Every Coulomb That Crosses a Bulb Hands Over Its Energy and Leaves
Derive the expression for electrical energy from the definition of potential difference, get the three forms of electrical power by bringing in Ohm's law, convert between joule and kilowatt hour, and work out a household electricity bill from appliance ratings.
Where does the energy in an electric circuit actually come from?
A bulb glows, a heater warms and a fan turns, all from a current that seems to pass straight through them and out the other side. Nothing is used up in the sense of being consumed — the same number of electrons leave the bulb as enter it. So what is being spent?
The energy each coulomb of charge carries.
That is what a potential difference measures. From the previous part, means that a potential difference of V across a bulb gives **every coulomb passing through it joules of energy — energy the supply put in and the bulb takes out.
So the energy delivered is simply the charge that passed multiplied by the volts each coulomb carried:**
and since , that becomes , which is the working form for every calculation in this part.
Bring Ohm's law into it and the same energy can be written three ways, because and are not independent once the resistance is fixed. Which form you use depends only on which two quantities the question gives you — and choosing well can turn three lines of working into one.
From energy to power is one division. Power is energy per second, so , and the three energy forms become three power forms.
And then the practical half. The joule is far too small a unit for a house, which is why the electricity board bills in kilowatt hours — the "units" printed on the bill. **One unit is million joules, and being able to move between the two is worth as many marks as any formula.
The chapter closes where every student has a personal interest: the bill itself.** Given the power ratings printed on your own appliances and the hours you use them, the monthly consumption and its cost are a single calculation.
This page covers the second part of the ICSE Class 10 Physics chapter on electricity and magnetism: electrical energy, the three forms of electrical power, the units of energy, and household consumption.
The energy each coulomb of charge carries.
That is what a potential difference measures. From the previous part, means that a potential difference of V across a bulb gives **every coulomb passing through it joules of energy — energy the supply put in and the bulb takes out.
So the energy delivered is simply the charge that passed multiplied by the volts each coulomb carried:**
and since , that becomes , which is the working form for every calculation in this part.
Bring Ohm's law into it and the same energy can be written three ways, because and are not independent once the resistance is fixed. Which form you use depends only on which two quantities the question gives you — and choosing well can turn three lines of working into one.
From energy to power is one division. Power is energy per second, so , and the three energy forms become three power forms.
And then the practical half. The joule is far too small a unit for a house, which is why the electricity board bills in kilowatt hours — the "units" printed on the bill. **One unit is million joules, and being able to move between the two is worth as many marks as any formula.
The chapter closes where every student has a personal interest: the bill itself.** Given the power ratings printed on your own appliances and the hours you use them, the monthly consumption and its cost are a single calculation.
This page covers the second part of the ICSE Class 10 Physics chapter on electricity and magnetism: electrical energy, the three forms of electrical power, the units of energy, and household consumption.
Formula
What are the formulas for electrical energy and power?
One energy formula in three forms, and one power formula in three forms — all obtained by substituting Ohm's law.
The derivation of the first one. From the definition of potential difference,
and since the charge that passes in time is ,
Substituting Ohm's law two ways gives the other forms. Putting ,
and putting ,
**Dividing each by the time gives the three power forms, since power is the rate of doing work. The SI unit of power is the watt (W)**, one joule per second, and kW W.
Which form to use, decided by what you are given:
- Voltage and current — use
- Current and resistance — use
- Voltage and resistance — use
Worked example 1 — energy from charge and voltage. A charge of C flows through a potential difference of V. Find the energy delivered.
Worked example 2 — energy from voltage, current and time. A current of A flows through an appliance across a V supply for one minute. Find the energy consumed.
The time had to be converted to seconds first. Using instead of would give an answer sixty times too small, and that is the commonest slip in this family of questions.
Worked example 3 — the same energy by two of the three forms. A current of A flows through a resistance of ohm for s. Find the heat produced.
Check with the voltage form. The potential difference across it is V, so
The two routes agree, which is the cleanest way of confirming that the right form was used.
Worked example 4 — the square in action. If the current in that same resistor were doubled to A for the same time, find the heat produced.
Four times as much heat for twice the current. The current appears squared, which is why overloading a circuit is so dangerous — a modest rise in current gives a steep rise in heating.
One point about the word "consumed". Electrical energy is not destroyed in an appliance; it is converted — into heat in a heater, into heat and light in a bulb, into mechanical energy in a fan. Saying the energy is used up is a marked error, and the correct phrasing names what it became.
The derivation of the first one. From the definition of potential difference,
and since the charge that passes in time is ,
Substituting Ohm's law two ways gives the other forms. Putting ,
and putting ,
**Dividing each by the time gives the three power forms, since power is the rate of doing work. The SI unit of power is the watt (W)**, one joule per second, and kW W.
Which form to use, decided by what you are given:
- Voltage and current — use
- Current and resistance — use
- Voltage and resistance — use
Worked example 1 — energy from charge and voltage. A charge of C flows through a potential difference of V. Find the energy delivered.
Worked example 2 — energy from voltage, current and time. A current of A flows through an appliance across a V supply for one minute. Find the energy consumed.
The time had to be converted to seconds first. Using instead of would give an answer sixty times too small, and that is the commonest slip in this family of questions.
Worked example 3 — the same energy by two of the three forms. A current of A flows through a resistance of ohm for s. Find the heat produced.
Check with the voltage form. The potential difference across it is V, so
The two routes agree, which is the cleanest way of confirming that the right form was used.
Worked example 4 — the square in action. If the current in that same resistor were doubled to A for the same time, find the heat produced.
Four times as much heat for twice the current. The current appears squared, which is why overloading a circuit is so dangerous — a modest rise in current gives a steep rise in heating.
One point about the word "consumed". Electrical energy is not destroyed in an appliance; it is converted — into heat in a heater, into heat and light in a bulb, into mechanical energy in a fan. Saying the energy is used up is a marked error, and the correct phrasing names what it became.
How do you use the three power forms on an appliance rating?
Read the rating as a pair of values — a power and a voltage — and use whichever form connects them to what is asked.
What a rating means. A bulb marked " W, V" consumes W **only when connected to a V supply. Its resistance is fixed by its construction, so those two numbers together determine everything else about it.
Worked example 1 — resistance and current from a rating.** A bulb is marked W, V. Find its resistance and the current it draws.
For the resistance, use the form containing the voltage and the power:
**For the current, use :**
Check with the third form:
All three agree, and running that check is worth one line whenever a rating is involved.
Worked example 2 — comparing two bulbs. Two bulbs are marked W, V and W, V. Which has the greater resistance?
**The W bulb has the greater resistance, which surprises most students. At a fixed voltage, a higher power means a lower resistance**, because puts in the denominator. The brighter bulb lets more current through, not less.
Worked example 3 — a heating element. An electric iron rated W works on a V supply. Find its resistance and the current it draws.
**Compare that ohm with the ohm of the bulb.** The iron uses ten times the power of a W bulb, so it must have about a tenth the resistance — **exactly what predicts.
Worked example 4 — energy from a rating.** A kW heater is used for minutes. Find the energy consumed in kilowatt hour and in joule.
Notice that the time was in hours for the first and the conversion factor did the rest. Working in kilowatt and hour gives kilowatt hour directly, and that is the easier route whenever the answer is wanted as a bill unit.
Worked example 5 — energy using the voltage form. A resistance of ohm is connected to a V supply for s. Find the heat produced.
And note what that tells you: the resistance is exactly that of a W bulb, so the answer had to be W for s. Recognising a rating hidden inside a numerical is a genuine time-saver.
A fuse rating, which uses the same reasoning. A kW appliance on a V supply draws
so it needs a fuse rated just above that — a A fuse. **A A fuse would melt in normal use and a A fuse would let a A fault current pass**, which is exactly what the fuse exists to stop.
What a rating means. A bulb marked " W, V" consumes W **only when connected to a V supply. Its resistance is fixed by its construction, so those two numbers together determine everything else about it.
Worked example 1 — resistance and current from a rating.** A bulb is marked W, V. Find its resistance and the current it draws.
For the resistance, use the form containing the voltage and the power:
**For the current, use :**
Check with the third form:
All three agree, and running that check is worth one line whenever a rating is involved.
Worked example 2 — comparing two bulbs. Two bulbs are marked W, V and W, V. Which has the greater resistance?
**The W bulb has the greater resistance, which surprises most students. At a fixed voltage, a higher power means a lower resistance**, because puts in the denominator. The brighter bulb lets more current through, not less.
Worked example 3 — a heating element. An electric iron rated W works on a V supply. Find its resistance and the current it draws.
**Compare that ohm with the ohm of the bulb.** The iron uses ten times the power of a W bulb, so it must have about a tenth the resistance — **exactly what predicts.
Worked example 4 — energy from a rating.** A kW heater is used for minutes. Find the energy consumed in kilowatt hour and in joule.
Notice that the time was in hours for the first and the conversion factor did the rest. Working in kilowatt and hour gives kilowatt hour directly, and that is the easier route whenever the answer is wanted as a bill unit.
Worked example 5 — energy using the voltage form. A resistance of ohm is connected to a V supply for s. Find the heat produced.
And note what that tells you: the resistance is exactly that of a W bulb, so the answer had to be W for s. Recognising a rating hidden inside a numerical is a genuine time-saver.
A fuse rating, which uses the same reasoning. A kW appliance on a V supply draws
so it needs a fuse rated just above that — a A fuse. **A A fuse would melt in normal use and a A fuse would let a A fault current pass**, which is exactly what the fuse exists to stop.
What are the units of electrical energy, and how do you convert between them?
**The SI unit is the joule; the commercial unit is the kilowatt hour, and one kilowatt hour is million joules.
The SI unit. One joule** is the energy delivered when one coulomb passes through a potential difference of one volt — or, equally, when a power of one watt acts for one second:
The commercial unit. One kilowatt hour is the energy consumed by an appliance of power kW working for hour. It is the "unit" on an electricity bill.
Why a second unit is needed at all. A single W bulb left on for one hour consumes J. A month's consumption for a household runs into hundreds of millions of joules, and a bill written in such numbers would be unreadable. The kilowatt hour is simply a practical size of unit, and nothing more.
The working formula for consumption in bill units:
Worked example 1 — a single bulb. Find the energy consumed by a W bulb used for hours, in kilowatt hour and in joule.
Worked example 2 — converting the other way. Express J in kilowatt hour.
Worked example 3 — a bulb for one hour in joule. How much energy does a W bulb consume in one hour?
or, in bill units, kWh. Both are correct; the second is what you would pay for.
Worked example 4 — reading a meter. A house meter reads units at the start of a month and units at the end. Find the energy consumed and the bill at ₹ per unit.
And in joule that consumption is J — over four hundred million joules, which is precisely why the meter counts in units.
The distinction that costs the most marks in this chapter. Power and energy are different quantities with different units. The kilowatt is a unit of power; the kilowatt hour is a unit of energy.
- **A kW heater** has a power of kW whether it is switched on for a minute or a week
- The energy it consumes depends on how long it runs, and is measured in kWh
- The bill charges for energy, not for power
Which means two appliances of very different powers can cost the same. A W appliance for one hour and a W appliance for two hours both consume kWh, so both cost the same — and reading the name "kilowatt hour" as a power times a time is the quickest way never to confuse the two again.
The SI unit. One joule** is the energy delivered when one coulomb passes through a potential difference of one volt — or, equally, when a power of one watt acts for one second:
The commercial unit. One kilowatt hour is the energy consumed by an appliance of power kW working for hour. It is the "unit" on an electricity bill.
Why a second unit is needed at all. A single W bulb left on for one hour consumes J. A month's consumption for a household runs into hundreds of millions of joules, and a bill written in such numbers would be unreadable. The kilowatt hour is simply a practical size of unit, and nothing more.
The working formula for consumption in bill units:
Worked example 1 — a single bulb. Find the energy consumed by a W bulb used for hours, in kilowatt hour and in joule.
Worked example 2 — converting the other way. Express J in kilowatt hour.
Worked example 3 — a bulb for one hour in joule. How much energy does a W bulb consume in one hour?
or, in bill units, kWh. Both are correct; the second is what you would pay for.
Worked example 4 — reading a meter. A house meter reads units at the start of a month and units at the end. Find the energy consumed and the bill at ₹ per unit.
And in joule that consumption is J — over four hundred million joules, which is precisely why the meter counts in units.
The distinction that costs the most marks in this chapter. Power and energy are different quantities with different units. The kilowatt is a unit of power; the kilowatt hour is a unit of energy.
- **A kW heater** has a power of kW whether it is switched on for a minute or a week
- The energy it consumes depends on how long it runs, and is measured in kWh
- The bill charges for energy, not for power
Which means two appliances of very different powers can cost the same. A W appliance for one hour and a W appliance for two hours both consume kWh, so both cost the same — and reading the name "kilowatt hour" as a power times a time is the quickest way never to confuse the two again.
How do you work out a month's electricity consumption and its cost?
Multiply each appliance's power by its daily hours, add them for a day's total, multiply by the number of days, and then by the rate.
The four-step layout, which is worth following exactly because each step is markable:
- For each appliance, compute power in watt hours per day, giving watt hour per day
- Add them all to get the daily total, and divide by for kilowatt hour
- Multiply by the number of days in the period
- Multiply by the rate per unit to get the cost
Worked example 1 — a whole house. A house uses bulbs of W each for hours a day, fans of W each for hours a day, and a television of W for hours a day. Find the monthly consumption over days and the cost at ₹ per unit.
The bulbs:
The fans:
The television:
The daily total:
The monthly total:
The cost:
Notice which appliance dominates. The five bulbs together consume more than the fans or the television, not because each bulb is powerful but because there are five of them running for six hours. Consumption is the product of power, number and time, and any of the three can be the largest factor.
Worked example 2 — a single heavy appliance. An electric iron of W is used for hours daily. Find the energy consumed in days and the cost at ₹ per unit.
Compare that with the whole lighting load above. One iron for two hours a day costs about as much as five bulbs running six hours a day — which is the real lesson of this section. A high-power appliance used briefly can outweigh many low-power ones used for long periods.
Worked example 3 — finding the hours from the bill. A kW geyser contributed kWh to a month's bill. For how many hours a day was it used over days?
Worked example 4 — comparing two lamps. A house replaces bulbs of W with light-emitting-diode lamps of W, each used hours a day. Find the monthly saving at ₹ per unit over days.
Before:
After:
The daily saving:
The monthly saving:
Why that saving is possible at all, and it links back to the earlier chapter. A filament lamp produces light by getting hot, so most of the energy it converts leaves as heat. A light-emitting diode produces light without the detour through high temperature, so a far smaller power gives a comparable brightness. **Same , much less needed for the same job.
One condition on all these calculations. They assume every appliance runs at its rated power, which needs the supply voltage to be at its rated value. If the supply voltage falls, the power consumed falls too**, since — which is why appliances run sluggishly during a voltage drop and why a bill can be lower than expected.
The four-step layout, which is worth following exactly because each step is markable:
- For each appliance, compute power in watt hours per day, giving watt hour per day
- Add them all to get the daily total, and divide by for kilowatt hour
- Multiply by the number of days in the period
- Multiply by the rate per unit to get the cost
Worked example 1 — a whole house. A house uses bulbs of W each for hours a day, fans of W each for hours a day, and a television of W for hours a day. Find the monthly consumption over days and the cost at ₹ per unit.
The bulbs:
The fans:
The television:
The daily total:
The monthly total:
The cost:
Notice which appliance dominates. The five bulbs together consume more than the fans or the television, not because each bulb is powerful but because there are five of them running for six hours. Consumption is the product of power, number and time, and any of the three can be the largest factor.
Worked example 2 — a single heavy appliance. An electric iron of W is used for hours daily. Find the energy consumed in days and the cost at ₹ per unit.
Compare that with the whole lighting load above. One iron for two hours a day costs about as much as five bulbs running six hours a day — which is the real lesson of this section. A high-power appliance used briefly can outweigh many low-power ones used for long periods.
Worked example 3 — finding the hours from the bill. A kW geyser contributed kWh to a month's bill. For how many hours a day was it used over days?
Worked example 4 — comparing two lamps. A house replaces bulbs of W with light-emitting-diode lamps of W, each used hours a day. Find the monthly saving at ₹ per unit over days.
Before:
After:
The daily saving:
The monthly saving:
Why that saving is possible at all, and it links back to the earlier chapter. A filament lamp produces light by getting hot, so most of the energy it converts leaves as heat. A light-emitting diode produces light without the detour through high temperature, so a far smaller power gives a comparable brightness. **Same , much less needed for the same job.
One condition on all these calculations. They assume every appliance runs at its rated power, which needs the supply voltage to be at its rated value. If the supply voltage falls, the power consumed falls too**, since — which is why appliances run sluggishly during a voltage drop and why a bill can be lower than expected.
Exam tip
Which steps protect the marks in an electrical energy numerical?
Write the given quantities with their units, decide whether the answer is wanted in joule or in kilowatt hour, and convert the time to match. Almost every lost mark here is a unit error.
- Convert time to seconds for an answer in joule, and to hours for an answer in kilowatt hour
- Pick the power form from what you are given — , or — rather than converting between them first
- Square the current in ; writing loses the whole numerical
- **From a rating, get the resistance as ** and the current as , using the rated values
- Remember that a higher power means a lower resistance at a fixed voltage
- **Learn kWh J** and kW W
- **Divide by to turn watt hours into kilowatt hours
- Check a rating-based answer with the third power form — all three must give the same figure
- Keep the bill in two steps: energy in kWh, then cost at the rate
- Give every answer its unit, and say which quantity it is
The misconception to name. A kilowatt and a kilowatt hour are not the same kind of quantity. The kilowatt measures power — how fast energy is converted; the kilowatt hour measures energy — how much was converted.** The bill charges for energy, so a kW heater costs nothing while it is switched off, and the same kW rating appears on the appliance whether it is used for a minute or a month. Writing that the bill is "in kilowatts" is a marked error.
A second trap. Assuming a higher-power appliance has a higher resistance. At a fixed voltage it has a lower one, so a W bulb at V has ohm while a W bulb has about ohm. The reasoning is , and the intuition that "more powerful means more resisting" is exactly backwards.
- Convert time to seconds for an answer in joule, and to hours for an answer in kilowatt hour
- Pick the power form from what you are given — , or — rather than converting between them first
- Square the current in ; writing loses the whole numerical
- **From a rating, get the resistance as ** and the current as , using the rated values
- Remember that a higher power means a lower resistance at a fixed voltage
- **Learn kWh J** and kW W
- **Divide by to turn watt hours into kilowatt hours
- Check a rating-based answer with the third power form — all three must give the same figure
- Keep the bill in two steps: energy in kWh, then cost at the rate
- Give every answer its unit, and say which quantity it is
The misconception to name. A kilowatt and a kilowatt hour are not the same kind of quantity. The kilowatt measures power — how fast energy is converted; the kilowatt hour measures energy — how much was converted.** The bill charges for energy, so a kW heater costs nothing while it is switched off, and the same kW rating appears on the appliance whether it is used for a minute or a month. Writing that the bill is "in kilowatts" is a marked error.
A second trap. Assuming a higher-power appliance has a higher resistance. At a fixed voltage it has a lower one, so a W bulb at V has ohm while a W bulb has about ohm. The reasoning is , and the intuition that "more powerful means more resisting" is exactly backwards.
Did you know
Why does the meter keep ticking when everything looks switched off?
Switch off every light and fan in a house and watch the meter. In most homes it does not stop. Something is still converting energy, and it is worth knowing what.
Many appliances are not really off when their front switch is off. A device on standby keeps part of its circuit powered, waiting for a remote control or a clock signal — so it continues to draw a small current. Any adapter left plugged in with nothing attached does the same, because its internal circuit is still energised.
Individually these are tiny powers. Collectively they run all day and all night, and it is the product of power and time that the meter records.
Run the arithmetic on a single example. **A device drawing W continuously for a whole month of days consumes**
**which is more than a W bulb used one hour a day for the same month**, at kWh. A small power running constantly beats a large power running briefly, and that is the whole point of the formula .
The same reasoning explains a puzzling feature of bills. A refrigerator has a modest power rating, yet it is often the largest single item on a household bill — because it is the only appliance that runs every hour of every day. Its compressor cycles on and off, so it does not draw its full rating continuously, but its total running time is enormous compared with anything else in the house.
And it explains why a geyser is so expensive to run despite being on for a short time. Its rating is large — often a kilowatt or two — so even an hour a day adds up quickly, as worked example 3 showed.
One more thing the meter is measuring that students often misread. The disc or display counts energy in kilowatt hours, not current and not power. So two houses drawing very different currents can have identical bills if their total energy over the month happens to match. The meter has no opinion about how fast you used the energy — only about how much.
A last connection back to the formulas. All of this comes from with the supply voltage fixed at V for every socket in the house. So the energy each appliance consumes is proportional to its current and to its running time, which is why the two questions worth asking about any appliance are how much current it draws and how long it stays on. Everything on the bill follows from those two numbers.
Many appliances are not really off when their front switch is off. A device on standby keeps part of its circuit powered, waiting for a remote control or a clock signal — so it continues to draw a small current. Any adapter left plugged in with nothing attached does the same, because its internal circuit is still energised.
Individually these are tiny powers. Collectively they run all day and all night, and it is the product of power and time that the meter records.
Run the arithmetic on a single example. **A device drawing W continuously for a whole month of days consumes**
**which is more than a W bulb used one hour a day for the same month**, at kWh. A small power running constantly beats a large power running briefly, and that is the whole point of the formula .
The same reasoning explains a puzzling feature of bills. A refrigerator has a modest power rating, yet it is often the largest single item on a household bill — because it is the only appliance that runs every hour of every day. Its compressor cycles on and off, so it does not draw its full rating continuously, but its total running time is enormous compared with anything else in the house.
And it explains why a geyser is so expensive to run despite being on for a short time. Its rating is large — often a kilowatt or two — so even an hour a day adds up quickly, as worked example 3 showed.
One more thing the meter is measuring that students often misread. The disc or display counts energy in kilowatt hours, not current and not power. So two houses drawing very different currents can have identical bills if their total energy over the month happens to match. The meter has no opinion about how fast you used the energy — only about how much.
A last connection back to the formulas. All of this comes from with the supply voltage fixed at V for every socket in the house. So the energy each appliance consumes is proportional to its current and to its running time, which is why the two questions worth asking about any appliance are how much current it draws and how long it stays on. Everything on the bill follows from those two numbers.
Exam relevance
How does electrical energy and power feed into JEE and NEET?
This is foundation work for Class 12 Current Electricity, examined in JEE Main, JEE Advanced and NEET Physics.
**Where leads. Class 12 keeps it unchanged and uses it to define the power delivered by a cell**, splitting it into the power dissipated in the external resistance and the power wasted in the internal resistance: while . **The condition for maximum power transfer, , is derived from exactly that split, and it is a recurring JEE Main result.
Where the three power forms lead. They become the deciding tool in the classic question of which bulb glows brightest when several rated bulbs are connected in series or in parallel. In series the current is common, so says the largest resistance dissipates most; in parallel the voltage is common, so says the smallest resistance dissipates most. That reversal is the whole point of the question, and it rests entirely on choosing the right form — the skill this chapter builds.
Where the rating idea leads.** Deriving from a marked rating and then combining rated bulbs in a network is a standard competitive item, often with the twist that a bulb run below its rated voltage no longer consumes its rated power. JEE Main sets it as a numerical, and the calculation is the one in this chapter with one extra step.
**Where the dependence leads. It is the reason transmission lines are run at very high voltage, which Class 12 treats quantitatively, and it reappears in Alternating Current** as the root-mean-square current — defined precisely so that gives the average power. The squared dependence you meet here is what makes that definition necessary.
Where the units lead. The kilowatt hour is used in every practical energy calculation, and the joule remains the SI unit throughout. NEET Physics uses power calculations in the questions on electrical instruments and on the heating effect.
Question types to expect. At this level: energy and power numericals, resistance and current from a rating, unit conversions, and household consumption. In competitive papers: brightest-bulb comparisons, maximum power transfer, power in network branches, bulbs operated off their rated voltage, and root-mean-square quantities in alternating-current circuits.
The single trap that costs marks. Forgetting to square the current. **, never — and since the square is what makes the heating rise so steeply, it is also the physically meaningful part. In JEE the same slip appears as taking the average of a current rather than the average of its square when computing average power.
A second trap. Treating the resistance as rising with the power rating. At a fixed voltage it falls**, so the W bulb has more resistance than the W one. In a competitive brightest-bulb question that single inversion reverses the whole answer.
Board versus competitive emphasis. The ICSE paper marks the derivation, the chosen form, the substitution, the unit and the conversion factor; a competitive paper marks a single power, a brightness comparison or an optimal resistance. The transferable habit is asking which quantity is common in the circuit — the current in series, the voltage in parallel — before choosing a power form, because that one question decides which of the three to use and therefore decides the answer.
**Where leads. Class 12 keeps it unchanged and uses it to define the power delivered by a cell**, splitting it into the power dissipated in the external resistance and the power wasted in the internal resistance: while . **The condition for maximum power transfer, , is derived from exactly that split, and it is a recurring JEE Main result.
Where the three power forms lead. They become the deciding tool in the classic question of which bulb glows brightest when several rated bulbs are connected in series or in parallel. In series the current is common, so says the largest resistance dissipates most; in parallel the voltage is common, so says the smallest resistance dissipates most. That reversal is the whole point of the question, and it rests entirely on choosing the right form — the skill this chapter builds.
Where the rating idea leads.** Deriving from a marked rating and then combining rated bulbs in a network is a standard competitive item, often with the twist that a bulb run below its rated voltage no longer consumes its rated power. JEE Main sets it as a numerical, and the calculation is the one in this chapter with one extra step.
**Where the dependence leads. It is the reason transmission lines are run at very high voltage, which Class 12 treats quantitatively, and it reappears in Alternating Current** as the root-mean-square current — defined precisely so that gives the average power. The squared dependence you meet here is what makes that definition necessary.
Where the units lead. The kilowatt hour is used in every practical energy calculation, and the joule remains the SI unit throughout. NEET Physics uses power calculations in the questions on electrical instruments and on the heating effect.
Question types to expect. At this level: energy and power numericals, resistance and current from a rating, unit conversions, and household consumption. In competitive papers: brightest-bulb comparisons, maximum power transfer, power in network branches, bulbs operated off their rated voltage, and root-mean-square quantities in alternating-current circuits.
The single trap that costs marks. Forgetting to square the current. **, never — and since the square is what makes the heating rise so steeply, it is also the physically meaningful part. In JEE the same slip appears as taking the average of a current rather than the average of its square when computing average power.
A second trap. Treating the resistance as rising with the power rating. At a fixed voltage it falls**, so the W bulb has more resistance than the W one. In a competitive brightest-bulb question that single inversion reverses the whole answer.
Board versus competitive emphasis. The ICSE paper marks the derivation, the chosen form, the substitution, the unit and the conversion factor; a competitive paper marks a single power, a brightness comparison or an optimal resistance. The transferable habit is asking which quantity is common in the circuit — the current in series, the voltage in parallel — before choosing a power form, because that one question decides which of the three to use and therefore decides the answer.
Key takeaways
What must you be able to do from this part?
One energy formula in three forms, one power formula in three forms, and one commercial unit.
- ****, from the definition , and with this becomes **
- Substituting Ohm's law** gives and
- Dividing by the time gives , in watt
- Choose the form from what is given: voltage and current, current and resistance, or voltage and resistance
- ** C through V delivers J; A at V for one minute delivers kJ
- A through ohm for s gives kJ**, confirmed by with V
- Doubling the current quadruples the heat — kJ instead of kJ
- From a rating: and , so W at V gives ohm and A
- At a fixed voltage a higher power means a lower resistance — W at V is about ohm against ohm for W
- **A W iron at V** has ohm and draws A; a kW appliance draws A and needs a A fuse
- SI unit the joule; commercial unit the kilowatt hour, with ** kWh J
- Energy in kWh** , so a kW heater for minutes uses kWh, or J
- **A meter reading rising from to ** means kWh, costing ₹ at ₹ a unit
- **Five W bulbs for h, two W fans for h and a W television for h** give kWh a day, kWh a month and ₹
- **A W iron for h a day for days** uses kWh, costing ₹
- **Replacing eight W bulbs with eight W lamps for h a day** saves kWh and ₹ a month
- The kilowatt is power; the kilowatt hour is energy — and the bill charges for energy
- A small power running constantly can beat a large power running briefly, since
The sharpest self-test is your own house. List four appliances with their ratings and the hours you use them, work out the monthly units, and then compare your figure with the last bill — and account for the difference before deciding the bill is wrong.
- ****, from the definition , and with this becomes **
- Substituting Ohm's law** gives and
- Dividing by the time gives , in watt
- Choose the form from what is given: voltage and current, current and resistance, or voltage and resistance
- ** C through V delivers J; A at V for one minute delivers kJ
- A through ohm for s gives kJ**, confirmed by with V
- Doubling the current quadruples the heat — kJ instead of kJ
- From a rating: and , so W at V gives ohm and A
- At a fixed voltage a higher power means a lower resistance — W at V is about ohm against ohm for W
- **A W iron at V** has ohm and draws A; a kW appliance draws A and needs a A fuse
- SI unit the joule; commercial unit the kilowatt hour, with ** kWh J
- Energy in kWh** , so a kW heater for minutes uses kWh, or J
- **A meter reading rising from to ** means kWh, costing ₹ at ₹ a unit
- **Five W bulbs for h, two W fans for h and a W television for h** give kWh a day, kWh a month and ₹
- **A W iron for h a day for days** uses kWh, costing ₹
- **Replacing eight W bulbs with eight W lamps for h a day** saves kWh and ₹ a month
- The kilowatt is power; the kilowatt hour is energy — and the bill charges for energy
- A small power running constantly can beat a large power running briefly, since
The sharpest self-test is your own house. List four appliances with their ratings and the hours you use them, work out the monthly units, and then compare your figure with the last bill — and account for the difference before deciding the bill is wrong.