Understanding Electrical Power and Circuit Efficiency

1. Introduction

Electrical power and efficiency are two key parameters that determine how effectively a circuit converts and uses energy. In real circuits, not all supplied energy reaches the load — part of it is inevitably lost as heat or electromagnetic interference (EMI).

2. Electrical Power

Symbol: P    Unit: Watt (W)

Power represents the rate at which electrical energy is consumed or converted. The basic formula is:

P = V × I

Other useful forms (when resistance is known):

P = I² × R
P = V² / R

3. Example Calculations

Example 1: A 10Ω resistor carries 2A of current. Calculate its power dissipation.

P = I² × R = 2² × 10 = 40 W

Example 2: A 12V–6W lamp. Find the current:

I = P / V = 6 / 12 = 0.5 A

4. Circuit Efficiency

Symbol: η (eta)    Unit: %

Efficiency measures the ratio of useful output power to total input power:

η = (P_out / P_in) × 100%

Example: A linear regulator with 12V input, 1A load, and 5V output:

P_in = 12 × 1 = 12W
P_out = 5 × 1 = 5W
η = (5 / 12) × 100% ≈ 41.7%

Over 7W are lost as heat — which is why linear regulators need large heatsinks.

5. Linear vs Switching Power Supplies

Criteria	Linear Supply	Switching Supply (SMPS)
Efficiency	40–60%	80–95%
Size	Bulky	Compact, lightweight
Noise	Low	High (EMI)
Applications	Audio, measurement devices	Computers, chargers, industrial systems

6. Factors Affecting Power and Efficiency

• Conductor resistance: Power loss increases with I²R.
• Temperature: High heat raises resistance, lowering efficiency.
• Design quality: Poor layout or cooling reduces performance.

7. Practical Design Tips

• Choose components rated at least 2× the expected power dissipation.
• Ensure proper airflow and spacing between heat-generating components.
• Use MOSFETs or high-efficiency ICs (low Rds-on, synchronous buck converters) to minimize losses.

8. Exercises

Exercise 1: A 24V source powers a 120Ω resistor. Calculate the current and power.

I = 24 / 120 = 0.2A
P = 24 × 0.2 = 4.8W

Exercise 2: A power supply input: 9V–2A; output: 5V–3A. Find its efficiency.

P_in = 9 × 2 = 18W
P_out = 5 × 3 = 15W
η = (15 / 18) × 100% = 83.3%

9. Conclusion

Understanding power and efficiency allows engineers to design safe, reliable, and energy-efficient circuits. Applying P = V×I and η = P_out/P_in helps optimize power consumption, reduce thermal losses, and improve circuit longevity.

References

• Thomas L. Floyd, Electronics Fundamentals: Circuits, Devices, and Applications, Pearson, 2014.
• Paul Horowitz & Winfield Hill, The Art of Electronics, Cambridge University Press, 2015.

Keywords: electrical power, efficiency, P=VI, power dissipation, SMPS, linear supply, circuit design, basic electronics