Socket Circuit Overload in a Small Home
At a small home in Nashik, the following appliances are connected to one 15 A, 230 V single-phase socket circuit: a 2,000 W storage geyser, a 1,000 W electric iron, a 750 W mixer, three ceiling fans of 70 W each and eight LED lamps of 9 W each. The owner wants to run the geyser and iron together and occasionally the mixer at the same time. The circuit is protected by a 15 A MCB and wired with 2.5 mm² copper. The wiring is about eight years old and the socket outlets are 6 A/16 A combined types. The geyser is used for 30 minutes a day; the tariff is ₹8.50 per unit. The owner complains that the MCB trips 'when everything is on' and asks whether a higher-rated MCB will solve the problem. Supply voltage at the meter is 230 V.
1. Calculate the total connected load and the current drawn if all appliances operate together. Is the 15 A MCB correctly sized? Explain what will happen if it is replaced with a 25 A MCB.
Read the worked answer
Connected load = 2000 + 1000 + 750 + (3 × 70) + (8 × 9) = 4032 W.
Current at 230 V: I = P/V = 4032/230 = 17.5 A.
The 15 A MCB is correctly sized for a 15 A socket circuit, but the connected load exceeds it. The tripping is a genuine overload warning, not a nuisance. If a 25 A MCB is fitted, the cable and sockets are no longer protected. A 2.5 mm² PVC copper cable is typically rated around 24 A in free air and less when enclosed, while the socket outlets are rated 16 A. A 25 A device would allow about 25 A through a 16 A socket and through old joints, causing overheating, insulation damage and possibly fire before the MCB trips.
The correct remedy is not a bigger MCB but load management and separate circuits: geyser on its own 16 A circuit, iron and mixer on a 16 A power circuit, and lighting on a 6 A circuit. Diversity does not make a single 15 A circuit safe for a 17.5 A connected load.
2. Calculate the geyser element resistance at 230 V and the current it takes. If the supply falls to 200 V, calculate the power delivered, assuming constant resistance, and comment on the heating time.
Read the worked answer
Element resistance: R = V²/P = (230)²/2000 = 52,900/2000 = 26.45 Ω.
Current: I = P/V = 2000/230 = 8.70 A, which also equals V/R = 230/26.45.
At 200 V, assuming the element resistance stays at 26.45 Ω:
P = V²/R = (200)²/26.45 = 40,000/26.45 = 1,512 W.
Current becomes 200/26.45 = 7.56 A.
The geyser now delivers about 24% less power, so the heating time increases by roughly the same ratio (about 1.32 times) if losses are ignored; standing losses make it worse in practice. This shows that a heating appliance's power rating is valid only at its rated voltage, and low supply voltage causes long heating times rather than indicating a faulty element.
3. Calculate the monthly energy consumption and cost of the geyser. Suggest one conservation measure and quantify its saving.
Read the worked answer
Energy per day = 2 kW × 0.5 h = 1 kWh.
For 30 days = 30 kWh (30 units).
Monthly cost = 30 × ₹8.50 = ₹255.
One practical measure is to reduce the daily run time to 20 minutes by using a timer or by scheduling bathing: saving = 2 kW × (10/60) h × 30 = 10 kWh = ₹85 per month. Alternatively, fit an insulated geyser jacket or keep the thermostat no higher than 60°C; a 20% reduction in standing loss saves about 6 kWh = ₹51 per month. The judgement is that the real saving comes from reducing run time and standing losses, not from fitting a higher-rated MCB.