Unit Test (Solutions): Electricity

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: When electric current is passed, electrons move from: (1 Mark)
(a) high potential to low potential
(b) low potential to high potential
(c) in the direction of the current
(d) against the direction of the current

Ans:  (d)

Electrons are negatively charged and move from the negative terminal (lower electric potential for positive charges) towards the positive terminal. This motion is opposite to the direction of conventional current, so electrons move against the direction of the current.

Q2: The instrument used for measuring electric current is: (1 Mark)
(a) Ammeter
(b) Galvanometer
(c) Voltmeter
(d) Potentiometer

Ans: (a)

An ammeter is designed to measure the electric current in a circuit. It must be connected in series with the circuit element so that the full current passes through the instrument and can be measured accurately.

Q3: The unit of potential difference is __________________. (1 Mark)

Ans: Volt

Q4: What is the maximum resistance which can be made using five resistors each of 1/5 Ω? (1 mark)
(a) 5 Ω
(b) 10 Ω
(c) 1/5 Ω
(d) 1 Ω

Ans: (d)

If all five resistors of value 1/5 Ω each are connected in series, the resistances add. Total resistance R = 5 × (1/5) Ω = 1 Ω. Series connection gives the maximum possible resistance, so the maximum is 1 Ω.

Q5: A piece of wire of resistance R is cut into five equal parts. These parts are then arranged in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is _____. (1 Mark)
(a)  5
(b) 1/5
(c)1/25
(d) 25

Ans: (d)

When the wire of resistance R is cut into five equal parts, each part has resistance r = R/5.

If these five equal resistances are connected in parallel, the equivalent resistance R′ is

R′ = r/5 = (R/5)/5 = R/25.

Therefore, R/R′ = R / (R/25) = 25.

Q6: What does an electric circuit mean? (2 Marks)

Ans: An electric circuit is a continuous, closed path made of conductors and electrical components through which electric current can flow. A simple circuit typically includes a source of emf (cell or battery), conducting wires, a switch to open or close the path, and a load (such as a bulb or resistor) that uses the electrical energy.

Q7: Write the relation between the resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it. (2 Marks)

Ans: Power P, voltage V and resistance R are related by

P = V²/R

Derivation (brief): P = VI and V = IR, so P = I×(IR) = I²R. Eliminating I using I = V/R gives P = V×(V/R) = V²/R. For a constant applied voltage V, the power of the filament varies inversely with R.

Q8: How does the use of a fuse wire protect electrical appliances? (2 Marks)

Ans: A fuse wire is a thin wire made of a material with a relatively low melting point and is placed in series with the appliance. When an excessive current flows (for example, due to a short circuit), the fuse wire heats up and melts quickly because of its small cross-sectional area. Melting of the fuse breaks the circuit and stops the current, preventing damage to the appliance and reducing the risk of fire.

Q9:  List the three factors on which the resistance of a conductor depends. (3 Marks)

Ans: A conductor's resistance depends on the following three factors:

(1) Length of the conductor - Resistance is directly proportional to the length. R ∝ l. A longer conductor offers more resistance.

(2) Cross-sectional area - Resistance is inversely proportional to the area of cross-section. R ∝ 1/A. A thicker conductor has lower resistance.

(3) Nature of the material - Different materials have different resistivities (ρ). For a given length and area, a material with higher resistivity has greater resistance. This relation is summarised by R = ρ l / A, where ρ is the resistivity of the material.

Q10: What is the commercial unit of electrical energy? Represent it in terms of joules. (3 marks)

Ans: The commercial unit of electrical energy is the kilowatt-hour (kWh).

1 kWh = 1 kW × 1 hour = 1000 W × 3600 s = 3.6 × 10⁶ J.

Q11: If an electric heater rated 800 W operates 6h/day. Find the Cost of energy to operate it for 30 days at ₹3.00 per unit of consumption. (3 Marks)

Ans: 

Power of heater, P = 800 W = 0.8 kW
Time per day, t = 6 h

Energy consumed per day = Power × Time = 0.8 kW × 6 h = 4.8 kWh/day

Energy consumed in 30 days = 4.8 kWh/day × 30 days = 144 kWh

Cost per unit (1 kWh) = ₹3.00

Therefore, total cost = 144 × 3 = ₹432

Q12: What is Joule's heating effect? How can it be demonstrated experimentally? List its four applications in daily life. (5 Marks)

Ans:

Joule's heating effect: When an electric current passes through a conductor, heat is produced in the conductor. The heat produced H is given by H = I²Rt, where I is current, R is resistance and t is time. Thus, heat produced is (i) directly proportional to I², (ii) directly proportional to R, and (iii) directly proportional to t.

Experiment to demonstrate Joule's law of heating:

1. Use an immersion heating rod connected to a mains supply through a regulator (or variable resistor) and place the rod in a fixed volume of water.
2. Set the regulator to a low setting, switch on and measure the time taken to raise the water temperature by a certain amount.
3. Increase the regulator setting (so that larger current flows), repeat the experiment for the same volume of water and measure the time again.
4. Compare the times: with larger current the water heats faster, showing more heat produced for higher current and illustrating H ∝ I²t.

Applications (four examples):

• Electric iron
• Electric heater and room heater elements
• Electric kettles and immersion rods
• Toasters and electric ovens

Q13: A hot plate of an electric oven connected to a 220 V supply line has two resistance coils such as A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What will be the currents in the three cases? (5 Marks)

Ans:

Case (i) When coils are used separately
For one coil: I = V/R = 220 V / 24 Ω = 9.17 A (approximately). Each coil, when used alone across 220 V, draws about 9.17 A.

Case (ii) When the coils are connected in series
Total resistance = 24 Ω + 24 Ω = 48 Ω.
Current through series connection: I = V / R = 220 V / 48 Ω = 4.58 A (approximately).

Case (iii) When the coils are in parallel
Equivalent resistance for two equal resistances in parallel: R_eq = (24 × 24) / (24 + 24) = 576 / 48 = 12 Ω.
Total current from the supply: I = V / R_eq = 220 V / 12 Ω = 18.33 A (approximately).