Case Study:

In Jaipur, Rajasthan, a family installed a rooftop solar panel system to reduce electricity costs and promote clean energy. The system consisted of solar panels with a total area of 10 m². On a sunny day, the average solar radiation received was about 1000 W/m².

However, the panels had an efficiency of only 18%, meaning only a fraction of the solar energy was converted into usable electrical energy. The rest was lost as heat and due to reflection.

The total power received by the panels was calculated as:
Power input = Area × Solar intensity = 10 × 1000 = 10,000 W

The useful electrical power output was:
Power output = 18% of 10,000 W = 1800 W

Over 5 hours of effective sunlight, the system generated:
Energy = 1800 × 5 = 9000 Wh = 9 kWh

Despite this, the family noticed a drop in efficiency during summer afternoons due to overheating and dust accumulation on panels. After regular cleaning and installing a cooling mechanism, efficiency improved slightly.

Meanwhile, a nearby commercial solar plant used tracking systems to align panels with the sun, increasing energy output by nearly 20%.

This case highlights the importance of efficiency, environmental factors, and technological improvements in maximizing solar energy output in India’s growing renewable energy sector.

Questions:

Section A - MCQs

1. What is the main reason only 18% of solar energy is converted into electricity?

A. Poor wiring
B. Energy losses due to heat and reflection
C. Low sunlight
D. High voltage

2. What is the total power input received by the panels?

A. 1000 W
B. 1800 W
C. 10,000 W
D. 9000 W

3. Why did efficiency decrease during summer afternoons?

A. Increased sunlight
B. Overheating and dust accumulation
C. Lower voltage
D. Shorter daylight hours

4. How do solar tracking systems improve efficiency?

A. Reduce panel size
B. Increase temperature
C. Align panels with sunlight for maximum exposure
D. Store energy

Section B - Short Answer Questions

1. Calculate the electrical power output if efficiency is 18% and input power is 10,000 W.

2. Explain one environmental factor affecting solar panel efficiency.

3. Why is cleaning solar panels important for maintaining efficiency?

Section C - Long Answer Question

1. If the efficiency of the solar panels increases from 18% to 22%, how will it affect energy production? Calculate the new output and analyze its significance for households.

Answer Key:

Section A - MCQs Answers:

1. B
2. C
3. B
4. C

Section B - Short Answers:

1.
Power output = 18% of 10,000 W
= 0.18 × 10,000
= 1800 W

2. High temperature reduces efficiency as it increases resistance and energy loss in the panels.

3. Dust blocks sunlight, reducing the amount of energy absorbed, thus lowering efficiency.

Section C - Long Answer:

• New efficiency = 22%
• Power output = 0.22 × 10,000 = 2200 W
• Energy for 5 hours = 2200 × 5 = 11,000 Wh = 11 kWh

Analysis:

• Increase of 2 kWh energy
• More cost savings for households
• Better utilization of renewable energy
• Encourages adoption of efficient technologies