A Geyser Running Half an Hour Costs More Than a Bulb All Day

Because the bill depends on power multiplied by time, and a geyser's power is enormous compared with a bulb's.

A $60\text{W}$ bulb burning for ten hours uses $0.6$ units of electricity. A $1500\text{W}$ geyser running for just half an hour uses $0.75$ units — more, in one twenty-fifth of the time.

So switching off a bulb saves far less than shortening a bath. This page covers the first part of the ICSE Class 8 Physics chapter on electricity: what current and potential difference measure, how power follows from them, and how power becomes a number on a bill.

Electric current is the rate of flow of charge through a conductor:

$$I=\frac{Q}{t}$$

Its SI unit is the ampere ($\text{A}$). One ampere is one coulomb of charge passing a point each second, so $1\text{A}=1\text{C/s}$. Current is measured with an ammeter, connected in series with the component.

Worked example. If $60\text{C}$ of charge passes through a wire in $20\text{s}$:

$$I=\frac{60}{20}=3\text{A}$$

Potential difference between two points is the work done in moving unit charge from one to the other:

$$V=\frac{W}{Q}$$

Its SI unit is the volt ($\text{V}$), and $1\text{V}=1\text{J/C}$. It is measured with a voltmeter, connected in parallel across the component.

What potential difference actually does. It is the electrical push that drives the current. A cell or a generator maintains that push; without it the free electrons in a wire drift about randomly and no current flows in any direction.

A helpful comparison. Water flows through a pipe because of a difference in pressure between the two ends. Charge flows through a wire because of a difference in potential. Remove the pressure difference and the water stops even though the pipe is still full — and the same is true of the wire.

The connection rule is worth memorising properly. Ammeter in series, voltmeter in parallel. An ammeter has very low resistance and would short out a component if connected across it; a voltmeter has very high resistance and would nearly stop the current if placed in the line.

A kilowatt-hour is the energy used by a $1\text{kW}$ appliance running for $1\text{hour}$. It is the commercial unit of electrical energy, and one kilowatt-hour is one unit on your bill.

$$1\text{kWh}=1000\text{W}\times 3600\text{s}=3.6\times 10^6\text{J}$$

So one unit is $3.6$ million joules — which is why the bill is written in units and not in joules.

The working formula:

$$\text{units (kWh)}=\frac{\text{power in watts}\times \text{time in hours}}{1000}$$

Worked example 1 — one appliance. An electric iron rated $1000\text{W}$ is used for $2$ hours a day for $30$ days. At a rate of $\text{₹}8$ per unit, find the cost.

$$\text{units}=\frac{1000\times 2\times 30}{1000}=60\text{kWh}$$

$$\text{cost}=60\times 8=\text{₹}480$$

Worked example 2 — several bulbs. Five bulbs of $60\text{W}$ each burn for $5$ hours:

$$\text{units}=\frac{5\times 60\times 5}{1000}=\frac{1500}{1000}=1.5\text{kWh}$$

$$\text{cost}=1.5\times 8=\text{₹}12$$

Worked example 3 — back to the opening claim. The $60\text{W}$ bulb for $10$ hours:

$$\frac{60\times 10}{1000}=0.6\text{kWh}$$

The $1500\text{W}$ geyser for $0.5$ hour:

$$\frac{1500\times 0.5}{1000}=0.75\text{kWh}$$

So the geyser really does consume more in thirty minutes than the bulb does in ten hours.

The division by 1000 is the step that gets forgotten. Power must be converted from watts to kilowatts, or equivalently the whole product divided by $1000$. Leaving it out inflates every answer by a factor of a thousand — and an electricity bill of lakhs of rupees should tell you immediately that something went wrong.

Two numbers: the voltage it is designed for and the power it consumes at that voltage. A plate reading $220\text{V},1000\text{W}$ means the appliance draws $1000\text{W}$ when connected to a $220\text{V}$ supply.

From those two you can work out everything else — the current it draws, the fuse it needs, and the cost of running it.

Typical household ratings, and what they imply. Ranking common appliances by power immediately ranks them by running cost for the same time:

- An LED bulb or a mobile charger — a few watts
- A ceiling fan — of the order of tens of watts
- A television or a refrigerator — around a hundred watts, though a fridge runs for many hours
- An electric iron, a microwave or a toaster — around a kilowatt
- A geyser, an air conditioner or a room heater — well over a kilowatt

The comparison that matters is power times time, not power alone. A refrigerator of modest power running all day may use more units than an iron of ten times the power used for twenty minutes. Work out the kilowatt-hours for each before deciding which is the expensive one.

Worked comparison. A $150\text{W}$ refrigerator running effectively $8$ hours a day:

$$\frac{150\times 8}{1000}=1.2\text{kWh per day}$$

An $1000\text{W}$ iron used $20$ minutes, which is $\frac{1}{3}$ hour:

$$\frac{1000\times \frac{1}{3}}{1000}\approx 0.33\text{kWh per day}$$

The refrigerator, despite its far lower power, uses roughly four times the energy — because it is on for twenty-four times as long.

A boundary case worth knowing. Run an appliance below its rated voltage and it draws less power and works poorly; run it above, and it draws more power than designed and may burn out. This is why an appliance meant for one country's supply voltage cannot simply be plugged in elsewhere, and why a voltage stabiliser is fitted for large appliances in areas where the supply fluctuates.