Class 10 Science Chapter 11 Electricity

Key Concept: Ohm’s Law Application

c) 0.5 A
[Solution Description] Using Ohm’s Law, $I = \frac{V}{R}$, where $V = 9 \, V$ and $R = 18 \, \Omega$:
$I = \frac{9 \, V}{18 \, \Omega} = 0.5 \, A$
Thus, the current flowing through the circuit is 0.5 A.

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Key Concept: Real-world Application

c) Rs 2.25
[Solution Description]
First, calculate the daily energy consumption in kilowatt-hours $(kWh)$:
$\text{Daily Energy} = 60 \text{ W} \times \frac{15}{60} \text{ h} = 15 \text{ Wh} = 0.015 \text{ kWh}$
Multiply by 30 days to get the monthly energy usage:
$\text{Monthly Energy} = 0.015 \text{ kWh/day} \times 30 \text{ days} = 0.45 \text{ kWh}$
Finally, multiply by the cost per kWh:
$\text{Cost} = 0.45 \text{ kWh} \times 5 \text{ Rs/kWh} = 2.25 \text{ Rs}$

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Key Concept: Resistance Calculation, Advanced Ohm’s Law

d) 9R
[Solution Description] The initial resistance of the wire is given by $R = \rho \frac{l}{A}$. When the wire is stretched to three times its original length, its new length becomes $l’ = 3l$. Since the volume remains constant, we have $l’A’ = lA$, which gives $A’ = \frac{A}{3}$. Substituting into the resistance formula, the new resistance R’ is $R’ = \rho \frac{l’}{A’} = \rho \frac{3l}{A/3} = 9 \cdot \rho \frac{l}{A} = 9R$.

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Key Concept: Comparative Analysis

c) Both have the same resistance
[Solution Description]
Resistance depends on the formula:
$R = \rho \frac{l}{A}$
where $\rho$ is the resistivity.

For Wire P:
$R_P = \rho \frac{5}{2}$

For Wire Q:
$R_Q = \rho \frac{10}{4} = \rho \frac{5}{2}$

Since both resistances are equal, both wires have the same resistance.

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Key Concept: Ohm’s Law Equation

c) $V = IR$
[Solution Description] According to Ohm’s law, the voltage (V) across a conductor is equal to the current (I) flowing through it multiplied by its resistance (R). Mathematically, this relationship is represented as $V = IR$. This equation shows how potential difference, current, and resistance are related in an electrical circuit.

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Key Concept: Conceptual Synthesis, Comparative Analysis

c) It stays the same
[Solution Description]
In a parallel circuit, each component has the same voltage across it. When one bulb in a parallel connection burns out, it leaves the circuit open at its branch, but does not affect the other branches. The remaining bulb still receives full voltage from the battery because the potential difference remains unchanged in a parallel configuration. This means the brightness of the remaining bulb will remain unchanged.

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Key Concept: Material Properties

c) High melting point
[Solution Description] Tungsten is used because it has a high melting point, which allows it to withstand the high temperatures needed to produce light without melting.

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Key Concept: Real-World Application

c) 50 W
[Solution Description]
The power consumed by an electrical component is given by the formula:
$P = I^2R$
Substituting the given values:
$$
P = (1 \, \text{A})^2 \times 50 \, \Omega
= 1 \, \text{A}^2 \times 50 \, \Omega
= 50 \, \text{W}
$$
So, the power consumed by the filament is 50 watts.

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Key Concept: Advanced Calculation, Problem Solving

c) $450 \, \text{kJ}$
[Solution Description]
Using Joule’s law of heating, the heat produced H can be calculated using the formula $H = I^2Rt$. Here, $I = 10 \, \text{A}$, $R = 15 \, \Omega$, and $t = 5 \times 60 \, \text{seconds} = 300 \, \text{s}$. Plugging in these values:

$H = (10)^2 \cdot 15 \cdot 300 = 100 \cdot 15 \cdot 300 = 450000 \, \text{J} = 450 \, \text{kJ}$

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Key Concept: Complex Calculation, Real-world Application

d) 7.06 A
[Solution Description]
To find the current, we first need to calculate the resistance using the formula:
$R = \rho \frac{L}{A}$
where $\rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m}$, $L = 100 \, \text{m}$, and $A = 1 \times 10^{-6} \, \text{m}^2$. Substituting these values in:
$R = 1.7 \times 10^{-8} \times \frac{100}{1 \times 10^{-6}} = 1.7 \, \Omega$
Now, use Ohm’s Law to find the current:
$I = \frac{V}{R} = \frac{12}{1.7} \approx 7.06 \, \text{A}$

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Key Concept: Basic Definition

c) Kilowatt-hour
[Solution Description] The commercial unit of electrical energy, often billed by electric companies, is known as the kilowatt-hour (kWh). It represents the amount of energy consumed when a device with a power rating of one kilowatt operates for one hour.

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Key Concepts: Complex Calculation, Advanced Application

c) 12.25 A

[Solution Description]

To find the current, we start with the power formula $P = I^2R$. Here, $P = 1500$ W and $R = 10$ ohms.

From the relation:

$P = I^2R$

We can solve for I as follows:

$I^2 = \frac{P}{R} = \frac{1500}{10} = 150 \quad \implies \quad I = \sqrt{150} \approx 12.25 \, \text{A}$

Therefore, the current through the heating element is approximately 12.25 A.

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Key Concept: Theoretical Application, Multi-Step Solutions

c) 8.15%
[Solution Description]
Let’s denote initial resistance as $R_0$. The new length $l_f$ becomes $1.05l_0$, and new resistivity $\rho_f$ becomes $1.03\rho_0$. Resistance is given by $R = \rho \frac{l}{A}$, so considering only changes in length and resistivity without any change in cross-section area:

New resistance $R_f$ can be expressed as

$R_f = \left(1.03\rho_0 \right) \frac{1.05l_0}{A} = 1.0815\left(\rho_0 \frac{l_0}{A}\right).$

The percentage change in resistance is

$$
\frac{R_f – R_0}{R_0} \times 100\%
= \left(1.0815 – 1\right) \times 100\%
= 8.15%.
$$

Hence, the resistance increases by 8.15%.

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Key Concept: Concept Integration, Problem Solving

c) 2 hours
[Solution Description]
Energy consumed in kilowatt-hours is related to power and time by the equation:
$\text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (h)}$
For a power of 1500 W, which is 1.5 kW, we have:
$3 \, \text{kWh} = 1.5 \, \text{kW} \times \text{Time (h)}$
Solving for time:
$\text{Time} = \frac{3 \, \text{kWh}}{1.5 \, \text{kW}} = 2 \, \text{hours}$

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Key Concept: Power and Energy Relationship

b) 3.3 kWh
[Solution Description]
The power P of the motor can be calculated using the formula P = VI, where $ V = 220 \, \text{V} $ and $ I = 3 \, \text{A} $. So,

$P = 220 \times 3 = 660 \, \text{W}$

To find the energy consumed over 5 hours, use the formula:

$W = P \times t$

Where $t = 5 \, \text{hours} = 5 \times 3600 \, \text{seconds}$. Therefore,

$W = 660 \times 5 = 3300 \, \text{Wh} = 3.3 \, \text{kWh}$

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Key Concepts: Power Formula Variations

c) 20 W

[Solution Description]

We use the formula $P = V^2 / R$ to find the power dissipated. Here, $V = 10$ V and $R = 5$ $\Omega$.

$P = \frac{10^2}{5} = \frac{100}{5} = 20 \text{ W}$

Therefore, the power dissipated by the device is 20 W.

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Key Concept: Advanced Calculation, Filament Life, Electric Heating Effect

c) 8707 J
[Solution Description]
First, calculate the resistance ( R ) of the filament using the resistivity formula:

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

Where:
$-  \rho = 5.20 \times 10^{-8} \, \Omega\, \text{m} $
$-  L = 0.4 \, \text{m} $
$-  A = \pi \left(\frac{d}{2}\right)^2 $, where $ d = 0.0002 \, \text{m} $

$A = \pi \left(\frac{0.0002}{2}\right)^2 = \pi \times 0.0001^2 = \pi \times 10^{-8} \, \text{m}^2$

$R = 5.20 \times 10^{-8} \times \frac{0.4}{\pi \times 10^{-8}} = \frac{5.20 \times 0.4}{\pi} \approx 6.62 \, \Omega$

Using the power formula for heat generation:

$P = \frac{V^2}{R} = \frac{240^2}{6.62} \approx 8707 \, \text{W} = 8707 \, \text{J/s}$

Thus, the heat produced per second is approximately 8707 J.

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Key Concept: Application Scenario

b) So that devices can be individually controlled without affecting others.
[Solution Description] When devices are connected in parallel, each device operates independently, meaning they each get the full voltage provided by the source. This configuration ensures that if one device fails, it does not affect the operation of others. Additionally, in parallel circuits, total resistance decreases, allowing the circuit to deliver adequate current to all devices based on their requirements.

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Key Concept: Basic Formula

b) $V = IR$
[Solution Description] Ohm’s law states that the potential difference across a conductor is directly proportional to the current flowing through it, provided temperature remains constant. The formula for Ohm’s Law is given by $V = I R$, where V  is the voltage, I is the current, and R is the resistance.

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Key Concept: Graph Interpretation

d) The resistance is constant
[Solution Description] A straight line through the origin on a V-I graph indicates that the potential difference across the conductor (V) is directly proportional to the current through it (I). This is consistent with Ohm’s law, $V = IR$, meaning the resistance (R) is constant. Therefore, the conductor behaves as an ohmic resistor.

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Key Concept: Complex Problem Solving, Multi-Step Calculation

a) $2.5 \Omega$

[Solution Description]

To find the resistance of the second wire, we use the formula $R = \rho \frac{l}{A}$. Since both wires are made of the same material, their resistivity $\rho$ remains constant.

For the first wire:

$R_1 = \rho \frac{l_1}{A_1} = 10 \, \Omega$

For the second wire:

Let $l_2 = \frac{1}{2} l_1$ and $A_2 = 2 A_1$, then

$R_2 = \rho \frac{l_2}{A_2} = \rho \frac{\frac{1}{2}l_1}{2A_1} = \rho \frac{l_1}{4A_1} = \frac{1}{4}R_1$

Therefore,

$R_2 = \frac{1}{4} \times 10 \, \Omega = 2.5 \, \Omega$

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Key Concept: Temperature Effect

c) It increases
[Solution Description] As the temperature of a metallic conductor increases, its atoms vibrate more vigorously, which causes more collisions with electrons. These collisions hinder the flow of current and increase the resistance.

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Key Concept: Circuit Connection

b) In series
[Solution Description]
An ammeter is always connected in series with the component or section of the circuit whose current you want to measure. This ensures that the full current flows through the ammeter for accurate measurement.

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Key Concept: Unit Identification

d) Ampere
[Solution Description] The SI unit of electric current is the ampere (A), named after the French scientist Andre-Marie Ampere. One ampere is equivalent to one coulomb of charge passing through a point in one second.

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Key Concept: Symbol Application

c) Current will stop flowing
[Solution Description]
An open plug key in a circuit diagram signifies that the circuit is incomplete because there is a break in the path of the electric current. As a result, no current will flow through the circuit until the plug key is closed to complete the circuit, allowing electricity to move freely through the components.

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Key Concept: Concept Integration, Multi-step Solutions

c) 4.8 V
[Solution Description]
First, calculate the total resistance in the series circuit:

$R_{\text{total}} = 4 + 6 = 10 \, \Omega$

Using Ohm’s Law, $I = \frac{V}{R_{\text{total}}}$:

$I = \frac{12}{10} = 1.2 \, \text{A}$

The potential difference across the $4 \Omega$ resistor is given by $V = I \cdot R$.

$V_{4 \, \Omega} = 1.2 \times 4 = 4.8 \, \text{V}$

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Key Concept: Conceptual Application

b) It triples
[Solution Description] According to Ohm’s Law, V = IR, where V is the potential difference, I is the current, and R is the resistance. If the potential difference V is tripled while the current I remains constant, then the new resistance $R’$ can be found using:
$V’ = 3V = IR’$
Since I is constant, solving for $R’$ gives:
$R’ = \frac{3V}{I} = 3R$
Therefore, the resistance triples.

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Key Concept: Heat Calculation

c) 3600 J
[Solution Description]
Using Joule’s law of heating, the heat produced is given by $H = I^2 R t$.

Substitute the known values into the equation:
$H = (3\, \text{A})^2 \times 10\, \Omega \times 40\, \text{s}$
$H = 9 \times 10 \times 40 = 3600\, \text{J}$
Therefore, the heat generated is 3600 J.

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Key Concept: Conceptual Integration, Advanced Reasoning

b) 2 A

[Solution Description]

In a parallel circuit, the voltage across each branch is equal to the voltage of the source.

1. Total voltage provided by the battery: $V = 18 V$

2. Resistance of the specified branch: $R = 9 \, \Omega$

3. Use Ohm’s Law to find the current (I) through this branch:

$I = \frac{V}{R}= \frac{18\, \text{V}}{9\, \Omega}= 2 \, \text{A}$

Hence, the current flowing through the branch with $9 \Omega$ resistance is 2 A.

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Key Concept: Comparative Analysis, Fuse Material Selection

c) Low melting point
[Solution Description] The primary consideration in selecting a fuse wire’s material is its melting point. A fuse wire should have a low melting point so that it melts quickly under excessive current flow, thereby breaking the circuit and preventing damage.

This ensures the circuit is interrupted swiftly before any significant overheating occurs. Therefore, the correct property for a fuse wire is a low melting point.

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Key Concept: Real-world Application

b) Because a thick wire allows more electrons to pass through due to a larger area
[Solution Description] According to Ohm’s Law $(I = V/R)$, the current I through a conductor is inversely proportional to its resistance R, for a constant potential difference V. Resistance R is inversely proportional to the area of cross-section $(R \propto 1/A)$. A thicker wire has a larger area, hence lower resistance, allowing more current to flow through.

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Key Concept: Circuit Diagram Symbols

b) A circle with the letter “A” inside it

[Solution Description]

An ammeter is represented by a circle with the letter “A” inside it in a circuit diagram. This symbol specifies a device used to measure the current flowing through a circuit.

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Key Concept: Series and Parallel Analysis, Complex Circuit Design

c) 0.625 A
[Solution Description]
Begin by calculating the equivalent resistance for the 4 Ω and 6 Ω resistors in parallel, denoted as $R_{46}$:

$\frac{1}{R_{46}} = \frac{1}{4} + \frac{1}{6}$

Find the least common denominator:

$\frac{1}{R_{46}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$

Thus:

$R_{46} = \frac{12}{5} = 2.4 \, \Omega$

Next, add the series 12 Ω resistor to get the total resistance $R_T$:

$R_T = R_{46} + 12 = 2.4 + 12 = 14.4 \, \Omega$

Using Ohm’s law, calculate the total current I:

$I = \frac{V}{R_T} = \frac{9}{14.4}$

Simplifying the calculation:

$I \approx 0.625 \, \text{A}$

So, the current flowing through the circuit is approximately 0.625 A.

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Key Concept: Series Circuit

b) The circuit stops working
[Solution Description] In a series circuit, all components are connected end-to-end. If one resistor fails, it breaks the entire circuit path, causing the current to stop flowing.

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Key Concept: Application of Formula

c) 27 J
[Solution Description] To find the work done, we use the formula for potential difference, which relates the work done (W), potential difference (V), and charge (Q):
$V = \frac{W}{Q} \implies W = V \times Q$
Given $V = 9 \, \text{V}$ and $Q = 3 \, \text{C}$, substitute these values into the equation:
$W = 9 \, \text{V} \times 3 \, \text{C} = 27 \, \text{J}$
Therefore, the work done is 27 Joules.

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Key Concept: Series Circuit Analysis

c) 4 V
[Solution Description]
The total resistance $(R_{\text{total}})$ in the series is the sum of all resistances:
$R_{\text{total}} = 1\, \Omega + 2\, \Omega + 3\, \Omega = 6\, \Omega$.
According to Ohm’s Law, the current (I) through the circuit is:
$I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{12\, \text{V}}{6\, \Omega} = 2\, \text{A}$.
The potential difference across the 2 Ω resistor is then:
$V = I \times R = 2\, \text{A} \times 2\, \Omega = 4\, \text{V}$.
Therefore, the potential difference across the 2 Ω resistor is 4 V.

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Key Concept: Material Impact

b) Decreases
[Solution Description]
To determine how replacing a nichrome wire with a copper wire affects the resistance, we must compare their resistivities. The resistivity of nichrome is $100 \times 10^{-6} \, \Omega \, \text{m}$, while that of copper is $1.62 \times 10^{-8} \, \Omega \, \text{m}$. Resistance R is given by the formula:

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

As both wires have the same length l and cross-sectional area A, the difference in resistance depends on the resistivities $(\rho)$. Since copper has a much smaller resistivity than nichrome, the resistance will decrease when using copper.

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Key Concept: Energy Cost Calculation

d) Rs 360
[Solution Description]
First calculate the daily energy consumption of the heater:

$\text{Daily Energy Consumption} = 1000\, \text{W} \times 3\, \text{hours} = 3000\, \text{Wh} = 3\, \text{kWh}$

Then, calculate the monthly energy consumption:

$\text{Monthly Energy Consumption} = 3\, \text{kWh/day} \times 30\, \text{days} = 90\, \text{kWh}$

Finally, compute the cost:

$\text{Cost} = 90\, \text{kWh} \times \text{Rs}\, 4/\text{kWh} = \text{Rs}\, 360$

Hence, the monthly cost of energy consumption is $\text{₹360}$

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Key Concept: Current Calculation Using Ohm’s Law

b) 2 A
[Solution Description] Using Ohm’s Law, the current I can be calculated using the formula:
$I = \frac{V}{R}$
where $V = 16 \, \text{V}$ and $R = 8 \, \Omega$. Substituting the values:
$I = \frac{16 \, \text{V}}{8 \, \Omega} = 2 \, \text{A}$
Therefore, the current flowing through the resistor is 2 A.

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Key Concept: Simple Calculation

c) 30 C
[Solution Description] To find the charge, we use the formula $I = \frac{Q}{t}$. Rearranging gives $Q = I \times t$. Substituting $I = 1.5 \text{ A}$ and $t = 20 \text{ s}$:
$Q = 1.5 \times 20 = 30 \text{ C}$.
Therefore, the charge that passes through the circuit is 30 coulombs.

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