ScienceCBSE Class 10 · Electricity

Electric Power and Energy for Class 10

Updated 14 October 2026 · Fact-checked

Electric power is the rate at which electrical energy is used in a circuit. It equals voltage times current, and also current squared times resistance or voltage squared divided by resistance. Its unit is the watt. The board asks you to use these forms, find power at a new voltage, and work out energy and cost in kilowatt hours.

Start from P = VI; Ohm's law gives two more forms, and energy is power times time, so 1 kWh equals 3.6 × 10⁶ J.

Illustration drawn for QuestionBuddy.

Understand Electric Power and Energy

Power means how fast work is done, or how fast energy is used. When current flows through a device, the source does work moving charge, so electric power tells you how quickly the device uses electrical energy.

Why is it $P = V I$ ? Moving a charge $Q$ through a potential difference $V$ needs work $V Q$ . If that happens in time $t$ , the power is $\frac{V Q}{t}$ . Since $\frac{Q}{t}$ is the current $I$ , you get $P = V I$ . So power depends on both the push (voltage) and the flow (current).

Ohm's law, $V = I R$ , lets you write power two more ways. Put $V = I R$ into $P = V I$ to get $P = I^{2} R$ . Put $I = \frac{V}{R}$ in to get $P = \frac{V ^{2}}{R}$ . All three give the same answer; choose the one that uses what you know. In series the current is common, so $I^{2} R$ is handy. In parallel, or for an appliance run at a new voltage, the voltage is known, so $\frac{V ^{2}}{R}$ is handy.

The SI unit is the watt (W). A device uses 1 W when 1 A flows through it at 1 V, so $1 W = 1 V A$ . The watt is small, so the kilowatt (1000 W) is often used.

Energy is power multiplied by time. In joules, use watts and seconds. For bills, the commercial unit is the kilowatt hour (kW h), called one 'unit': energy used by 1 kW for 1 hour. Since $1 k W h = 1000 W \times 3600 s$ , it equals $3.6 \times 1 0^{6} J$ .

A rating such as '220 V, 100 W' tells you the power only at 220 V. Resistance actually changes with temperature, but board questions ask you to assume the resistance stays the same. So at another voltage, find R from the rating and use $P = \frac{V ^{2}}{R}$ .

Power also links straight to the heating effect. In a purely resistive device such as a heater, all the electrical energy supplied is turned into heat. So the heat produced in time t is $H = P t = V I t = I^{2} R t$ , and power is simply the rate of producing heat, $P = \frac{H}{t}$ .

Electric Power and Energy: formulas and key facts

Electric power: $P = V I$; From work VQ done in time t, with I = Q/t. P in watt, V in volt, I in ampere.
Power with current and resistance: $P = I^{2} R$; Best when current is known or common (series circuits).
Power with voltage and resistance: $P = \frac{V ^{2}}{R}$; Best when voltage is common (parallel) or for a rated device at a new voltage.
Unit of power: $1 W = 1 V \times 1 A$; 1 kW = 1000 W.
Electrical energy: $E = P \times t = V I t$; Joule if P in W and t in s; kW h if P in kW and t in h.
Commercial unit: $1 k W h = 3.6 \times 1 0^{6} J$; 1 kW h is one 'unit' on the bill. Cost = units × rate per unit.
Resistance from rating: $R = \frac{V ^{2}}{P}$; Use the rated V and P; R stays nearly the same at other voltages.

How to solve Electric Power and Energy questions

This method works for any power, energy or bill question.

1List the given values with units and what is asked.
2Convert units: mA to A, minutes to s or h, W to kW for bills.
3Pick the power form that uses known values: VI, $I^{2} R$ or $V^{2} / R$ .
4For a rated appliance at a new voltage, first find $R = V^{2} / P$ from the rating.
5Substitute with units and find power.
6For energy, multiply power by time; for a bill, kW × h × number of days, then × rate.
7Write the answer with its unit and check it makes sense.

Electric Power and Energy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Electric power and electrical energy

Imagine your family runs a 1.5 kW electric kettle for 20 minutes and a 100 W ceiling fan for 6 hours each day. Suppose the bill charges Rs 6 per unit.

Find the energy used by each appliance in 30 days and the total cost. Which appliance uses more energy?

Try it first, then show the solutionSolution

Kettle per day: $1.5 k W \times \frac{1}{3} h = 0.5 k W h$ .
Kettle in 30 days: $0.5 \times 30 = 15 k W h$ .
Fan per day: $0.1 k W \times 6 h = 0.6 k W h$ .
Fan in 30 days: $0.6 \times 30 = 18 k W h$ .
Total: $33 k W h$ ; cost $= 33 \times 6 =$ Rs 198.

Answer: Kettle 15 kW h, fan 18 kW h, total cost Rs 198. The fan uses more energy despite its lower power.

Memory hook: Energy = power × time: a small power running long can beat a big power running short.

Power with voltage and resistance

Suppose a bulb rated 220 V, 60 W is used in a room where the supply drops to 180 V during the evening.

Find the resistance of the bulb and the power it uses at 180 V, assuming its resistance stays the same. By what percentage has the power fallen?

Try it first, then show the solutionSolution

$R = \frac{V ^{2}}{P} = \frac{220 \times 220}{60} = \frac{48400}{60} \approx 807 Ω$ .
At 180 V: $P = \frac{V ^{2}}{R} = 60 \times (\frac{180}{220})^{2}$ .
$(\frac{180}{220})^{2} = (\frac{9}{11})^{2} = \frac{81}{121} \approx 0.669$ , so $P \approx 60 \times 0.669 \approx 40 W$ .
Fall $= 60 - 40 = 20$ W, which is about $\frac{20}{60} \times 100 \approx 33%$ .

Answer: About 807 Ω; at 180 V the bulb uses about 40 W, roughly a third less than its rating, so it glows dimmer.

Memory hook: Half the voltage, a quarter of the power.

Tricks to remember Electric Power and Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Three faces of power
"VI, I-squared-R, V-squared-over-R": start from VI and swap in Ohm's law.
Use it when: Rebuild any form in the exam if you forget it.
Unit means kilo-hour
Bill unit = kilowatts × hours. Watts? divide by 1000. Minutes? divide by 60.
Use it when: Avoid unit errors in cost questions.
36 with five zeros
1 kW h = 1000 × 3600 = 3,600,000 J.
Use it when: Converting units to joules.

Electric Power and Energy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Current drawn from a rating · 1 mark · MCQBoard level

An appliance rated 1100 W is run on a 220 V supply. The current it draws is:
(a) 0.2 A
(b) 2 A
(c) 5 A
(d) 50 A

Try it first, then show the solutionSolution

$I = \frac{P}{V} = \frac{1100}{220} = 5 A$ (1)

Answer: (c) 5 A

Example 2: Power at a lower voltage · 3 marks · Short answerBoard level

A heater is rated 220 V, 1000 W. (a) Find its resistance. (b) Find the power it uses on 110 V, assuming resistance stays constant. (c) State how much less heat it gives in the same time.

Try it first, then show the solutionSolution

$R = \frac{V ^{2}}{P} = \frac{220 \times 220}{1000} = 48.4 Ω$ (1)
$P^{'} = \frac{11 0 ^{2}}{48.4} = \frac{12100}{48.4} = 250 W$ (1)
Heat in the same time is proportional to power: 250 W is one quarter of 1000 W, so the heat is one quarter, that is three quarters (75%) less, or 750 J less every second (1)

Answer: R = 48.4 Ω; P = 250 W; the heater gives one quarter of the heat, i.e. 75% less (750 J less every second).

Example 3: Which uses more energy? · 3 marks · Short answerBoard level

Which uses more energy: a 400 W television run for 3 hours, or a 1500 W toaster run for 40 minutes? Give the energy of each in kW h and in joules.

Try it first, then show the solutionSolution

TV: $0.4 k W \times 3 h = 1.2 k W h$ (1)
Toaster: $1.5 k W \times \frac{2}{3} h = 1.0 k W h$ (1)
In joules: TV $1.2 \times 3.6 \times 1 0^{6} = 4.32 \times 1 0^{6} J$ , toaster $3.6 \times 1 0^{6} J$ ; the TV uses more (1)

Answer: TV uses more: 1.2 kW h (4.32 × 10⁶ J) against 1.0 kW h (3.6 × 10⁶ J).

Example 4: Assertion-reason: bulb on low voltage · 1 mark · Assertion-reasonBoard level

Assertion (A): A bulb rated 220 V, 100 W uses only 25 W when it is run on a 110 V supply.
Reason (R): For a fixed resistance, the power used is proportional to the square of the voltage.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$R = \frac{22 0 ^{2}}{100} = 484 Ω$ ; at 110 V, $P = \frac{11 0 ^{2}}{484} = 25 W$ , so A is true; since $P = \frac{V ^{2}}{R}$ , R is true and explains A (1)

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 5: Linking power and heat · 3 marks · Short answerBoard level

An electric heater of resistance 55 Ω draws a current of 4 A from the mains. (a) Find the rate at which heat is produced in it. (b) Find the heat produced in 10 minutes. (c) Find the supply voltage and show that $H = V I t$ gives the same heat.

Try it first, then show the solutionSolution

(a) Rate of heating = power: $P = I^{2} R = 4^{2} \times 55 = 880 W$ (1)
(b) $t = 600 s$ ; $H = P t = 880 \times 600 = 5.28 \times 1 0^{5} J$ (1)
(c) $V = I R = 4 \times 55 = 220 V$ ; $H = V I t = 220 \times 4 \times 600 = 5.28 \times 1 0^{5} J$ , the same (1)

Answer: Heat is produced at 880 W (880 J every second); 5.28 × 10⁵ J in 10 minutes; V = 220 V, and VIt gives the same 5.28 × 10⁵ J.

Example 6: Long answer: power in a mixed circuit · 5 marks · Long answerHarder board level

A 12 V battery, a plug key and a 4 Ω resistor are joined in series with a parallel combination of a 6 Ω and a 3 Ω resistor. (In words: from the positive terminal, current passes through the 4 Ω resistor to point A; between A and B the 6 Ω and 3 Ω resistors are connected side by side; from B the wire returns through the key to the negative terminal.)
(a) Find the total resistance of the circuit.
(b) Find the current drawn from the battery.
(c) Find the power used in the 4 Ω resistor.
(d) Find the potential difference across A and B and the power used in the 3 Ω and the 6 Ω resistors.
(e) Show that the total power equals VI for the battery, and find the energy supplied by the battery in 10 minutes.

Try it first, then show the solutionSolution

(a) $\frac{1}{R _{p}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ , so $R_{p} = 2 Ω$ ; total $R = 4 + 2 = 6 Ω$ (1)
(b) $I = \frac{V}{R} = \frac{12}{6} = 2 A$ (1)
(c) $P_{4} = I^{2} R = 2^{2} \times 4 = 16 W$ (1)
(d) $V_{A B} = 2 \times 2 = 4 V$ ; $P_{3} = \frac{4 ^{2}}{3} = \frac{16}{3} \approx 5.33 W$ ; $P_{6} = \frac{4 ^{2}}{6} = \frac{8}{3} \approx 2.67 W$ (1)
(e) Total $= 16 + \frac{16}{3} + \frac{8}{3} = 24 W = V I = 12 \times 2$ ; energy $= 24 \times 600 = 14400 J$ (1)

Answer: R = 6 Ω; I = 2 A; 16 W in the 4 Ω resistor; V_AB = 4 V, about 5.33 W in 3 Ω and 2.67 W in 6 Ω; total 24 W = 12 V × 2 A; energy in 10 minutes = 14400 J (1.44 × 10⁴ J).

Example 7: Rated bulbs in parallel and in series · 3 marks · Short answerHarder board level

Bulb P is rated 220 V, 200 W and bulb Q is rated 220 V, 50 W. Assume their resistances stay constant. (a) Find the resistance of each. (b) Both are joined in parallel across 220 V. Find the power of each and say which glows brighter. (c) Both are joined in series across 220 V. Find the power of each and say which glows brighter.

Try it first, then show the solutionSolution

(a) $R_{P} = \frac{22 0 ^{2}}{200} = 242 Ω$ ; $R_{Q} = \frac{22 0 ^{2}}{50} = 968 Ω$ (1)
(b) In parallel each gets 220 V, so P uses 200 W and Q uses 50 W; P glows brighter (1)
(c) Series: $R = 1210 Ω$ , $I = \frac{220}{1210} = \frac{2}{11} A$ ; $P_{P} = \frac{4}{121} \times 242 = 8 W$ , $P_{Q} = \frac{4}{121} \times 968 = 32 W$ ; Q glows brighter (1)

Answer: 242 Ω and 968 Ω. Parallel: 200 W and 50 W, P brighter. Series: 8 W and 32 W, Q brighter, because the same current flows and $I^{2} R$ is larger for the larger resistance.

Example 8: Case: the hostel room · 4 marks · Case-basedHOTS / competency

Read and answer. Imagine a hostel room on 220 V has a 100 W lamp and an 800 W heater in parallel, and the circuit has a 5 A fuse.
(a) Find the current through each appliance. (1)
(b) Will the fuse blow when both run? Justify. (1)
(c) Find the energy used if both run for 2 hours, in kW h. (2)
OR
(c) A student adds a second 800 W heater. Find the new total current and say what happens. (2)

Try it first, then show the solutionSolution

(a) Lamp $\frac{100}{220} = \frac{5}{11}$ A ≈ 0.45 A; heater $\frac{800}{220} = \frac{40}{11}$ A ≈ 3.64 A (1)
(b) Total = $\frac{900}{220}$ ≈ 4.09 A, clearly less than 5 A, so the fuse does not blow (1)
(c) Total power 900 W = 0.9 kW (1); energy $0.9 \times 2 = 1.8 k W h$ (1)
OR (c) Total power 1700 W, current $\frac{1700}{220}$ ≈ 7.73 A (1); this is larger than 5 A, so the fuse wire heats up, melts and breaks the circuit (1)

Answer: Lamp ≈ 0.45 A, heater ≈ 3.64 A; total ≈ 4.09 A, so the fuse holds; 1.8 kW h. With a second heater about 7.73 A flows and the fuse blows.

Where students lose marks in Electric Power and Energy

The slips examiners see most often, and what to write instead.

Mistake: Using watts and hours together to get kW h
Forgetting to divide by 1000.
Wrong working
400 W × 8 h = 3200 units
Right working
0.4 kW × 8 h = 3.2 kW h
Do this: Convert W to kW before multiplying by hours.
Usually costs: Final answer mark.
Mistake: Using rated power at a new voltage
Thinking power stays fixed.
Wrong working
100 W bulb on 110 V uses 100 W
Right working
R = 484 Ω; P = 110²/484 = 25 W
Do this: Power changes; resistance stays. Find R first.
Usually costs: 1-2 marks.
Mistake: Minutes left as minutes
Not converting time to hours or seconds.
Wrong working
1.5 kW × 20 min = 30 kW h
Right working
1.5 kW × (1/3) h = 0.5 kW h
Do this: Write time in hours for kW h, seconds for joules.
Usually costs: Substitution and answer marks.
Mistake: Writing IR² for power
Mixing up the square.
Wrong working
P = IR²
Right working
P = I²R
Do this: Derive from P = VI with V = IR each time.
Usually costs: Formula mark.

Exam tips

Always write the formula first, then substitution with units; formula and substitution carry marks even if arithmetic slips.
For any '220 V, P W' rating question, find R from the rating first.
State which form of P you use and why (common current or common voltage) in long answers.
In bill questions show daily units, monthly units and cost as separate lines.
Know 1 kW h = 3.6 × 10⁶ J and be able to show 1000 × 3600.

Practice questions on Electric Power and Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerHarder board level
A bulb draws 0.25 A from a 220 V supply. (a) Find its power and resistance. (b) It is used 4 hours a day for 30 days; find the units used. (c) If the supply falls to 200 V, find the power it uses, assuming its resistance stays the same.
Hint
Find R first; it is the quantity you keep fixed when the voltage changes.
Answer
(a) $P = V I = 220 \times 0.25 = 55$ W; $R = \frac{220}{0.25} = 880 Ω$ (1 mark). (b) $0.055 k W \times 4 \times 30 = 6.6$ kW h (1 mark). (c) $P = \frac{20 0 ^{2}}{880} = \frac{40000}{880} \approx 45.5$ W (1 mark).
Q2 · 3 marks · Short answerBoard level
An electric motor takes 4 A from a 220 V line. Find its power and the energy it uses in 3 hours, in kW h and in joules.
Hint
Find P in kW first, then multiply by hours.
Answer
P = 880 W = 0.88 kW; E = 0.88 × 3 = 2.64 kW h = 2.64 × 3.6 × 10⁶ = 9.504 × 10⁶ J.
Q3 · 3 marks · Short answerBoard level
Two 24 Ω coils are used on 220 V. Find the power drawn when (a) one coil is used, (b) both are in series, (c) both are in parallel. Which setting heats fastest?
Hint
Use P = V²/R with the equivalent resistance each time.
Answer
(a) 48400/24 ≈ 2016.7 W; (b) R = 48 Ω, ≈ 1008.3 W; (c) R = 12 Ω, ≈ 4033.3 W. Parallel heats fastest.
Q4 · 3 marks · Short answerBoard level
A home uses a 2 kW air heater for 3 h, four 50 W bulbs for 5 h and a 500 W pump for 1 h daily. Find the units used in 30 days and the cost at Rs 5 per unit.
Hint
Find daily kW h for each, add, then multiply by 30.
Answer
Heater 6, bulbs 0.2 × 5 = 1, pump 0.5; daily 7.5 kW h; 30 days = 225 kW h; cost Rs 1125.
Q5 · 3 marks · Short answerHOTS / competency
Bulbs A (220 V, 100 W) and B (220 V, 25 W) are joined in series to 220 V. A student says A glows brighter because it has higher power. Find the error and the correct answer.
Hint
Find each resistance, then compare $I^{2} R$ with the same current.
Answer
R_A = 484 Ω, R_B = 1936 Ω. In series the current is the same, so P = I²R is larger for B; B glows brighter. The rating applies only when each runs alone at 220 V.
Q6 · 3 marks · Short answerHarder board level
A 3 Ω resistor is joined in series with a parallel combination of two 6 Ω resistors, across a 12 V battery. Find the current from the battery, the power used in each resistor, and show that the total equals VI.
Hint
Two equal 6 Ω resistors in parallel give 3 Ω; the main current then splits equally.
Answer
Parallel part = 3 Ω, total = 6 Ω, I = 12/6 = 2 A. Power in 3 Ω = 2² × 3 = 12 W. P.d. across the parallel part = 2 × 3 = 6 V, so each 6 Ω carries 1 A and uses 6² /6 = 6 W. Total = 12 + 6 + 6 = 24 W = 12 V × 2 A.
Q7 · 3 marks · Short answerHarder board level
A heater coil is rated 220 V, 1100 W. It is cut into two equal halves and only one half is used on 220 V. Find the new resistance, power and current. Is it safe to use this on a circuit with a 5 A fuse? Give a reason.
Hint
Find R from the rating; half the length means half the resistance.
Answer
Full coil R = 220²/1100 = 44 Ω; half coil = 22 Ω. P = 220²/22 = 2200 W; I = 220/22 = 10 A. Not safe: 10 A is larger than 5 A, so the fuse wire melts and breaks the circuit.
Q8 · 3 marks · Short answerHarder board level
An electric kettle is rated 220 V, 1100 W. One day the supply is only 200 V. Assuming its resistance stays the same, find (a) its resistance, (b) its power and current at 200 V, and (c) how long it now takes to supply the $3.3 \times 1 0^{5}$ J that it supplies in 5 minutes at 220 V. Is a 5 A fuse still enough?
Hint
Keep R fixed; heat needed is the same, so time = heat ÷ new power.
Answer
(a) $R = \frac{22 0 ^{2}}{1100} = 44 Ω$ (1 mark). (b) $P = \frac{20 0 ^{2}}{44} \approx 909$ W; $I = \frac{200}{44} \approx 4.55$ A (1 mark). (c) $t = \frac{3.3 \times 1 0 ^{5}}{909} \approx 363$ s, about 6 minutes. The current 4.55 A is below 5 A, so the fuse is enough (at 220 V it draws exactly 5 A) (1 mark).
Q9 · 3 marks · Short answerHarder board level
Two identical bulbs, each rated 220 V, 100 W, are used 5 hours a day for 30 days on a 220 V supply. Compare the monthly bill at Rs 5 per unit when they are joined (a) in parallel and (b) in series. Find the power of each bulb in each case and say which arrangement gives more light.
Hint
Each bulb has $R = \frac{22 0 ^{2}}{100} = 484 Ω$ . In series each gets half the voltage.
Answer
(a) Parallel: each bulb gets 220 V and uses 100 W; total 200 W; energy $= 0.2 \times 5 \times 30 = 30$ kW h; bill Rs 150 (1 mark). (b) Series: total $R = 968 Ω$ , each bulb has 110 V, so each uses $\frac{11 0 ^{2}}{484} = 25$ W; total 50 W; energy $= 0.05 \times 150 = 7.5$ kW h; bill Rs 37.50 (1 mark). The parallel arrangement gives much more light (each bulb at full 100 W) but costs four times as much (1 mark).
Q10 · 4 marks · Case-basedHOTS / competency
Case: Suppose a room on a 220 V supply has a 5 A fuse. A 750 W electric iron is already running, and you want to switch on some 100 W bulbs in parallel with it.
(a) What is the largest total power the circuit can carry without the current going above 5 A? (1)
(b) Find the current drawn by the iron alone. (1)
(c) Find the largest number of 100 W bulbs you can switch on with the iron without blowing the fuse. Justify with the current in each case. (2)
Hint
Maximum power = V × fuse current; then check the current with n and n + 1 bulbs.
Answer
(a) 220 × 5 = 1100 W. (b) 750/220 ≈ 3.41 A. (c) Power left = 350 W, so at most 3 bulbs: with 3 bulbs total = 1050 W, I = 1050/220 ≈ 4.77 A, below 5 A; with 4 bulbs total = 1150 W, I ≈ 5.23 A, above 5 A, so the fuse would blow. Answer: 3 bulbs.
Q11 · 4 marks · Case-basedHOTS / competency
Case study: A model railway lamp is rated 6 V, 3 W. Only a 12 V battery is available, so a student joins a resistor in series with the lamp so that the lamp gets exactly its rated voltage.
(i) Find the resistance of the lamp and the current it needs. (1 mark)
(ii) Find the resistance of the series resistor. (1 mark)
(iii) Find the power wasted in the resistor and the total power drawn from the battery. What fraction of the battery's power reaches the lamp? (2 marks)
Hint
Use the rating to get R and I for the lamp; the resistor must take the other 6 V.
Answer
(i) $R_{l am p} = \frac{6 ^{2}}{3} = 12 Ω$ ; $I = \frac{3}{6} = 0.5$ A. (1)
(ii) The resistor takes $12 - 6 = 6$ V with 0.5 A through it, so $R = \frac{6}{0.5} = 12 Ω$ . (1)
(iii) Power in the resistor $= 6 \times 0.5 = 3$ W; battery power $= 12 \times 0.5 = 6$ W. Only half the battery's power reaches the lamp; the other half is wasted as heat in the resistor. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define electric power.Show answer
The rate at which electrical energy is used or dissipated in a circuit.
2.Three forms of power?Show answer
$P = V I = I^{2} R = V^{2} / R$ .
3.Define 1 watt.Show answer
Power used by a device carrying 1 A at a p.d. of 1 V.
4.Commercial unit of energy?Show answer
Kilowatt hour (kW h), called a 'unit'.
5.1 kW h in joules?Show answer
$3.6 \times 1 0^{6} J$ .
6.Energy in kW h used by a 60 W bulb running for 5 h?Show answer
$0.06 k W \times 5 h = 0.3 k W h$ .

Electric Power and Energy: frequently asked questions

Which power formula should I use in a numerical?

Use the one that needs only the quantities you know. Use P = I squared R when current is common, as in series, and P = V squared over R when voltage is common, as in parallel or for a rated device at a new voltage.

Is kilowatt hour a unit of power?

No. The kilowatt hour is a unit of energy, since it is power multiplied by time. The kilowatt is the unit of power.

Why does a 100 W bulb give only 25 W on 110 V?

Board questions assume its resistance stays the same, so power varies with the square of the voltage. Halving the voltage then makes the power one quarter. (A real filament is cooler at the lower voltage, so its resistance is a little lower, but you ignore this in exam questions.)