Case Study Questions for Class 10 Science Chapter 12 Electricity

In CBSE Class 10 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Here, we have provided case based/passage based questions for Class 10 Science Chapter 12 Electricity.

Case Study Questions for Class 10 Science Chapter 12 Electricity

Case Study Questions

Question 1:

Read the following and answer the questions from (i) to (v) given below:
The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule (as shown in figure).

Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved.

(i) The SI unit of electric energy per unit time is
(a) joule
(b) joule-second
(c) watt
(d) watt-second

(ii) Kilowatt-hour is equal to
(a) 3.6 ×10⁴ J
(b) 3.6 ×10⁶ J
(c) 36 ×10⁶ J
(d) 36 ×10⁴ J

(iii) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is
(a) doubled
(b) half
(c) remains same
(d) four times

(iv) The power of a lamp is 60 W. The energy consumed in 1 minute is
(a) 360 J
(b) 36 J
(c) 3600 J
(d) 3.6 J

(v) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 Ω for 30 minutes.
(a) 40 kJ
(b) 60 kJ
(c) 10 kJ
(d) 90 kJ

Answers:
(i) (c) Watt
(ii) (b) 3.6 x 10⁶ J
(iii) (a) doubled
(iv) (a) 360 J
(v) (d) 90 kJ

---

Case Study Questions

Question 2:

Ravi noticed a power outage in one of the rooms. Upon inspection, he found that the fuse had melted. His father explained that this happened because too many devices were connected to a single socket, leading to an overload. Ravi also observed that the wiring system was a parallel connection, which helped isolate the problem to one room.

Q1. Why is parallel connection used in household circuits?
(a) It is cheaper
(b) It allows equal current through all devices
(c) It ensures that each device gets the same voltage
(d) It reduces energy consumption

Q2. What is the role of a fuse in an electric circuit?
(a) To complete the circuit
(b) To increase current
(c) To protect the circuit from excess current
(d) To supply constant voltage

Q3. What is meant by overloading?
(a) Using a thick wire
(b) Using low voltage appliances
(c) Connecting too many devices to a single socket
(d) Switching off all appliances

Q4. A heater (1500 W), a fan (100 W), and a bulb (60 W) are used together for 2 hours. What is the total energy consumed?
(a) 1.66 kWh
(b) 3.32 kWh
(c) 2.66 kWh
(d) 4.2 kWh

---

Case Study Question 3:

Electric Kettle vs Iron

An electric kettle rated 1500 W and an iron rated 1000 W are used for 30 minutes each. Raj wants to compare their power consumption and current draw. He also learns that power consumption depends on the product of voltage and current and affects the electricity bill.

Q1. Which appliance consumes more energy?
(a) Iron
(b) Kettle
(c) Both equally
(d) Cannot be determined

Q2. How much energy does the kettle consume in 30 minutes?
(a) 1.5 kWh
(b) 0.75 kWh
(c) 0.5 kWh
(d) 1.25 kWh

Q3. What is the current drawn by the kettle? (Supply voltage = 220 V)
(a) 5.5 A
(b) 6.8 A
(c) 8.2 A
(d) 4.5 A

Q4. Which formula correctly represents power?
(a) P = I × R
(b) P = I / V
(c) P = V × I
(d) P = V / I

---

Case Study Question 4:

Series and Parallel Circuits

In a school exhibition, students present two setups: one with bulbs in series and the other in parallel. When a bulb in the series circuit fused, all others went off. In the parallel setup, others stayed on. Their teacher explained how total resistance and brightness differ in both arrangements.

Q1. What happens to total resistance in a series circuit when more resistors are added?
(a) Increases
(b) Decreases
(c) Remains the same
(d) Becomes zero

Q2. What is the brightness of bulbs in parallel compared to series?
(a) Less in parallel
(b) Same in both
(c) Greater in series
(d) Greater in parallel

Q3. In a parallel circuit with three equal bulbs, the total current is:
(a) Equal to current through one bulb
(b) Less than current through one bulb
(c) Sum of individual currents
(d) Zero

Q4. Which of the following is correct for resistors in parallel?
(a) $\frac{1}{R}=R_1+R_2+R_3$
(b) $R=R_1+R_2+R_3$
(c) $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
(d) $R=\frac{R_1+R_2+R_3}{3}$

---

Case Study Question 5:

Street Lights and Power Loss

To save power, the municipality installs energy-efficient LED lights instead of old sodium lamps. A technician explains that LEDs draw less current and that electricity transmission is more efficient at higher voltages because it reduces power loss due to heating.

Q1. What is the formula for power loss due to heating?
(a) $P=I^2 R$
(b) $P=V^2 / R$
(c) $P=I / R$
(d) $P=V \times I$

Q2. Why is high voltage used in transmission lines?
(a) To reduce voltage fluctuations
(b) To increase current
(c) To reduce current and hence heat loss
(d) To increase resistance

Q3. An LED bulb of 10 W is used for 5 hours daily. Energy consumed in 30 days is:
(a) 1.5 kWh
(b) 3 kWh
(c) 0.75 kWh
(d) 2 kWh

Q4. Which measure does not help in saving electricity?
(a) Using energy-efficient appliances
(b) Turning off unused devices
(c) Using appliances during peak hours
(d) Fixing faulty wiring

---

Also Check

• Case study questions for class 6 to 12
• Assertion reason questions for class 6 to 12

Related Posts

Category Lists (All Posts)