CBSE Class 10 Science: Chapter 12 Electricity Case Study Questions
Welcome to Part 6 of our 8-part series on Chapter 12, Electricity. This post features Case Study based questions, also known as passage-based or source-based questions. These questions test your ability to read and analyze a situation and apply your conceptual knowledge.
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Part 6: Case Study Based Questions
Ohm's law states that the electric current (I) flowing through a metallic conductor is directly proportional to the potential difference (V) across its ends, provided the physical conditions (like temperature) remain constant. Mathematically, V ∝ I or V = RI, where R is the constant of proportionality called resistance. The resistance of a conductor depends on its length (L), area of cross-section (A), and the nature of its material (resistivity, ρ). The formula is R = ρ(L/A). Materials with low resistivity, like copper, are good conductors, while materials with high resistivity, like nichrome, are used as heating elements.
R = ρ(L/A) = 4Ω
R' = ρ(L'/A') = ρ(2L / (A/2)) = ρ(4L/A) = 4 * [ρ(L/A)] = 4 * 4Ω = 16Ω.
(i) High resistivity (which produces more heat).
(ii) High melting point (which prevents them from melting).
(iii) They do not oxidize (burn) easily at high temperatures.
Resistors can be combined in two ways: in series and in parallel. In a series circuit, the current through each component is the same, and the total voltage is the sum of the voltages across each component (V = V1 + V2 + ...). The equivalent resistance (Rs) is the sum of individual resistances (Rs = R1 + R2 + ...).
In a parallel circuit, the voltage across each component is the same, and the total current is the sum of the currents through each component (I = I1 + I2 + ...). The reciprocal of the equivalent resistance (Rp) is the sum of the reciprocals of individual resistances (1/Rp = 1/R1 + 1/R2 + ...).
For the 4Ω resistor: I = V / R = 6V / 4Ω = 1.5 A.
Reasons:
(i) Each appliance gets the full mains voltage (e.g., 220V).
(ii) Each appliance can be operated independently with its own switch.
(iii) If one appliance fails, the others are not affected.
When an electric current (I) flows through a conductor of resistance (R) for a time (t), the electric energy consumed is converted into heat. This is known as the heating effect of electric current. The heat produced (H) is given by Joule's Law of Heating: H = I²Rt. This law explains why a fuse wire melts or why an electric iron gets hot.
Electric power (P) is the rate at which this energy is consumed (P = H/t). Therefore, P = I²R, or P = V²/R, or P = VI. The SI unit of power is the Watt (W). The commercial unit of electrical energy is the kilowatt-hour (kWh), where 1 kWh = 3.6 × 10⁶ J.
Time per day (t) = 2 hours
Number of days in September = 30
Total Energy (E) = P × t × days = 2 kW × 2 h × 30 = 120 kWh (or 120 units).
Total Cost = Total Energy × Cost per unit = 120 × ₹6 = ₹720.
Since V (220V) is constant for both, resistance (R) is inversely proportional to power (P).
The bulb with lower power (60W) will have higher resistance.
A student sets up an experiment to verify Ohm's law. They connect a nichrome wire (resistor R) to a 1.5V battery, an ammeter (A), a voltmeter (V), and a rheostat (Rh) to vary the current. The student records the following readings:
| S. No. | Voltmeter (V) | Ammeter (A) |
|---|---|---|
| 1 | 0.3 V | 0.1 A |
| 2 | 0.6 V | 0.2 A |
| 3 | 0.9 V | 0.3 A |
Reading 1: R = 0.3V / 0.1A = 3 Ω
Reading 2: R = 0.6V / 0.2A = 3 Ω
Reading 3: R = 0.9V / 0.3A = 3 Ω
The average resistance is 3 Ω.
V = 1.2 V
I = V / R = 1.2 V / 3 Ω = 0.4 A.
Every electrical appliance has a power rating, usually given in Watts (W), and a voltage rating, given in Volts (V). For example, a bulb might be rated "60W - 220V". This means the bulb will consume 60 Watts of power (or 60 Joules of energy per second) when operated at a potential difference of 220 Volts. Using these two values, one can calculate the "safe" operating current (P = VI) and the resistance (P = V²/R) of the appliance. If the voltage supplied is different from the rated voltage, the power consumed will also be different.
R = (220V)² / 100W = (220 × 220) / 100 = 48400 / 100 = 484 Ω.
I = 100W / 220V ≈ 0.4545 A.
The new voltage (V') is 110V.
New Power (P') = (V')² / R = (110V)² / 484Ω = (110 × 110) / 484 = 12100 / 484 = 25 W.
In a series circuit, the total 220V voltage will be *divided* between the 484Ω bulb and the 300Ω bulb. The voltage across the 484Ω bulb will be less than 220V, so its power consumption will be much less than 100W.
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Resistivity (ρ) is a fundamental property of a material that measures how strongly it resists electric current. A low resistivity indicates a material that readily allows the flow of current. The resistivity of pure metals (like copper, silver) is very low and increases with an increase in temperature. In contrast, insulators (like glass, rubber) have extremely high resistivity. Alloys, which are mixtures of metals, generally have higher resistivity than their constituent pure metals. For example, nichrome (an alloy of nickel, chromium, iron, and manganese) has a resistivity about 60 times higher than that of copper. This property makes alloys useful for specific applications.
(i) It has very high resistivity (produces more heat).
(ii) It has a high melting point.
(iii) It does not oxidize easily at high temperatures.
Since length (L) and area (A) are the same for both wires, R is directly proportional to resistivity (R ∝ ρ).
As the passage states, nichrome has much higher resistivity (ρ) than copper, so it will have higher resistance.
