C) due to Joule's heating and emits light. The filament is made of tungsten (high melting point) and is enclosed in a bulb filled with inactive gases (nitrogen, argon) to prevent oxidation and prolong its life. Most energy is dissipated as heat, only a small part is light.
- Electric Fuse: A safety device that protects circuits and appliances from damage due to excessive current. It contains a wire made of a material with a low melting point (e.g., alloy of lead and tin). If the current exceeds a safe limit, the fuse wire melts and breaks the circuit, stopping the current flow. Fuses are rated based on the maximum current they can safely carry (e.g., 1 A, 5 A, 10 A). A fuse is always connected in series with the appliance.
Electric Power
Electric power ($P$) is the rate at which electric energy is consumed or dissipated in an electric circuit. It is also the rate at which work is done by the electric current.
From the work done ($W = VIt$) or heat dissipated ($H = VIt$) in time $t$, the power is:
$P = \frac{\text{Energy consumed}}{\text{Time}} = \frac{VIt}{t} = VI$.
Using Ohm's Law, we can express power in terms of current and resistance, or voltage and resistance:
- $P = VI$
- Using $V=IR$, $P = (IR)I = I^2R$.
- Using $I=V/R$, $P = V(V/R) = V^2/R$.
So, the electric power is given by:
$$ P = VI = I^2R = \frac{V^2}{R} $$The SI unit of electric power is the watt (W). One watt is the power consumed by a device carrying 1 ampere of current when operated at a potential difference of 1 volt ($1 \text{ W} = 1 \text{ V} \times 1 \text{ A}$).
The watt is a small unit. For practical purposes, larger units are used:
- Kilowatt (kW): $1 \text{ kW} = 1000 \text{ W}$.
- Megawatt (MW): $1 \text{ MW} = 10^6 \text{ W}$.
Electrical energy consumed is often expressed in watt-hour (W h) or kilowatt-hour (kW 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 the kilowatt-hour (kW h), commonly called a 'unit' of electricity.
$1 \text{ kW h} = 1 \text{ kW} \times 1 \text{ h} = 1000 \text{ W} \times 3600 \text{ s} = 3,600,000 \text{ J}$.
$1 \text{ kW h} = 3.6 \times 10^6 \text{ J}$.
We pay electricity companies for the electrical energy we use, measured in kilowatt-hours.
Example 12.12. An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Answer:
Given: Voltage, $V = 220$ V. Current, $I = 0.50$ A.
Using the power formula $P = VI$:
$P = 220 \text{ V} \times 0.50 \text{ A} = 110 \text{ W}$.
The power of the bulb is 110 watts.
Example 12.13. An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at $\textsf{₹}$ 3.00 per kW h?
Answer:
Given: Power rating of refrigerator, $P = 400$ W. Operating time per day = 8 hours.
Number of days = 30. Cost per kW h = $\textsf{₹}$ 3.00.
Calculate the total energy consumed in 30 days:
Daily energy consumption = Power $\times$ Time = $400 \text{ W} \times 8 \text{ h} = 3200 \text{ W h}$.
Total energy consumption in 30 days = Daily consumption $\times$ Number of days = $3200 \text{ W h/day} \times 30 \text{ days} = 96000 \text{ W h}$.
Convert energy from W h to kW h: $96000 \text{ W h} = 96000 / 1000 \text{ kW h} = 96 \text{ kW h}$.
Calculate the total cost:
Total cost = Total energy consumed (in kW h) $\times$ Cost per kW h.
Total cost = $96 \text{ kW h} \times \textsf{₹}$ 3.00/kW h = $\textsf{₹}$ 288.00.
The cost of operating the refrigerator for 30 days is $\textsf{₹}$ 288.00.
Intext Questions
Page No. 200
Question 1. What does an electric circuit mean?
Answer:
Question 2. Define the unit of current.
Answer:
Question 3. Calculate the number of electrons constituting one coulomb of charge.
Answer:
Page No. 202
Question 1. Name a device that helps to maintain a potential difference across a conductor.
Answer:
Question 2. What is meant by saying that the potential difference between two points is 1 V?
Answer:
Question 3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Page No. 209
Question 1. On what factors does the resistance of a conductor depend?
Answer:
Question 2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Question 3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
Question 4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
Question 5. Use the data in Table 12.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
Page No. 213
Question 1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 $\Omega$ resistor, an 8 $\Omega$ resistor, and a 12 $\Omega$ resistor, and a plug key, all connected in series.
Answer:
Question 2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 $\Omega$ resistor. What would be the readings in the ammeter and the voltmeter?
Answer:
Page No. 216
Question 1. Judge the equivalent resistance when the following are connected in parallel –
(a) 1 $\Omega$ and $10^6 \Omega$,
(b) 1 $\Omega$ and $10^3 \Omega$, and $10^6 \Omega$.
Answer:
Question 2. An electric lamp of 100 $\Omega$, a toaster of resistance 50 $\Omega$, and a water filter of resistance 500 $\Omega$ are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
Question 3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
Question 4. How can three resistors of resistances 2 $\Omega$, 3 $\Omega$, and 6 $\Omega$ be connected to give a total resistance of
(a) 4 $\Omega$,
(b) 1 $\Omega$?
Answer:
Question 5. What is
(a) the highest,
(b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 $\Omega$, 8 $\Omega$, 12 $\Omega$, 24 $\Omega$?
Answer:
Page No. 218
Question 1. Why does the cord of an electric heater not glow while the heating element does?
Answer:
Question 2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
Question 3. An electric iron of resistance 20 $\Omega$ takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Page No. 220
Question 1. What determines the rate at which energy is delivered by a current?
Answer:
Question 2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Exercises
Question 1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is –
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer:
Question 2. Which of the following terms does not represent electrical power in a circuit?
(a) $I^2R$
(b) $IR^2$
(c) VI
(d) $V^2/R$
Answer:
Question 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
Question 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer:
Question 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
Question 6. A copper wire has diameter 0.5 mm and resistivity of $1.6 \times 10^{–8} \Omega m$. What will be the length of this wire to make its resistance 10 $\Omega$? How much does the resistance change if the diameter is doubled?
Answer:
Question 7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –
| I (amperes) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
|---|---|---|---|---|---|
| V (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
Question 8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Question 9. A battery of 9 V is connected in series with resistors of 0.2 $\Omega$, 0.3 $\Omega$, 0.4 $\Omega$ , 0.5 $\Omega$ and 12 $\Omega$, respectively. How much current would flow through the 12 $\Omega$ resistor?
Answer:
Question 10. How many 176 $\Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Question 11. Show how you would connect three resistors, each of resistance 6 $\Omega$, so that the combination has a resistance of (i) 9 $\Omega$, (ii) 4 $\Omega$.
Answer:
Question 12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Question 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 $\Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
Question 14. Compare the power used in the 2 $\Omega$ resistor in each of the following circuits:
(i) a 6 V battery in series with 1 $\Omega$ and 2 $\Omega$ resistors, and
(ii) a 4 V battery in parallel with 12 $\Omega$ and 2 $\Omega$ resistors.
Answer:
Question 15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Question 16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
Question 17. An electric heater of resistance 8 $\Omega$ draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
Question 18. Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer: