CBSE Class 10 Science: Chapter 12 Electricity SA Questions (with Answers)
Welcome to Part 3 of our 8-part series on Chapter 12, Electricity. This post focuses on Short Answer (SA) Questions, typically worth 2 or 3 marks. These questions require conceptual understanding, definitions with explanations, and simple numerical calculations.
Part 3: Short Answer (SA) Questions
State Ohm's Law. Draw a V-I graph for an ohmic conductor. What does the slope of this graph represent?
Ohm's Law: It states that the potential difference (V) across the ends of a given metallic conductor in an electric circuit is directly proportional to the current (I) flowing through it, provided its temperature remains the same. (V ∝ I, or V = IR)
V-I Graph: For an ohmic conductor, the V-I graph is a straight line passing through the origin.
(A simple graph should be drawn by the student showing V on the y-axis and I on the x-axis, with a straight line from (0,0)).
Slope: The slope of the V-I graph (with V on y-axis and I on x-axis) represents the resistance (R) of the conductor. (Slope = ΔV / ΔI = R).
On what factors does the resistance of a conductor depend? Write the mathematical expression for it.
- Length of the conductor (L): Resistance is directly proportional to length (R ∝ L).
- Area of cross-section (A): Resistance is inversely proportional to the area of cross-section (R ∝ 1/A).
- Nature of the material: Different materials have different resistivities.
- Temperature: Resistance of pure metals generally increases with an increase in temperature.
Where R is resistance, ρ (rho) is the electrical resistivity, L is length, and A is the area of cross-section.
State Joule's Law of Heating. List two practical applications of this effect in daily life.
(i) directly proportional to the square of the current (I²) for a given resistance,
(ii) directly proportional to the resistance (R) for a given current, and
(iii) directly proportional to the time (t) for which the current flows.
Formula: H = I²Rt
Practical Applications:
- Electric Heating Devices: Devices like an electric iron, toaster, oven, and room heater work on this principle.
- Electric Bulb/Lamp: The filament (made of tungsten) gets heated to a very high temperature and starts glowing to produce light.
Why are domestic electric circuits connected in parallel and not in series? Give two reasons.
Domestic appliances are connected in parallel for the following main reasons:
1. Constant Voltage: In a parallel connection, each appliance receives the full mains voltage (e.g., 220 V). In a series circuit, the voltage gets divided among the appliances, so they would not operate properly.
2. Independent Operation: Each appliance can be turned on or off independently with its own switch, without affecting the others. In a series circuit, if one appliance fails (or is switched off), the circuit breaks and all other appliances stop working.
Define 1 Watt of electric power. An electric bulb is rated 40 W, 220 V. Find the current drawn by it when it is connected to a 220 V supply.
1 Watt: Electric power is said to be 1 Watt if 1 Ampere of current flows through a circuit having 1 Volt of potential difference. (P = VI, so 1 W = 1 V × 1 A).
Numerical Problem:
Given:
Power (P) = 40 W
Voltage (V) = 220 V
We know the formula: P = V × I
We need to find Current (I).
I = P / V
I = 40 W / 220 V
I = 4 / 22 A = 2 / 11 A
I ≈ 0.18 A
The current drawn by the bulb is 0.18 A.
Why are heating elements (like in toasters and electric irons) made of an alloy rather than a pure metal? Give two reasons.
Alloys (like Nichrome) are used for making heating elements for two main reasons:
1. High Resistivity: Alloys have a much higher resistivity than their constituent pure metals. This high resistance allows them to produce a large amount of heat (H ∝ R) when current passes through them.
2. High Melting Point & Low Oxidation: Alloys have a high melting point and do not oxidize (or burn) easily, even at very high (red-hot) temperatures. Pure metals would melt or corrode quickly under such conditions.
Distinguish between Resistance and Resistivity (at least two points).
1. Definition:
- Resistance: It is the property of a conductor to oppose the flow of electric current through it.
- Resistivity: It is the resistance of a conductor of unit length (1 m) and unit cross-sectional area (1 m²).
- Resistance: It depends on the length, area of cross-section, temperature, and nature of the material.
- Resistivity: It is a characteristic property of the material and depends only on the nature of the material and its temperature.
- Resistance: The SI unit is Ohm (Ω).
- Resistivity: The SI unit is Ohm-meter (Ω m).
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An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of the energy to operate it for 30 days at ₹ 5.00 per kWh?
Step 1: Find total energy consumed in kWh.
Power (P) = 400 W = 0.4 kW (since 1000 W = 1 kW)
Time per day (t) = 8 hours
Number of days = 30
Total time = 8 hours/day × 30 days = 240 hours
Total Energy (E) = Power (P) × Total Time (t)
E = 0.4 kW × 240 h
E = 96 kWh
Step 2: Calculate the total cost.
Cost per kWh = ₹ 5.00
Total Cost = Total Energy (kWh) × Cost per kWh
Total Cost = 96 × 5.00
Total Cost = ₹ 480
The total cost to operate the refrigerator for 30 days is ₹ 480.
Derive the expression for the equivalent resistance (R_s) when three resistors (R1, R2, R3) are connected in series.
When resistors are connected in series:
1. The current (I) flowing through each resistor is the same.
2. The total potential difference (V) across the combination is the sum of the potential differences across each resistor (V = V1 + V2 + V3).
Using Ohm's Law (V = IR):
V1 = IR1
V2 = IR2
V3 = IR3
Let R_s be the equivalent resistance of the circuit. The total potential difference is:
V = I * R_s
Substituting these values into V = V1 + V2 + V3:
I * R_s = IR1 + IR2 + IR3
I * R_s = I (R1 + R2 + R3)
Cancelling 'I' from both sides, we get:
R_s = R1 + R2 + R3
Derive the expression for the equivalent resistance (R_p) when three resistors (R1, R2, R3) are connected in parallel.
When resistors are connected in parallel:
1. The potential difference (V) across each resistor is the same.
2. The total current (I) from the source is the sum of the currents flowing through each resistor (I = I1 + I2 + I3).
Using Ohm's Law (I = V/R):
I1 = V / R1
I2 = V / R2
I3 = V / R3
Let R_p be the equivalent resistance of the circuit. The total current is:
I = V / R_p
Substituting these values into I = I1 + I2 + I3:
V / R_p = (V / R1) + (V / R2) + (V / R3)
V / R_p = V (1/R1 + 1/R2 + 1/R3)
Cancelling 'V' from both sides, we get:
1/R_p = 1/R1 + 1/R2 + 1/R3
Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
(Students should draw a circuit diagram with the following components connected in a single loop):
1. A battery (symbol) consisting of 3 cells, labelled as 6V (or 3 x 2V).
2. Three resistors (symbols) labelled 5 Ω, 8 Ω, and 12 Ω.
3. A plug key (symbol) in the closed position.
4. Connecting wires (straight lines) forming a closed loop, with current direction (arrow) shown from positive to negative terminal.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 seconds.
We will use Joule's Law of Heating: H = I²Rt
Given:
Resistance (R) = 20 Ω
Current (I) = 5 A
Time (t) = 30 s
Heat (H) = (5 A)² × 20 Ω × 30 s
H = 25 × 20 × 30
H = 500 × 30
H = 15000 J (or 15 kJ)
The heat developed is 15,000 Joules.
A wire of resistivity 'ρ' is stretched to double its length. What will be its new resistivity?
The new resistivity will still be 'ρ' (it remains unchanged).
Reason: Resistivity (ρ) is a characteristic property of the *material* itself. It does not depend on the physical dimensions (like length or area) of the wire. Stretching the wire changes its length and area, which will change its *resistance*, but its *resistivity* stays the same.
Differentiate between an Ammeter and a Voltmeter (3 points).
1. Purpose:
- Ammeter: Measures electric current.
- Voltmeter: Measures potential difference (voltage).
- Ammeter: Always connected in series in the circuit.
- Voltmeter: Always connected in parallel across the component.
- Ammeter: Has very low (ideally zero) resistance.
- Voltmeter: Has very high (ideally infinite) resistance.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Step 1: Find current for Lamp 1 (I1).
P1 = 100 W, V = 220 V
P1 = V × I1 => I1 = P1 / V = 100 / 220 A
Step 2: Find current for Lamp 2 (I2).
P2 = 60 W, V = 220 V
P2 = V × I2 => I2 = P2 / V = 60 / 220 A
Step 3: Find total current (I).
In a parallel circuit, the total current is the sum of the individual currents:
I = I1 + I2
I = (100 / 220) + (60 / 220)
I = (100 + 60) / 220 = 160 / 220 A
I = 16 / 22 A = 8 / 11 A
I ≈ 0.73 A
The total current drawn from the line is 0.73 A.
What is the advantage of connecting electrical appliances in parallel with the mains?
The main advantage is that the potential difference (voltage) across each appliance is the same as the mains supply (e.g., 220 V). This allows each appliance to operate at its intended voltage and power.
(Other valid points: If one appliance fails, others keep working. The total resistance of the circuit is decreased.)
A wire of resistance 'R' is cut into three equal pieces. These pieces are then connected in parallel. What is the equivalent resistance of the combination?
Let the original resistance be R.
Step 1: When cut into 3 equal pieces, the resistance of each new piece is R/3 (since R ∝ L).
Step 2: These 3 pieces are connected in parallel. Let the equivalent resistance be R_p.
1/R_p = 1/(R/3) + 1/(R/3) + 1/(R/3)
1/R_p = 3/R + 3/R + 3/R
1/R_p = (3 + 3 + 3) / R
1/R_p = 9 / R
R_p = R / 9
The equivalent resistance is one-ninth (1/9) of the original resistance.
Why does the cord of an electric heater not glow while its heating element does?
The heating element and the cord are connected in series, so they carry the same current (I).
According to Joule's Law (H = I²Rt), the heat produced is directly proportional to resistance (H ∝ R).
Heating Element: It is made of an alloy (like Nichrome) which has a very high resistance. This produces a large amount of heat, causing it to glow.
Cord: It is made of a good conductor (like copper) which has very low resistance. Therefore, very little heat is produced in the cord, and it does not glow.
An electric iron draws a current of 3.4 A from the 220 V supply line. What current will this electric iron draw when connected to a 110 V supply line?
Step 1: Find the resistance (R) of the iron (which is constant).
Given: V1 = 220 V, I1 = 3.4 A
From Ohm's Law, R = V1 / I1
R = 220 V / 3.4 A
R = 2200 / 34 Ω ≈ 64.7 Ω
Step 2: Find the new current (I2) at 110 V.
Given: V2 = 110 V
I2 = V2 / R
I2 = 110 V / (220 / 3.4 A)
I2 = 110 * (3.4 / 220) A
I2 = (1 / 2) * 3.4 A
I2 = 1.7 A
The iron will draw 1.7 A from a 110 V supply line.
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A. What current will the heater draw if the potential difference is increased to 120 V?
Method 1: Using Ratios
Assuming the resistance (R) of the heater is constant, by Ohm's Law, V ∝ I.
This means if the voltage (V) is doubled, the current (I) will also double.
The voltage is increased from 60 V to 120 V (it is doubled).
Therefore, the new current will be 4 A × 2 = 8 A.
Method 2: Calculating Resistance
R = V1 / I1 = 60 V / 4 A = 15 Ω.
New current I2 = V2 / R = 120 V / 15 Ω = 8 A.
A wire of resistance 10 Ω is stretched to double its original length. What will be its new resistance?
The new resistance will be 40 Ω.
When a wire is stretched, its volume (V) remains constant.
Let original length = L, original area = A. So, V = L × A.
New length L' = 2L.
New volume V' = L' × A' = (2L) × A'.
Since V = V', we have L × A = (2L) × A', which means the new area A' = A/2.
Original Resistance R = ρ(L/A) = 10 Ω.
New Resistance R' = ρ(L'/A')
R' = ρ(2L / (A/2))
R' = ρ(4L / A)
R' = 4 × [ρ(L/A)]
R' = 4 × R
R' = 4 × 10 Ω = 40 Ω.
Explain why copper and aluminium wires are usually employed for electricity transmission.
Copper and aluminium are used for electricity transmission for two main reasons:
1. Low Resistivity: They are excellent conductors of electricity and have very low resistivity. This minimizes the loss of electrical energy in the form of heat (H = I²Rt) during long-distance transmission.
2. Ductility: They are highly ductile metals, which means they can be easily drawn into thin wires required for cables.
(Aluminium is often preferred for overhead lines as it is much lighter and cheaper than copper).
What is an electric fuse? What material is it made of and why?
Electric Fuse: A fuse is a safety device used in electric circuits to protect the circuit and appliances from damage caused by overloading or short-circuiting.
Material: It is made of a wire of a pure metal (like tin or zinc) or an alloy (like tin-lead) which has a:
1. Low Melting Point: It must melt and break the circuit quickly when a current higher than the safe limit flows through it.
2. High Resistivity: (Compared to the copper wires in the circuit) This ensures it heats up significantly (H = I²Rt) and melts when a large current flows.
Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. Find the equivalent resistance of the combination.
For resistors in parallel, the formula for equivalent resistance (R_p) is:
1/R_p = 1/R1 + 1/R2 + 1/R3
1/R_p = 1/2 + 1/3 + 1/6
To add these fractions, find a common denominator (which is 6):
1/R_p = (3/6) + (2/6) + (1/6)
1/R_p = (3 + 2 + 1) / 6
1/R_p = 6 / 6
1/R_p = 1
Therefore, R_p = 1 Ω.
The equivalent resistance of the combination is 1 Ω.
Define electric current and potential difference. State their SI units.
- Definition: It is defined as the rate of flow of electric charge (Q) through a cross-section of a conductor. (Formula: I = Q/t).
- SI Unit: The SI unit of electric current is the Ampere (A).
- Definition: It is defined as the work done (W) to move a unit charge (Q) from one point to another in an electric field. (Formula: V = W/Q).
- SI Unit: The SI unit of potential difference is the Volt (V).
