Transmission Loss, Efficiency, and Fuse Rating
A line of resistance R loses I squared R. Efficiency is useful output divided by input. For loads on a common voltage, total current is total rated power divided by supply voltage, then a suitable standard fuse is selected above normal current.
Why this shows up in the exam
Power transmission · Household load planning · Fuse capacity estimation
Learn the idea
Line losses scale as current squared, while a fuse must safely carry the expected total current. For fixed delivered power, raising transmission voltage lowers current and sharply reduces heating loss. In a building, parallel appliance powers add before current and fuse capacity are chosen.
🧠 Memory hook: High voltage means low current and low I-squared-R loss.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- P_loss = I² R_line — power lost as heat in the resistive line
- eta = P_load/(P_load+P_loss) — transmission efficiency in the simple model
- I_total = sum P_i/V_supply — parallel rated loads at common voltage
How to approach it
- 1Find load current from delivered conditions
- 2Compute line loss and input
- 3For fuse questions, sum simultaneous loads and choose the next safe rating
Common slip-ups that cost marks
- •Dividing delivered power by sending voltage inconsistently
- •Adding appliance currents from different voltages blindly
- •Choosing a fuse below normal operating current
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 2 ohm resistor. Find the circuit current.
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