Electricity Class 10 Science — Competency Based Questions

20 Questions | 100 Marks

A group of students is conducting an experiment to verify Ohm's Law. They set up a circuit with a battery, an ammeter, a voltmeter, and a resistor. They vary the voltage across the resistor and record the corresponding current. They observe that as the voltage increases, the current also increases proportionally. They plot a graph of voltage (V) on the y-axis versus current (I) on the x-axis and obtain a straight line passing through the origin.(a) What does the straight line graph indicate about the relationship between voltage and current for the resistor? [1 mark](b) Based on their observations, what physical law are the students verifying? State this law. [2 marks](c) If the resistor used in the experiment has a resistance of 20 Ω, calculate the current flowing through it when a potential difference of 12 V is applied across it. 2marks

Rohan is trying to understand how household electrical appliances are connected. He notices that when he switches on the main switch, all the appliances in his room (light, fan, TV) turn on simultaneously. He also observes that if one appliance is switched off, the others continue to work.(a) What type of electrical connection is typically used for household appliances? [1 mark](b) Explain the advantage of this type of connection over a series connection, considering the operation of individual appliances. [2 marks](c) If a 100 W bulb and a 50 W fan are connected to a 220 V supply line in parallel, calculate the current drawn by each appliance and the total current drawn from the line. 3marks

A wire of resistance

R

is stretched so that its length becomes

2 l

and its area of cross-section becomes

A / 2

.(a) How does the resistance of the wire change? Justify your answer. [2 marks](b) If the original wire had a resistance of 10 Ω, what would be the new resistance after stretching? 3marks

Two resistors,

R_{1}

and

R_{2}

, are connected in a circuit. In the first case, they are connected in series to a battery, and in the second case, they are connected in parallel to the same battery.(a) State the relationship between the equivalent resistance in the series combination (

R_{s}

) and the individual resistances (

R_{1}, R_{2}

). [1 mark](b) State the relationship between the equivalent resistance in the parallel combination (

R_{p}

) and the individual resistances (

R_{1}, R_{2}

). [1 mark](c) If

R_{1} = 5

Ω and

R_{2} = 10

Ω, calculate the ratio of heat produced when these resistors are connected first in series and then in parallel to the same battery. 3marks

Assertion (A): The heating element of an electric heater is made of an alloy like nichrome, not pure copper.Reason (R): Alloys have higher resistivity and do not oxidize easily at high temperatures compared to pure metals.(a) If both A and R are true and R is the correct explanation of A. [1 mark](b) If both A and R are true but R is not the correct explanation of A. [1 mark](c) If A is true but R is false. [1 mark](d) If A is false but R is true. 1mark

A family is concerned about their electricity bill. They have a 100 W bulb that is used for 10 hours daily, and a 2 kW electric heater used for 1 hour daily. The electricity board charges Rs 5 per kWh.(a) Calculate the energy consumed by the bulb in a day in kWh. [2 marks](b) Calculate the energy consumed by the electric heater in a day in kWh. [2 marks](c) Calculate the total cost of operating these appliances for a month (30 days). 1mark

An electric bulb is rated 200 W at 250 V.(a) What is the resistance of the bulb filament? [2 marks](b) How much current does it draw when connected to a 250 V supply? 2marks

A circuit contains resistors connected in a specific arrangement. When a current flows through the circuit, it is observed that the same current flows through each resistor.(a) What type of combination is this? [1 mark](b) If three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in this manner, what is the total resistance of the combination? [2 marks](c) If a 6 V battery is connected to this combination, calculate the current flowing through each resistor and the potential difference across each resistor. 2marks

Priya is learning about electric current and its effects. She uses a battery, an ammeter, and a wire to form a simple circuit. She notices that the wire heats up when current flows through it.(a) What phenomenon is responsible for the heating of the wire? [1 mark](b) State Joule's law of heating. [2 marks](c) If a current of 2 A flows through a resistance of 10 Ω for 5 minutes, calculate the heat produced. 2marks

A fuse wire of a specific rating is used in household electrical circuits.(a) What is the primary function of a fuse wire in an electrical circuit? [1 mark](b) Why is it important for a fuse wire to have a low melting point? [2 marks](c) If a circuit is designed to handle a maximum current of 3 A, what rating of fuse wire should be used? Explain your choice. 2marks

A conductor has resistance R. If the conductor is melted and recast such that its length is doubled, how will its resistance change?(a) How does the resistance of a conductor depend on its length? [1 mark](b) If the length is doubled, what happens to the area of cross-section? Justify. [2 marks](c) Calculate the new resistance in terms of the original resistance R. 2marks

Assertion (A): In a parallel circuit, the total current drawn from the source is the sum of the currents flowing through each branch.Reason (R): The potential difference across each branch of a parallel circuit is the same.(a) If both A and R are true and R is the correct explanation of A. [1 mark](b) If both A and R are true but R is not the correct explanation of A. [1 mark](c) If A is true but R is false. [1 mark](d) If A is false but R is true. 1mark

A device has a resistance of 50 Ω. It is connected to a 12 V battery.(a) Calculate the current flowing through the device. [2 marks](b) If the battery voltage is increased to 24 V, while the resistance remains the same, what will be the new current? 2marks

A student sets up a circuit with two resistors,

R_{1} = 2

Ω and

R_{2} = 4

Ω, connected in series to a 6 V battery.(a) Calculate the total resistance of the circuit. [1 mark](b) Calculate the current flowing through the circuit. [1 mark](c) Calculate the power dissipated in the 2 Ω resistor. 2marks

Identify the type of electricity that drives the glowing of an LED street light. 1mark(a) Static electricity(b) Current electricity(c) Induced electricity(d) None of the above

A student uses a voltmeter to measure the potential difference across a resistor.(a) How should a voltmeter be connected in a circuit? [1 mark](b) Why is this specific connection used for a voltmeter? 2marks

An electric appliance draws a current of 1.5 A when connected to a 220 V source.(a) Calculate the power consumed by the appliance. [2 marks](b) Calculate the energy consumed by the appliance in 10 minutes. 2marks

Four resistors of resistances 10 Ω, 20 Ω, 30 Ω, and 40 Ω are connected in parallel.(a) State the formula for the equivalent resistance of this parallel combination. [1 mark](b) Calculate the equivalent resistance of the combination. 3marks

Assertion (A): Tungsten is used for making filaments of electric bulbs.Reason (R): Tungsten has a very high melting point and does not oxidize easily at high temperatures.(a) If both A and R are true and R is the correct explanation of A. [1 mark](b) If both A and R are true but R is not the correct explanation of A. [1 mark](c) If A is true but R is false. [1 mark](d) If A is false but R is true. 1mark

Consider a typical household circuit where multiple appliances are connected.(a) Why is it important that the wiring in household circuits is done with low resistance materials like copper? [2 marks](b) What happens if a very high current flows through household wiring? Explain the consequence. [2 marks](c) How can you prevent excessive current flow in household circuits? 1mark

Answers

(a) The proportional increase of current with voltage indicates that the resistance of the resistor is constant. 1mark(b) The students are verifying Ohm's Law. Ohm's Law states that the potential difference (V) across the ends of a conductor is directly proportional to the current (I) flowing through it, provided its temperature remains the same. Mathematically,

V \propto I

V = I R

, where R is the constant of proportionality (resistance). [2 marks](c) Given:

V = 12

R = 20

Ω. Using Ohm's Law,

I = V / R = 12

V / 20 Ω = 0.6 A. [2 marks]

(a) Household appliances are typically connected in parallel. 1mark(b) In a parallel connection, each appliance receives the full mains voltage independently. If one appliance fails or is switched off, the circuit for other appliances remains complete, and they continue to operate. In a series connection, the current would be the same through all appliances, and if one appliance fails, the entire circuit breaks, and all appliances stop working. Also, different appliances require different currents to operate, which is easily achieved in parallel connections. [2 marks](c) For the 100 W bulb:

I_{b u l b} = P / V = 100

W / 220 V = 0.45 A (approx). For the 50 W fan:

I_{f a n} = P / V = 50

W / 220 V = 0.23 A (approx). Total current

I_{t o t a l} = I_{b u l b} + I_{f a n} = 0.45

A + 0.23 A = 0.68 A (approx). [3 marks]

(a) Let the original length be

l

, area of cross-section be

A

, and resistance be

R

. So,

R = ρ \frac{l}{A}

. When the wire is stretched so that its length becomes

2 l

, its area of cross-section becomes

A / 2

(since the volume remains constant). The new resistance

R^{'}

will be

R^{'} = ρ \frac{2 l}{A / 2} = ρ \frac{4 l}{A} = 4 (ρ \frac{l}{A}) = 4 R

. Thus, the resistance becomes four times. 2marks(b) If the original resistance

R = 10

Ω, then the new resistance

R^{'} = 4 \times R = 4 \times 10

Ω = 40 Ω. [3 marks]

(a) When resistors are connected in series, the equivalent resistance is the sum of individual resistances:

R_{s} = R_{1} + R_{2}

. 1mark(b) When resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances:

\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

. [1 mark](c) For series connection:

R_{s} = 5

Ω + 10 Ω = 15 Ω. Heat produced

H_{s} \propto \frac{V^{2}}{R_{s}} t

. For parallel connection:

\frac{1}{R_{p}} = \frac{1}{5} + \frac{1}{10} = \frac{2 + 1}{10} = \frac{3}{10}

, so

R_{p} = \frac{10}{3}

Ω. Heat produced

H_{p} \propto \frac{V^{2}}{R_{p}} t

.The ratio of heat produced is

\frac{H_{s}}{H_{p}} = \frac{V^{2} / R_{s} \times t}{V^{2} / R_{p} \times t} = \frac{R_{p}}{R_{s}} = \frac{10 / 3}{15} = \frac{10}{45} = \frac{2}{9}

. So the ratio is 2:9. [3 marks]

(a) If both A and R are true and R is the correct explanation of A. 1mark

(a) Energy consumed by bulb = Power × Time = 100 W × 10 h = 1000 Wh = 1 kWh. 2marks(b) Energy consumed by heater = Power × Time = 2 kW × 1 h = 2 kWh. [2 marks](c) Total energy consumed in 30 days = (1 kWh + 2 kWh) × 30 days = 3 kWh × 30 = 90 kWh. Cost = 90 kWh × Rs 5/kWh = Rs 450. [1 mark]

(a)

P = V^{2} / R \Rightarrow R = V^{2} / P = (250 V)^{2} / 200 W = 62500 / 200 = 312.5

Ω. 2marks(b)

I = P / V = 200

W / 250 V = 0.8 A. [2 marks]

(a) Series combination. 1mark(b) For resistors in series,

R_{t o t a l} = R_{1} + R_{2} + R_{3} = 2

Ω + 3 Ω + 6 Ω = 11 Ω. [2 marks](c) Current through each resistor is the same as the total current:

I = V / R_{t o t a l} = 6

V / 11 Ω = 0.55 A (approx). Potential difference across each resistor:

V_{1} = I R_{1} = 0.55 \times 2 = 1.1

V_{2} = I R_{2} = 0.55 \times 3 = 1.65

V_{3} = I R_{3} = 0.55 \times 6 = 3.3

V. (Note:

1.1 + 1.65 + 3.3 = 6.05

V, which is approximately 6 V due to rounding). [2 marks]

(a) The phenomenon responsible is the heating effect of electric current (Joule's heating). 1mark(b) Joule's law of heating states that the heat produced in a resistor is directly proportional to the square of the current flowing through it, directly proportional to its resistance, and directly proportional to the time for which the current flows. Mathematically,

H = I^{2} R t

. [2 marks](c) Given:

I = 2

R = 10

Ω,

t = 5

minutes = 5 × 60 = 300 s. Heat produced

H = I^{2} R t = (2 A)^{2} \times 10

Ω × 300 s = 4 × 10 × 300 = 12000 J. [2 marks]

(a) The primary function of a fuse wire is to protect electrical circuits and appliances from damage due to overcurrent or short circuits by breaking the circuit when the current exceeds a safe limit. 1mark(b) A fuse wire needs to have a low melting point so that it melts quickly and breaks the circuit when an excessive current flows, thus preventing damage to other components. [2 marks](c) A 3 A fuse wire should be used. This is because the fuse rating should be slightly higher than the normal operating current of the circuit (1.5 A in this case) to avoid nuisance tripping, but low enough to blow off during an overcurrent situation. A 5 A fuse might not blow off in case of a moderate overcurrent, potentially damaging the appliances. [2 marks]

(a) The resistance of a conductor is directly proportional to its length (

R \propto l

). 1mark(b) When the length of the conductor is doubled, its volume remains constant. If the length becomes

2 l

, the area of cross-section must become

A / 2

to keep the volume constant (

V = l \times A = (2 l) \times (A / 2)

). [2 marks](c) Original resistance

R = ρ \frac{l}{A}

. New resistance

R^{'} = ρ \frac{2 l}{A / 2} = ρ \frac{4 l}{A} = 4 (ρ \frac{l}{A}) = 4 R

. The new resistance is 4 times the original resistance. [2 marks]

(a) If both A and R are true and R is the correct explanation of A. 1mark

(a) Using Ohm's Law,

I = V / R = 12

V / 50 Ω = 0.24 A. 2marks(b) With increased voltage, the new current

I^{'} = V^{'} / R = 24

V / 50 Ω = 0.48 A. [2 marks]

(a) For series resistors,

R_{t o t a l} = R_{1} + R_{2} = 2

Ω + 4 Ω = 6 Ω. 1mark(b) Current

I = V / R_{t o t a l} = 6

V / 6 Ω = 1 A. [1 mark](c) Power dissipated in the 2 Ω resistor

P_{1} = I^{2} R_{1} = (1 A)^{2} \times 2

Ω = 2 W. [2 marks]

(b) Current electricity 1mark

(a) A voltmeter should be connected in parallel across the component for which the potential difference is to be measured. 1mark(b) A voltmeter has a very high resistance. When connected in parallel, it draws a negligible current from the circuit, ensuring that the current distribution and potential difference across the component remain largely unaffected. [2 marks]

(a) Power

P = V I = 1.5

A × 220 V = 330 W. 2marks(b) Energy

E = P \times t = 330

W × (10 minutes × 60 seconds/minute) = 330 W × 600 s = 198000 J. [2 marks]

(a) For resistors in parallel,

\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}

. 1mark(b)

\frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} + \frac{1}{40} = \frac{12 + 6 + 4 + 3}{120} = \frac{25}{120}

R_{p} = \frac{120}{25} = \frac{24}{5} = 4.8

Ω. [3 marks]

(a) If both A and R are true and R is the correct explanation of A. 1mark

(a) Copper is a good conductor with very low resistivity. Using it for wiring minimizes energy loss as heat (

H = I^{2} R t

), ensuring that most of the electrical energy is delivered to the appliances. Low resistance also means less voltage drop across the wires. 2marks(b) If a very high current flows through household wiring, it can cause significant heating of the wires due to

H = I^{2} R t

. This excessive heat can melt the insulation, leading to short circuits, fire hazards, and damage to appliances. [2 marks](c) Excessive current flow can be prevented by using fuses or circuit breakers of appropriate ratings. These devices are designed to interrupt the circuit when the current exceeds a safe limit, thus protecting the wiring and appliances. Also, avoiding overloading by connecting too many high-power devices simultaneously is crucial. [1 mark]

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Electricity — Competency Based Questions

Competency based questions

Answers

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