Electricity (Class 10): Notes और MCQ Practice

आसान भाषा में

Electricity असल में charges (इलेक्ट्रॉन्स) का flow है। जब किसी बंद circuit में Potential Difference (Voltage) अप्लाई किया जाता है, तो Current बहता है। इस current के रास्ते में आने वाली रुकावट को Resistance कहते हैं, और current flow होने से Heat और Power उत्पन्न होती है।

Important Formulas & Definitions

• Electric Current (I): Charges के बहने की दर।

Formula: I = Q ÷ t (Unit: Ampere, A)

• Potential Difference (V): Work done per unit charge.

Formula: V = W ÷ Q (Unit: Volt, V)

• Ohm's Law: V = I × R (Constant temperature पर, V ∝ I)
• Resistance (R): R = ρ × (l ÷ A)

(ρ = Resistivity, Unit: Ω m। यह केवल material और temperature पर निर्भर करती है, length या area पर नहीं।)

• Series Combination: Rs = R1 + R2 + R3 (Current समान, Voltage divide)
• Parallel Combination: 1 ÷ Rp = (1 ÷ R1) + (1 ÷ R2) + (1 ÷ R3) (Voltage समान, Current divide)
• Joule's Law of Heating: H = I² × R × t (Unit: Joule, J)
• Electric Power (P): P = V × I = I² × R = V² ÷ R (Unit: Watt, W)
• Commercial Unit: 1 kWh = 3.6 × 10⁶ Joules

2 Memory Tricks

1. Ohm's Law Triangle (VIR): याद रखें "VIP = I × R"। V हमेशा top पर रहता है, यानी V = I × R, I = V ÷ R, और R = V ÷ I।
2. Device Connection: Voltmeter = Parallel (Trick: Vice President) और Ammeter = Series (Trick: All Students)।

2 Common Board Mistakes

1. Parallel Resistance का Invert न करना: छात्र 1 ÷ Rp निकाल लेते हैं, लेकिन final answer लिखते समय उसे पलटकर Rp लिखना भूल जाते हैं।
2. Time को Seconds में न बदलना: Numerical solve करते समय time अगर minutes या hours में दिया हो, तो उसे seconds में बदले बिना solve करने पर गलत answer आता है (जब तक kWh न पूछा गया हो)।

Class 10 Science: Electricity

1. Chapter in Short

This chapter explains how electricity flows and how we use it to run appliances. It covers fundamental concepts such as electric charge, electric current, potential difference, Ohm's law, electrical resistance, series and parallel circuits, the heating effect of electric current, and electric power. By understanding these concepts, you will learn how electrical circuits work in our homes and how electricity bills are calculated.

2. Key Concepts Explained Simply

Electric Current and Circuit

• Electric Charge ($Q$): Matter is made of atoms containing protons and electrons. Electrons carry a negative charge. Charge is measured in Coulombs (C).
• Electric Current ($I$): The rate of flow of electric charges through a conductor. Think of it like water flowing through a pipe: the flow of water is the water current; the flow of charges is the electric current.
• Direction of Current: By convention, current flows from the positive terminal to the negative terminal, which is opposite to the actual direction of electron flow.

Electric Potential and Potential Difference ($V$)

• Charges do not move on their own. Water flows in a pipe only if there is a difference in pressure (like water flowing from a higher tank to a lower tap).
• Similarly, electrons move in a wire only when there is an electric pressure difference, known as the potential difference, provided by a battery or cell.
• It is defined as the work done to move a unit positive charge from one point to another.

Ohm's Law

• Ohm's law states that at a constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends ($V \propto I$).
• Resistance ($R$): The property of a conductor to oppose the flow of charges. Think of it like traffic or friction on a road that slows cars down.

Factors Affecting Resistance of a Conductor

The resistance of a uniform wire depends on:

1. Length ($l$): Resistance is directly proportional to length ($R \propto l$). A longer wire offers more resistance (like a longer road has more traffic obstacles).
2. Area of cross-section ($A$): Resistance is inversely proportional to thickness ($R \propto 1 ÷ A$). A thicker wire offers less resistance (like a wider highway allows traffic to move easily).
3. Nature of the material: Measured by resistivity ($\rho$). Metals like copper and silver have very low resistivity, while rubber and glass have very high resistivity.
4. Temperature: For most metals, resistance increases with an increase in temperature.

Resistors in Series and Parallel

• Series Combination:
• Resistors are connected end-to-end in a single line.
• The current ($I$) remains the same through all resistors.
• The total potential difference is split: $V = V_1 + V_2 + V_3$.
• Disadvantage: If one component fails, the entire circuit breaks.
• Parallel Combination:
• Resistors are connected across two common points.
• The potential difference ($V$) remains the same across each resistor.
• The total current divides among branches: $I = I_1 + I_2 + I_3$.
• Advantage: If one appliance stops working, others keep working. Appliances also get full voltage. This is why our household circuits are wired in parallel.

Heating Effect of Electric Current (Joule's Law)

• When current passes through a purely resistive wire (like nichrome), electrical energy gets converted into heat.
• Applications: Electric iron, electric heater, toaster, and electric fuse.
• Electric Fuse: A safety device with a low melting point that breaks the circuit if excessive current flows, protecting appliances from damage.

Electric Power

• The rate at which electrical energy is consumed in an electric circuit.
• The commercial unit of electrical energy is the kilowatt-hour (kWh), commonly called a "unit" on electricity bills.

3. Important Definitions, Formulas, and Facts

Definitions and Standard Units

• Electric Current ($I$): $I = Q ÷ t$
• SI Unit: Ampere (A)
• $1\text{ A} = 1\text{ Coulomb} ÷ 1\text{ second}$
• Potential Difference ($V$): $V = W ÷ Q$
• SI Unit: Volt (V)
• $1\text{ V} = 1\text{ Joule} ÷ 1\text{ Coulomb}$
• Resistance ($R$): $R = V ÷ I$
• SI Unit: Ohm ($\Omega$)
• Resistivity ($\rho$): The resistance of a conductor of unit length and unit cross-sectional area.
• SI Unit: Ohm-metre ($\Omega\cdot\text{m}$)
• Electric Power ($P$): The rate of electrical energy consumption.
• SI Unit: Watt (W)

Core Formulas

• Charge: $Q = n × e$ (where $n$ is number of electrons, $e = 1.6 × 10^{-19}\text{ C}$)
• Ohm's Law: $V = I × R$
• Resistance of a wire: $R = \rho × (l ÷ A)$
• Resistors in Series: $R_s = R_1 + R_2 + R_3$
• Resistors in Parallel: $1 ÷ R_p = (1 ÷ R_1) + (1 ÷ R_2) + (1 ÷ R_3)$
• Joule's Law of Heating: $H = I² × R × t = V × I × t = (V² ÷ R) × t$
• Electric Power: $P = V × I = I² × R = V² ÷ R$
• Commercial Unit of Energy: $1\text{ kWh} = 1000\text{ W} × 3600\text{ s} = 3.6 × 10^6\text{ Joules}$

Important Facts

• Ammeter: Always connected in series in a circuit; has very low resistance.
• Voltmeter: Always connected in parallel across the component; has very high resistance.
• Alloys like nichrome have higher resistivity than pure metals and do not oxidize (burn) easily at high temperatures, making them ideal for heating elements.
• Tungsten is used almost exclusively for filament bulbs because of its very high melting point.

4. Solved Examples and Board-Style Questions

Question 1 (Current & Charge)

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

• Given:

$I = 0.5\text{ A}$

$t = 10\text{ minutes} = 10 × 60\text{ s} = 600\text{ s}$

• Formula:

$Q = I × t$

• Calculation:

$Q = 0.5 × 600 = 300\text{ C}$

• Answer: The amount of electric charge is 300 Coulombs.

Question 2 (Resistance & Dimensions)

A wire of given material having length $l$ and area of cross-section $A$ has a resistance of $4\ \Omega$. What would be the resistance of another wire of the same material having length $l ÷ 2$ and area of cross-section $2A$?

• Given:

Original resistance $R_1 = \rho × (l ÷ A) = 4\ \Omega$

• New Wire:

Length $l' = l ÷ 2$

Area $A' = 2A$

• Calculation:

$R_2 = \rho × (l' ÷ A') = \rho × [(l ÷ 2) ÷ (2A)] = \rho × [l ÷ (4A)]$

$R_2 = (1 ÷ 4) × [\rho × (l ÷ A)] = (1 ÷ 4) × R_1$

$R_2 = 4 ÷ 4 = 1\ \Omega$

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

Question 3 (Series and Parallel Combination)

An electric lamp of resistance $20\ \Omega$ and a conductor of $4\ \Omega$ resistance are connected in series to a $6\text{ V}$ battery. Calculate:

(a) The total resistance of the circuit

(b) The current flowing through the circuit

(c) The potential difference across the electric lamp and the conductor

• Given:

$R_1 = 20\ \Omega$, $R_2 = 4\ \Omega$, $V = 6\text{ V}$

• Calculations:
• (a) Total resistance in series:

$R_s = R_1 + R_2 = 20 + 4 = 24\ \Omega$

• (b) Current:

$I = V ÷ R_s = 6 ÷ 24 = 0.25\text{ A}$

• (c) Potential difference across the lamp ($V_1$):

$V_1 = I × R_1 = 0.25 × 20 = 5\text{ V}$

Potential difference across the conductor ($V_2$):

$V_2 = I × R_2 = 0.25 × 4 = 1\text{ V}$

• Answer:

Total resistance = $24\ \Omega$, Current = $0.25\text{ A}$, Voltage across lamp = $5\text{ V}$, Voltage across conductor = $1\text{ V}$.

Question 4 (Electric Energy & Cost)

An electric refrigerator rated $400\text{ W}$ operates 8 hours per day. What is the cost of energy to operate it for 30 days at Rs 3.00 per kWh?

• Given:

Power $P = 400\text{ W} = 400 ÷ 1000\text{ kW} = 0.4\text{ kW}$

Time per day = $8\text{ hours}$

Total days = $30$

Total time $t = 8 × 30 = 240\text{ hours}$

• Total Energy Consumed ($E$):

$E = P × t = 0.4\text{ kW} × 240\text{ h} = 96\text{ kWh}$

• Total Cost:

$\text{Cost} = 96\text{ kWh} × \text{Rs } 3.00\text{ per kWh} = \text{Rs } 288$

• Answer: The total cost of energy is Rs 288.

5. Common Mistakes and Quick Revision Points

Common Mistakes

1. Forgetting to convert time to seconds: In $I = Q ÷ t$ and $H = I² × R × t$, time must be in seconds. In energy bill problems ($E = P × t$), time is usually in hours and power in kilowatts.
2. Inverting the parallel resistance: When using $1 ÷ R_p = (1 ÷ R_1) + (1 ÷ R_2)$, students calculate the sum of fractions and forget to invert the answer to find $R_p$.
3. Connecting meters incorrectly:

• Ammeter goes in series (never parallel, or it may burn out due to low resistance).
• Voltmeter goes in parallel (never series, or negligible current will flow due to high resistance).

1. Confusing current and voltage rules:

• In series: Current is constant; Voltage divides.
• In parallel: Voltage is constant; Current divides.

Quick Revision Points

• Doubling length doubles resistance; doubling area halves resistance.
• Combining resistors in series increases the overall resistance (greater than the highest resistor).
• Combining resistors in parallel decreases the overall resistance (smaller than the smallest resistor).
• Fuse wire is always connected in the live wire before the appliance.
• $1\text{ unit of electricity} = 1\text{ kWh} = 3.6 × 10^6\text{ Joules}$.