CBSE
Class 10 Electricity Numericals With Full Solutions
Class 10 electricity numericals come down to a short toolkit: V = IR, series resistances add, parallel resistances combine by reciprocals, heat H = I^2Rt, power P = VI = V^2/R, and 1 unit = 1 kWh for energy costs. Below are six numericals solved in the CBSE layout of given, formula, substitution, answer.
Electricity numericals are the most predictable marks in the Class 10 science paper: the same handful of formulas gets dressed up in different household settings. This page works through six numericals covering Ohm's law, series and parallel combinations, the heating effect of current, power ratings, and the cost of electrical energy in units. Each one is set out the way CBSE marking schemes reward: state what is given, write the formula, substitute with units, then give the final answer with its unit. The unit conversion traps are flagged as they appear, minutes to seconds and watts to kilowatts, because that is where most of the lost marks actually go.
The complete formula kit for electricity numericals
Six relationships cover every numerical in this chapter. Ohm's law: V = IR. Series circuits: resistances simply add, and the current is the same everywhere. Parallel circuits: 1/Rp = 1/R1 + 1/R2, and the answer is always smaller than the smallest branch. Heating: H = I^2Rt joules, with time in seconds. Power: P = VI, with the variants P = I^2R and P = V^2/R derived by substituting Ohm's law. Energy billing: units on the meter are kilowatt-hours, energy in kWh = power in kW × time in hours, and cost = units × tariff. Notice the two different worlds here: heat in joules lives in SI seconds, while billing lives in kilowatts and hours. Numericals mix the worlds deliberately, and choosing the right world before substituting is most of the skill.
Unit conversions are where the marks actually go
Almost every lost mark in electricity numericals is a unit slip rather than a physics misunderstanding. The big four: minutes must become seconds before H = I^2Rt (5 minutes is 300 s); watts must become kilowatts before counting units (400 W is 0.4 kW); commercial units are kilowatt-hours, and one unit equals 3.6 × 10^6 J if a question bridges the two systems; and milliamperes must become amperes before Ohm's law. Build the habit of converting in a labelled line of its own, t = 5 × 60 = 300 s, before touching the formula. It is also worth writing the unit next to every substituted number, because that makes a mismatched unit look wrong on the page while there is still time to fix it.
How a 3-mark numerical is actually marked
CBSE science marking schemes for numericals usually split three ways: the formula, the substitution, and the answer with its unit. That split has practical consequences. Writing H = I^2Rt earns its mark even before any numbers appear, so never skip straight to arithmetic. Substituting with visible values, H = 4 × 20 × 300, earns the second even if a multiplication then goes wrong. The final mark needs both the number and the unit, and 24000 on its own does not get it; 24,000 J does. This is why the solved numericals above all follow the same skeleton of given, formula, substitution, answer. It looks slightly ceremonial, but the ceremony is literally where the marks are attached, and it takes maybe twenty extra seconds per question.
Worked questions, step by step
An electric iron draws a current of 4 A when plugged into a 220 V household supply. Calculate the resistance of its heating element.
- Given: V = 220 V and I = 4 A. To find: the resistance R.
- Formula: Ohm's law, V = IR, rearranged to R = V/I.
- R = 220/4 = 55.
- Answer: the resistance of the element is 55 ohm.
Answer: R = 55 ohm
Where marks slip: A surprising number of scripts compute I/V here instead of V/I; picture the V-I-R triangle with V on top and I, R below, and the rearrangement comes out right every time.
Try one yourself: A filament bulb draws 0.5 A from a 220 V supply. Find its resistance. (Answer: 440 ohm)
Resistors of 5 ohm, 10 ohm and 15 ohm are joined in series across a 6 V battery. Find the total resistance of the circuit and the current flowing through it.
- Given: R1 = 5 ohm, R2 = 10 ohm and R3 = 15 ohm in series, with V = 6 V. To find: total resistance and current.
- Formula: in series, total resistance Rs = R1 + R2 + R3.
- Rs = 5 + 10 + 15 = 30 ohm.
- Current from Ohm's law: I = V/Rs = 6/30 = 0.2 A.
- The same 0.2 A flows through every resistor, because current is common in a series circuit.
- Answer: total resistance 30 ohm and current 0.2 A.
Answer: Total resistance = 30 ohm; current = 0.2 A
Where marks slip: If a follow-up asks for the potential difference across one resistor, use V = IR with that resistor alone, for example 0.2 × 10 = 2 V; applying the full 6 V to a single resistor is the standard giveaway error.
Try one yourself: Resistors of 2 ohm, 3 ohm and 7 ohm are connected in series with a 6 V battery. Find the current. (Answer: 0.5 A)
A 6 ohm resistor and a 3 ohm resistor are connected in parallel to a 4 V battery. Work out the equivalent resistance of the pair and the total current drawn from the battery.
- Given: R1 = 6 ohm and R2 = 3 ohm in parallel, with V = 4 V. To find: equivalent resistance and total current.
- Formula: 1/Rp = 1/R1 + 1/R2.
- 1/Rp = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so Rp = 2 ohm.
- Total current: I = V/Rp = 4/2 = 2 A.
- Notice Rp = 2 ohm is smaller than either resistor, which is always true for a parallel pair.
- Answer: equivalent resistance 2 ohm and total current 2 A.
Answer: Equivalent resistance = 2 ohm; total current = 2 A
Where marks slip: The classic mistake is stopping at 1/Rp = 1/2 and writing the resistance as 1/2 ohm; you must flip the reciprocal at the end, and the quick sanity check is that a parallel answer is always below the smallest branch.
Try one yourself: A 4 ohm and a 12 ohm resistor are in parallel across a 6 V battery. Find the equivalent resistance and the total current. (Answer: 3 ohm and 2 A)
A current of 2 A passes through a 20 ohm resistor for 5 minutes. How much heat is produced in the resistor?
- Given: I = 2 A, R = 20 ohm and t = 5 minutes. To find: the heat produced H.
- Convert time to SI units first: t = 5 × 60 = 300 s.
- Formula: Joule's law of heating, H = I^2Rt.
- H = (2)^2 × 20 × 300 = 4 × 20 × 300.
- H = 24000 J = 24 kJ.
- Answer: the heat produced is 24,000 J, that is 24 kJ.
Answer: H = 24,000 J (24 kJ)
Where marks slip: Leaving the time in minutes is the single biggest mark-loser on heating numericals; convert to seconds before substituting, and squaring only the current, not IR, is the other trap.
Try one yourself: A 3 A current flows through a 10 ohm resistor for 2 minutes. Find the heat produced. (Answer: 10,800 J)
An electric kettle is rated 2200 W at 220 V. Find the current it draws in normal use and the resistance of its heating element.
- Given: P = 2200 W and V = 220 V. To find: the current I and the resistance R.
- Formula: P = VI, so I = P/V.
- I = 2200/220 = 10 A.
- Resistance from Ohm's law: R = V/I = 220/10 = 22 ohm, or directly R = V^2/P = 48400/2200 = 22 ohm.
- Answer: the kettle draws 10 A and its element has a resistance of 22 ohm.
Answer: I = 10 A; R = 22 ohm
Where marks slip: The rating plate gives P and V, never I, so start from P = VI; students who begin with V = IR and invent a current get no method marks.
Try one yourself: A room heater is rated 1100 W at 220 V. Find the current it draws and the element's resistance. (Answer: 5 A and 44 ohm)
A refrigerator rated at 400 W runs for 8 hours every day. If electricity costs Rs 6 per unit, what is the cost of running it for a 30-day month?
- Given: P = 400 W = 0.4 kW, running 8 hours daily for 30 days, tariff Rs 6 per unit. To find: the monthly cost.
- Formula: energy in units (kWh) = power in kW × time in hours.
- Energy per day = 0.4 × 8 = 3.2 kWh.
- Energy for the month = 3.2 × 30 = 96 kWh, which is 96 units.
- Cost = 96 × 6 = Rs 576.
- Answer: running the refrigerator costs Rs 576 for the month.
Answer: Monthly cost = Rs 576
Where marks slip: The whole question hinges on converting 400 W to 0.4 kW before multiplying; if your unit count comes out in the thousands you have forgotten the kilo, so pause and check the order of magnitude.
Try one yourself: A 100 W television runs 6 hours a day for 30 days in a home where a unit costs Rs 5. Find the monthly cost. (Answer: Rs 90)
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Solve my question →Quiz me on this topicQuestions students ask
What is 1 unit of electricity?
One unit is one kilowatt-hour, the energy a 1 kW appliance uses in one hour, and it is what domestic meters count. In SI terms 1 kWh = 3.6 × 10^6 J. For billing questions, convert power to kilowatts, multiply by hours to get units, then multiply by the tariff to get the cost in rupees.
Which formulas cover all Class 10 electricity numericals?
Six: V = IR; series resistance Rs = R1 + R2 + R3; parallel resistance 1/Rp = 1/R1 + 1/R2; heat H = I^2Rt with time in seconds; power P = VI along with the derived forms I^2R and V^2/R; and energy in kWh = kW × hours for cost questions. Every numerical on this page uses only these.
Will I lose marks for missing units?
Usually yes: the final mark in a numerical is generally tied to the answer with its correct unit, so 55 alone earns less than 55 ohm. Get in the habit of carrying units through the substitution line as well, because a stray minute or watt spotted there can be fixed before it corrupts the final answer.
Why does a parallel combination have less resistance than either resistor?
Adding a branch gives the current an extra path, so more total current flows for the same voltage, which by R = V/I means a lower effective resistance. This is also your best sanity check: if your parallel answer comes out bigger than the smallest branch resistance, the reciprocal step went wrong, most often by forgetting to flip 1/Rp at the end.