ElectricityClass 10 Physics NCERT Solutions

Answer: (d) 25

Explanation: Let the initial resistance of the wire be $R$ and its length be $l$. Resistance is directly proportional to length. When the wire is cut into five equal parts, the length of each part becomes $l/5$.

Therefore, the resistance of each of the five parts is: $R_{\text{part}} = \frac{R}{5}$

These five parts are then connected in parallel. The equivalent resistance $R'$ of this parallel combination is given by the formula: $\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$

Substituting the resistance of each part: $\frac{1}{R'} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5}$ $\frac{1}{R'} = 5 \times \frac{1}{R/5} = 5 \times \frac{5}{R} = \frac{25}{R}$

Therefore, the equivalent resistance is: $R' = \frac{R}{25}$

The question asks for the ratio $R / R'$. $\frac{R}{R'} = \frac{R}{R/25} = R \times \frac{25}{R} = 25$

Thus, the ratio $R / R'$ is 25.

Answer: (b) $I R^{2}$

Explanation: Electrical power ($P$) is the rate at which electrical energy is consumed in a circuit. The basic formula for power is: $P = V I$ where $V$ is the potential difference and $I$ is the current.

Using Ohm's law, $V = IR$, we can derive other forms of the power formula:

Option (c) is the fundamental definition of electrical power.

The term $I R^2$ does not represent electrical power. Therefore, it is the correct answer.

Answer: (d) 25 W

Explanation: The resistance of the electric bulb's filament is constant. We can calculate this resistance from its power rating.

Given (Rating): Rated Voltage, $V = 220 \text{ V}$ Rated Power, $P = 100 \text{ W}$

Formula: The relationship between power, voltage, and resistance is $P = \frac{V^2}{R}$. We can find the resistance $R$: $R = \frac{V^2}{P}$

Calculation of Resistance: $R = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} \Omega = 484 \Omega$

Now, the bulb is operated at a new voltage.

Given (Operation): Operating Voltage, $V' = 110 \text{ V}$ Resistance, $R = 484 \Omega$ (this remains constant)

Calculation of Power Consumed: Using the same formula, we calculate the new power consumed, $P'$: $P' = \frac{(V')^2}{R} = \frac{(110 \text{ V})^2}{484 \Omega} = \frac{12100}{484} \text{ W} = 25 \text{ W}$

Final Answer: The power consumed will be 25 W.

Answer: (c) 1:4

Explanation: Let the resistance of each of the two conducting wires be $R$. The potential difference across the combination is $V$.

1. Series Combination: When the wires are connected in series, the total resistance is: $R_s = R + R = 2R$ The heat produced ($H_s$) in time $t$ is given by the formula $H = \frac{V^2}{R_{\text{eq}}} t$. $H_s = \frac{V^2}{R_s} t = \frac{V^2}{2R} t \quad \dots (1)$

2. Parallel Combination: When the wires are connected in parallel, the total resistance $R_p$ is given by: $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$ $R_p = \frac{R}{2}$ The heat produced ($H_p$) in time $t$ is: $H_p = \frac{V^2}{R_p} t = \frac{V^2}{R/2} t = \frac{2V^2}{R} t \quad \dots (2)$

Ratio of Heat Produced: Now, we find the ratio of heat produced in the series combination to that in the parallel combination. $\frac{H_s}{H_p} = \frac{\frac{V^2}{2R} t}{\frac{2V^2}{R} t}$ $\frac{H_s}{H_p} = \frac{V^2}{2R} \times \frac{R}{2V^2} = \frac{1}{4}$

Thus, the ratio $H_s : H_p$ is $1:4$.

A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured.

A voltmeter has a very high internal resistance. When connected in parallel, it draws a negligible amount of current from the main circuit. This ensures that the potential difference being measured is not significantly altered by the presence of the voltmeter, leading to an accurate measurement.

Given: Diameter, $d = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$ Resistivity, $\rho = 1.6 \times 10^{-8} \Omega \text{ m}$ Resistance, $R = 10 \Omega$

To Find: Length of the wire, $l$

Formula: The resistance of a wire is given by $R = \rho \frac{l}{A}$, where $A$ is the cross-sectional area. The area is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$. Rearranging for length, $l = \frac{R A}{\rho}$.

Calculation: First, calculate the area $A$: $A = \frac{\pi (0.5 \times 10^{-3})^2}{4} = \frac{\pi \times 0.25 \times 10^{-6}}{4} \approx 0.19635 \times 10^{-6} \text{ m}^2$ Now, calculate the length $l$: $l = \frac{10 \Omega \times (0.19635 \times 10^{-6} \text{ m}^2)}{1.6 \times 10^{-8} \Omega \text{ m}}$ $l = \frac{1.9635 \times 10^{-6}}{1.6 \times 10^{-8}} \text{ m} = 1.227 \times 10^2 \text{ m} = 122.7 \text{ m}$

Final Answer (Part 1): The length of the wire will be approximately $122.7$ m.

Resistance $R$ is inversely proportional to the area of cross-section $A$. $R \propto \frac{1}{A}$ Since $A = \frac{\pi d^2}{4}$, the area is proportional to the square of the diameter ($A \propto d^2$). Therefore, resistance is inversely proportional to the square of the diameter: $R \propto \frac{1}{d^2}$

Let the initial resistance be $R_1$ and the initial diameter be $d_1$. The new diameter is $d_2 = 2d_1$. Let the new resistance be $R_2$.

The ratio of the new resistance to the old resistance is: $\frac{R_2}{R_1} = \frac{1/d_2^2}{1/d_1^2} = \left(\frac{d_1}{d_2}\right)^2 = \left(\frac{d_1}{2d_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ So, $R_2 = \frac{1}{4} R_1$.

The new resistance will be one-fourth of the original resistance.

Calculation of new resistance: $R_2 = \frac{1}{4} \times 10 \Omega = 2.5 \Omega$

Final Answer (Part 2): If the diameter is doubled, the resistance becomes one-fourth of its original value, changing from $10 \Omega$ to $2.5 \Omega$.

Plotting the Graph: To find the resistance, we plot a graph of potential difference $V$ (on the y-axis) against current $I$ (on the x-axis). The data points are (0.5, 1.6), (1.0, 3.4), (2.0, 6.7), (3.0, 10.2), and (4.0, 13.2).

The graph will be a straight line passing through the origin, which verifies Ohm's law ($V \propto I$).

Calculating the Resistance: According to Ohm's law, $V = IR$. The resistance $R$ is the slope of the V-I graph. $R = \text{Slope} = \frac{\Delta V}{\Delta I}$

We can calculate the slope by taking any two points on the line of best fit. Let's use the first and last data points for an approximate calculation: Point 1: ($I_1 = 0.5$ A, $V_1 = 1.6$ V) Point 2: ($I_2 = 4.0$ A, $V_2 = 13.2$ V)

Calculation: $\text{Slope} = \frac{V_2 - V_1}{I_2 - I_1} = \frac{13.2 - 1.6}{4.0 - 0.5} = \frac{11.6}{3.5} \approx 3.31 \Omega$

Alternatively, we can find the average of the resistance values calculated from each data pair ($R = V/I$): $R_1 = 1.6 / 0.5 = 3.20 \Omega$ $R_2 = 3.4 / 1.0 = 3.40 \Omega$ $R_3 = 6.7 / 2.0 = 3.35 \Omega$ $R_4 = 10.2 / 3.0 = 3.40 \Omega$ $R_5 = 13.2 / 4.0 = 3.30 \Omega$

Average Resistance: $R_{\text{avg}} = \frac{3.20 + 3.40 + 3.35 + 3.40 + 3.30}{5} = \frac{16.65}{5} = 3.33 \Omega$

Both methods give a similar result.

Final Answer: The resistance of the resistor is approximately $3.3 \Omega$.

Given: Potential difference, $V = 12 \text{ V}$ Current, $I = 2.5 \text{ mA} = 2.5 \times 10^{-3} \text{ A}$

To Find: Resistance of the resistor, $R$

Formula: According to Ohm's law: $V = IR$ Therefore, the resistance $R$ can be calculated as: $R = \frac{V}{I}$

Calculation: $R = \frac{12 \text{ V}}{2.5 \times 10^{-3} \text{ A}}$ $R = \frac{12}{2.5} \times 10^3 \Omega$ $R = 4.8 \times 10^3 \Omega = 4800 \Omega$

Final Answer: The value of the resistance of the resistor is $4800 \Omega$ or $4.8 \text{ k}\Omega$.

Given: Voltage of the battery, $V = 9 \text{ V}$ Resistances connected in series: $R_1 = 0.2 \Omega, R_2 = 0.3 \Omega, R_3 = 0.4 \Omega, R_4 = 0.5 \Omega, R_5 = 12 \Omega$.

To Find: Current flowing through the $12 \Omega$ resistor.

Concept: In a series circuit, the current is the same through every component. Therefore, we first need to calculate the total current flowing in the circuit by finding the total equivalent resistance.

Formula: The equivalent resistance ($R_s$) for resistors in series is the sum of their individual resistances: $R_s = R_1 + R_2 + R_3 + R_4 + R_5$ By Ohm's law, the total current $I$ is: $I = \frac{V}{R_s}$

Calculation: First, calculate the total resistance $R_s$: $R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \Omega$ Now, calculate the total current $I$ in the circuit: $I = \frac{9 \text{ V}}{13.4 \Omega} \approx 0.6716 \text{ A}$

Since all the resistors are connected in series, the same current flows through each of them.

Final Answer: The current flowing through the $12 \Omega$ resistor is approximately $0.67$ A.

Given: Voltage of the line, $V = 220 \text{ V}$ Total current required, $I = 5 \text{ A}$ Resistance of each resistor, $R = 176 \Omega$

To Find: The number of resistors required, $n$.

Formula: First, we find the total equivalent resistance ($R_p$) required for the circuit using Ohm's law: $R_p = \frac{V}{I}$ Next, if $n$ identical resistors of resistance $R$ are connected in parallel, their equivalent resistance $R_p$ is given by: $\frac{1}{R_p} = \frac{n}{R} \quad \text{or} \quad R_p = \frac{R}{n}$

Final Answer: 4 resistors of $176 \Omega$ each are required to be connected in parallel.

Let the three resistors be $R_1 = 6 \Omega$, $R_2 = 6 \Omega$, and $R_3 = 6 \Omega$.

(i) To get a total resistance of $9 \Omega$: This can be achieved by connecting two resistors in parallel and then connecting the third resistor in series with this parallel combination.

Arrangement for (i): Connect two $6 \Omega$ resistors in parallel, and connect this pair in series with the third $6 \Omega$ resistor.

(ii) To get a total resistance of $4 \Omega$: This can be achieved by connecting two resistors in series and then connecting the third resistor in parallel with this series combination.

Arrangement for (ii): Connect two $6 \Omega$ resistors in series, and connect this pair in parallel with the third $6 \Omega$ resistor.

Given: Voltage of the supply line, $V = 220 \text{ V}$ Power rating of each bulb, $P = 10 \text{ W}$ Maximum allowable current, $I_{\text{max}} = 5 \text{ A}$

To Find: The number of lamps ($n$) that can be connected in parallel.

Formula: First, we calculate the current drawn by a single bulb using the formula $P = VI$. $I_{\text{bulb}} = \frac{P}{V}$ When $n$ bulbs are connected in parallel, the total current drawn from the supply is the sum of the currents drawn by each bulb. $I_{\text{total}} = n \times I_{\text{bulb}}$ The total current must not exceed the maximum allowable current, so $I_{\text{total}} \leq I_{\text{max}}$.

Final Answer: A maximum of 110 lamps can be connected in parallel.

Given: Voltage of the supply line, $V = 220 \text{ V}$ Resistance of coil A, $R_A = 24 \Omega$ Resistance of coil B, $R_B = 24 \Omega$

Formula: The current $I$ is calculated using Ohm's law: $I = \frac{V}{R_{\text{eq}}}$, where $R_{\text{eq}}$ is the equivalent resistance for each case.

Case 1: Coils used separately When only one coil is used, the resistance in the circuit is $R_{\text{eq}} = 24 \Omega$. $I_1 = \frac{V}{R_A} = \frac{220 \text{ V}}{24 \Omega} \approx 9.17 \text{ A}$

Case 2: Coils used in series When the coils are connected in series, the equivalent resistance is: $R_s = R_A + R_B = 24 \Omega + 24 \Omega = 48 \Omega$ The current in the circuit is: $I_2 = \frac{V}{R_s} = \frac{220 \text{ V}}{48 \Omega} \approx 4.58 \text{ A}$

Case 3: Coils used in parallel When the coils are connected in parallel, the equivalent resistance is: $\frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12} \Omega^{-1}$ $R_p = 12 \Omega$ The current in the circuit is: $I_3 = \frac{V}{R_p} = \frac{220 \text{ V}}{12 \Omega} \approx 18.33 \text{ A}$

Final Answer: The currents in the three cases are:

Given: Voltage, $V = 6 \text{ V}$ Resistors are in series: $R_1 = 1 \Omega$, $R_2 = 2 \Omega$.

Given: Voltage, $V = 4 \text{ V}$ Resistors are in parallel: $R_1 = 12 \Omega$, $R_2 = 2 \Omega$.

Comparison: The power used in the $2 \Omega$ resistor in circuit (i) is $P_1 = 8 \text{ W}$. The power used in the $2 \Omega$ resistor in circuit (ii) is $P_2 = 8 \text{ W}$.

The ratio of the power used is $\frac{P_1}{P_2} = \frac{8 \text{ W}}{8 \text{ W}} = 1$.

Final Answer: The power used in the $2 \Omega$ resistor is the same in both circuits ($8$ W).

Given: Lamp 1 rating: Power $P_1 = 100 \text{ W}$, Voltage $V_1 = 220 \text{ V}$ Lamp 2 rating: Power $P_2 = 60 \text{ W}$, Voltage $V_2 = 220 \text{ V}$ Supply voltage, $V_{\text{supply}} = 220 \text{ V}$

To Find: Total current ($I_{\text{total}}$) drawn from the line.

Final Answer: The current drawn from the line is $\frac{8}{11}$ A, which is approximately $0.73$ A.

Given: For the TV set: Power, $P_{TV} = 250 \text{ W}$ Time, $t_{TV} = 1 \text{ hr} = 3600 \text{ s}$

For the toaster: Power, $P_{\text{toaster}} = 1200 \text{ W}$ Time, $t_{\text{toaster}} = 10 \text{ minutes} = 10 \times 60 \text{ s} = 600 \text{ s}$

To Find: Which appliance uses more energy.

Formula: Electrical energy ($E$) is the product of power ($P$) and time ($t$): $E = P \times t$

Comparison: Comparing the energy consumed by both appliances: $900,000 \text{ J} > 720,000 \text{ J}$ Therefore, $E_{TV} > E_{\text{toaster}}$.

Final Answer: The 250 W TV set used for 1 hour consumes more energy than the 1200 W toaster used for 10 minutes.

Given: Resistance of the heater, $R = 44 \Omega$ Current drawn, $I = 5 \text{ A}$ Time, $t = 2 \text{ hours}$ (This information is not needed to find the rate).

To Find: The rate at which heat is developed in the heater.

Concept: The rate at which heat is developed is the electrical power ($P$).

Formula: Power can be calculated using Joule's law of heating. The rate of heat production (power) is given by: $P = I^2 R$

Calculation: $P = (5 \text{ A})^2 \times 44 \Omega$ $P = 25 \text{ A}^2 \times 44 \Omega$ $P = 1100 \text{ W}$

Since 1 Watt is equal to 1 Joule per second (1 J/s), the rate of heat development is 1100 J/s.

Final Answer: The rate at which heat is developed in the heater is $1100$ W.

(a) Why is the tungsten used almost exclusively for filament of electric lamps? Tungsten is used for the filaments of electric lamps due to the following properties:

(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? Alloys (like Nichrome) are used in heating devices for two main reasons:

(c) Why is the series arrangement not used for domestic circuits? Series arrangements are unsuitable for domestic circuits due to several disadvantages:

(d) How does the resistance of a wire vary with its area of cross-section? The resistance ($R$) of a wire is inversely proportional to its area of cross-section ($A$). This relationship is expressed by the formula: $R = \rho \frac{l}{A}$ This means that if the area of cross-section increases (i.e., the wire becomes thicker), its resistance decreases. Conversely, if the area of cross-section decreases (i.e., the wire becomes thinner), its resistance increases. A thicker wire provides more pathways for the electric charge to flow, thus offering less opposition.

(e) Why are copper and aluminium wires usually employed for electricity transmission? Copper and aluminium are used for electricity transmission lines for the following reasons: