Ramesh's Repair Shop: Is a 15 A Connection Enough?
Ramesh runs a small fabrication-cum-repair shop in Ranchi on a single-phase 230 V domestic connection protected by a 15 A MCB. His connected equipment is: one arc-welding transformer drawing 1.5 kW at a power factor of 0.60; one 1 hp single-phase motor (746 W output) with 80 % efficiency and 0.80 power factor; four LED tube-lights of 40 W each at 0.95 power factor; and a 2 kW induction hot plate at unity power factor. He plans to run everything together for eight hours a day, 26 days a month. The tariff is flat at ₹7.50 per unit and his licence carries no power-factor penalty clause. The local electrician has told him that a 15 A connection "means about 3.5 kW, so it will be fine". Ramesh now also believes that correcting the motor's power factor alone will bring his monthly bill down, and he wants the cheapest change that keeps him legal.
1. Compute the total current drawn when all four loads run together. Is the 15 A MCB adequate? Show the apparent-power arithmetic and state precisely what the electrician got wrong.
Read the worked answer
Work in volt-amperes, because the MCB and the cable respond to current, not to watts.
- Welding transformer: 1500 / 0.60 = 2500 VA
- Motor: input power = 746 / 0.80 = 932.5 W; apparent power = 932.5 / 0.80 = 1165.6 VA
- Lights: 4 × 40 = 160 W; 160 / 0.95 = 168.4 VA
- Hot plate: 2000 / 1.00 = 2000 VA
Total S ≈ 2500 + 1165.6 + 168.4 + 2000 = 5834 VA, so I = 5834 / 230 ≈ 25.4 A.
The MCB passes only 15 A ≈ 3450 VA at 230 V, so the shop demands about 70 % more current than the service allows; the breaker will trip, and if it is a rewireable fuse instead, the internal wiring of a 15 A service (typically 2.5–4 mm², roughly 20–25 A) will run hot with no margin.
The electrician's error is equating 230 V × 15 A = 3450 VA with 3450 W of usable load. That identity holds only at unity power factor. At the shop's combined power factor (4592.5 W / 5834 VA ≈ 0.79), the same 3450 VA can carry only about 2.7 kW of real load.
2. Compute the monthly energy charge with all loads running as planned. Would improving the motor's power factor to unity reduce this bill? Support the answer with numbers.
Read the worked answer
Energy is billed on real power, so add the watts: 1500 + 932.5 + 160 + 2000 = 4592.5 W = 4.5925 kW.
- Daily energy = 4.5925 × 8 = 36.74 kWh
- Monthly energy = 36.74 × 26 = 955.2 kWh
- Monthly bill = 955.2 × 7.50 ≈ ₹7,165
Power-factor correction changes the apparent power drawn from the supply, not the active power converted into heat, light and mechanical work. With the motor corrected to unity, total S falls to 2500 + 932.5 + 168.4 + 2000 ≈ 5601 VA (current 24.4 A instead of 25.4 A), but the wattage stays 4592.5 W and the kWh bill is unchanged.
The correction would matter only if the utility billed kVAh or imposed a penalty below a threshold power factor (usually 0.90). Here the licence has no such clause, so the honest advice is: pf correction buys cable and breaker relief, not a rupee of saving.
3. Recommend the cheapest workable remedy for Ramesh, using load arithmetic rather than general advice.
Read the worked answer
Test the options against the 15 A (≈3450 VA) ceiling.
- Motor + lights + hot plate (welding off) = 1165.6 + 168.4 + 2000 = 3334 VA = 14.5 A - fits.
- Welding + lights (everything else off) = 2500 + 168.4 = 2668 VA = 11.6 A - fits.
- Welding + motor + lights = 2500 + 1165.6 + 168.4 = 3834 VA = 16.7 A - does not fit.
- Welding + hot plate = 4668 VA = 20.3 A - does not fit.
So no arrangement that puts the welding set on with the motor or the hot plate stays within 15 A. The interim fix is a hard rule (ideally an interlock or a changeover switch rather than a verbal one): the welding transformer runs only with the lights on, and the hot plate and motor run only when welding is off. That satisfies the existing connection but throttles his work.
The durable fix is a service upgrade: 25.4 A requires a 32 A MCB and a 6 mm² service cable, applied for on the basis of 5834 VA, not 4.6 kW. Power-factor correction alone cannot rescue the 15 A connection - even at unity pf the combined load would draw about 20 A.