Class 10 Electricity Numericals: Ohm’s Law, Resistance, Power and Worked Examples

Class 10 Electricity numericals become easier when students know which quantity is given, which formula applies and how to check the final unit. This detailed CBSE-aligned practice guide explains current, potential difference, Ohm’s law, resistance in series and parallel, power, energy and heating effects using clear worked examples. It is meant for learners and parents seeking a practical revision resource, not predictions of examination questions.

Focus keyword: Class 10 Electricity numericals. Use this guide with the school’s current prescribed textbook and teacher instructions. For related chapters, start with the Classes 9–10 Science guide. Questions are educational illustrations; never experiment with household mains electricity.

Why Class 10 Electricity numericals feel difficult

Class 10 Electricity numericals often feel harder than the underlying science. A question might mention a lamp, a resistor, a battery and several numbers. Students then rush to choose an equation without identifying what the question actually asks. The better approach is to translate the words into known quantities, draw a small sketch when a circuit is involved, choose a formula, substitute values with units and inspect the answer.

For example, “a potential difference of 12 volts drives a current of 3 amperes” gives V = 12 V and I = 3 A. If the question asks for resistance, the required relationship is R = V/I. Nothing about the wording requires a complicated calculation. This guide builds that habit through fully worked examples, not a list of formulas to memorise blindly.

Students should also practise explaining what the numerical result means: a resistance of 4 ohms describes opposition to current, while a power of 24 watts describes the rate of energy transfer. If a student’s basic multiplication and division need practice, short, regular sessions are more effective than solving twenty questions in a hurry. The Classes 9–10 Maths and Science study guide offers a broader revision routine.

Start with a quantities-and-units checklist

Before solving any question, write a small “given / find” table. Useful symbols include electric current I in amperes (A), charge Q in coulombs (C), time t in seconds (s), potential difference V in volts (V), resistance R in ohms (Ω), power P in watts (W), and energy E in joules (J). Those labels prevent confusing a measured quantity with the unit used to express it.

Conversion is part of the solution, not a small detail to skip. Divide milliamperes by 1,000 to obtain amperes: 500 mA = 0.5 A. Convert minutes into seconds when a formula expects time in seconds: 3 minutes = 180 seconds. Convert kilowatts to watts by multiplying by 1,000: 1.5 kW = 1,500 W. When working with the household electricity unit, 1 kilowatt-hour represents 3.6 million joules.

Try this warm-up: a device draws 250 mA for 2 minutes. The useful entries are I = 0.25 A and t = 120 s. Once these are written correctly, the student can find charge using Q = It. Keep a margin in the notebook for units and circle the requested quantity; this is a practical habit for school tests.

Electric current and charge: Q = It

Electric current tells us how much electric charge passes a point each second. The relationship is I = Q/t, or equivalently Q = It. A current of 1 ampere corresponds to 1 coulomb of charge passing a point in 1 second. A student does not need to imagine individual electrons to solve the basic exercise; identifying the rate and the duration is enough.

Worked example: A wire carries a steady current of 0.4 A for 5 minutes. Find the charge that passes through the wire. First convert time: t = 5 × 60 = 300 s. Write Q = It. Substitute: Q = 0.4 × 300 = 120 C. The answer is charge, so the final unit is coulombs, not amperes.

Reverse example: If 90 C passes a point in 30 s, then I = Q/t = 90/30 = 3 A. Notice that a given charge and a given time do not require resistance or Ohm’s law. Ask the student to explain why the answer should become larger when the same charge passes in less time: because the rate of charge flow increases. This verbal check builds conceptual understanding alongside arithmetic.

Potential difference and electrical work

Potential difference is defined in terms of work done per unit charge: V = W/Q. Here W means work or energy transferred, measured in joules, and Q is the charge in coulombs. Because the letter W also represents the unit watt in other questions, write a short label beside each symbol. In a definition, 1 volt equals 1 joule per coulomb.

Worked example: A source does 48 J of work in moving 8 C of charge between two points. Potential difference V = W/Q = 48/8 = 6 V. Now reverse the relationship. If 5 C of charge moves through a potential difference of 12 V, then W = VQ = 12 × 5 = 60 J.

A frequent error is to use I = Q/t when the question contains neither time nor current. Encourage students to underline what is given and what is asked before selecting a formula. In a real circuit, the potential difference across a device can be measured by placing a voltmeter in parallel with that device. The associated circuit rule becomes easier to remember once the meaning of voltage is clear.

Ohm’s law numericals: V = IR

Ohm’s law states that, for an ohmic conductor under appropriate constant physical conditions such as temperature, current is directly proportional to the potential difference across it. The working equation is V = IR. Its two rearrangements are I = V/R and R = V/I. The formula triangle can help memory, but it cannot replace knowing what each quantity means.

Example 1: The potential difference across a resistor is 18 V and its resistance is 6 Ω. Find current. I = V/R = 18/6 = 3 A. Example 2: A 0.75 A current flows through a 16 Ω resistor. Find voltage. V = IR = 0.75 × 16 = 12 V.

Example 3: A current of 200 mA flows when the applied voltage is 10 V. Convert 200 mA to 0.2 A. Then R = V/I = 10/0.2 = 50 Ω. This last question is where many students lose marks because they substitute 200 instead of 0.2. After solving, perform a quick check: 0.2 A × 50 Ω does indeed equal 10 V. That substitution back into the original equation verifies the answer without learning another formula.

How to read a current–voltage graph

A graph question tests whether the learner understands the relationship, rather than simply remembers three symbols. First read the labels on both axes. If voltage V is on the vertical axis and current I on the horizontal axis, the slope ΔV/ΔI gives resistance for an ohmic conductor. When the axes are reversed, the slope ΔI/ΔV gives the reciprocal of resistance; it would be wrong to call that slope resistance.

Example: Suppose a straight line passes through (I = 0.5 A, V = 4 V) and (I = 1.5 A, V = 12 V). Compute the change in voltage, 12 − 4 = 8 V, and the change in current, 1.5 − 0.5 = 1 A. Resistance R = ΔV/ΔI = 8 Ω. Using either point also gives V/I = 8 Ω because the line passes through the origin.

If the graph is curved, the V/I ratio may vary and a single constant resistance should not be assumed across all conditions. Students should avoid saying “the steeper line always has more resistance” without stating which quantity is on each axis. Practise drawing simple, clearly labelled graphs by hand, and write the unit beside the calculated slope. A correct method is more valuable than a graph copied without understanding.

Resistance, length and wire thickness

For a uniform wire of a particular material at a fixed temperature, resistance can be expressed as R = ρL/A, where ρ is resistivity, L is the wire’s length and A is its cross-sectional area. This explains why a longer wire offers more resistance when other conditions stay the same, while a thicker wire has less resistance. Students should distinguish resistance (which also depends on a wire’s dimensions) from resistivity (a material property at a given temperature).

Comparison example: A wire has resistance 5 Ω. A second wire is made of the same material, has the same cross-sectional area and is twice as long. R is proportional to L, so its resistance is 10 Ω. If instead the original wire keeps its length but its area becomes twice as large, the resistance becomes 2.5 Ω.

Do not assume that doubling the wire’s diameter doubles the cross-sectional area. A circular wire’s area depends on the square of its radius, so doubling the diameter makes the area four times as large and, at constant length and material, reduces resistance to one quarter. For instance, 8 Ω would become 2 Ω. These are comparison exercises; if numbers are given for resistivity, convert lengths and areas into consistent SI units first.

Resistors in series: add resistance correctly

In a series combination, resistors lie on a single path, so the same current passes through each resistor. The equivalent resistance is R_s = R₁ + R₂ + …. The total potential difference equals the sum of the voltage drops across individual resistors. This makes the first step of a series numerical easy: add the resistor values, then use Ohm’s law on the whole circuit.

Worked example: A 2 Ω resistor and a 4 Ω resistor are in series across a 12 V ideal supply. Equivalent resistance R_s = 2 + 4 = 6 Ω. Current I = V/R_s = 12/6 = 2 A. The voltage across 2 Ω is 2 × 2 = 4 V; across 4 Ω it is 2 × 4 = 8 V. Check: 4 + 8 = 12 V, matching the supply.

Common mistakes are using a different current in each component or adding the voltage drop to the battery voltage. Draw two rectangles for resistors in one loop, place a single arrow for current and label both resistances. That small sketch helps distinguish a series connection from a parallel branch and reduces errors when three or more resistors appear in the question.

Resistors in parallel: use reciprocal resistance

Resistors in parallel share the same two connection points, so they have the same potential difference across them. The current can split between branches. For two resistors, 1/R_p = 1/R₁ + 1/R₂, or R_p = R₁R₂/(R₁ + R₂). The equivalent resistance is less than the smallest individual branch resistance for ordinary positive resistor values.

Worked example: Connect a 6 Ω resistor and a 3 Ω resistor in parallel across 12 V. R_p = (6 × 3)/(6 + 3) = 18/9 = 2 Ω. The total current is 12/2 = 6 A. In the 6 Ω branch, I = 12/6 = 2 A; in the 3 Ω branch, I = 12/3 = 4 A. Check: 2 + 4 = 6 A.

Students sometimes add 6 and 3 to get 9 Ω, which would describe series resistance instead. The easiest diagnostic is to draw the junction where current splits. Each branch receives the same 12 V in this idealised example, not half the source voltage. A clear diagram followed by a sanity check will usually expose the wrong calculation before the final answer is written.

Electrical power: P = VI, I²R and V²/R

Power is the rate of transfer of energy. In an electrical circuit the useful relationships include P = VI, P = I²R and P = V²/R, for the relevant circuit conditions. Power is measured in watts. Choose the form that uses the quantities provided rather than inventing extra steps.

Worked example: A device operates at 12 V and takes a current of 2 A. Its power P = VI = 12 × 2 = 24 W. If a 4 Ω resistor carries 3 A, P = I²R = 3² × 4 = 36 W. If a 20 Ω resistor has 10 V across it, P = V²/R = 100/20 = 5 W.

Do not confuse energy and power. A 100 W device transfers energy at a rate of 100 joules per second while operating at that power; it does not automatically use 100 joules in an hour. Another frequent error is forgetting to square the current or voltage in the relevant formulas. Writing 3² as 3 × 3 first is a safe habit. Students can use these examples to verify their answers by calculating current or voltage through a second route.

Electrical energy and the kilowatt-hour

Electrical energy depends on power and the duration of use: E = Pt. When P is in watts and t is in seconds, energy is obtained in joules. For electricity-consumption examples, power in kilowatts multiplied by time in hours produces kilowatt-hours, abbreviated kWh. One kWh is an energy unit, not a unit of power, and equals 3.6 × 10⁶ joules.

Worked example: A 1,200 W heater operates for 2.5 hours. Convert 1,200 W into 1.2 kW. Energy used = 1.2 × 2.5 = 3 kWh. In joules this is 3 × 3.6 million = 10.8 million J. If an exercise supplies an illustrative tariff of ₹6 per kWh, the energy-only cost would be 3 × 6 = ₹18. This is a classroom calculation, not a quotation of any local electricity tariff or complete real-world bill.

Compare two appliances: a 100 W lamp used for 5 hours uses 0.5 kWh, while a 500 W device used for 1 hour also uses 0.5 kWh. The appliance with higher power does not always use more energy unless the operating duration is also considered. Write the conversion step visibly so the teacher can see why the final number is reasonable.

Joule heating: why wires get warm

When current passes through a resistor, electrical energy can turn into thermal energy. For a resistor carrying steady current under the relevant conditions, the Joule heating relation is H = I²Rt. Here H is heat produced in joules, I is the current in amperes, R is resistance in ohms and t is time in seconds. The squared current matters: doubling current while holding R and t constant makes calculated heating four times larger.

Worked example: Find the heat generated by a 5 Ω resistor carrying 2 A for 3 minutes. Convert the time into seconds: t = 180 s. Substitute: H = 2² × 5 × 180 = 4 × 5 × 180 = 3,600 J. You can check the rate of heating through power P = I²R = 20 W. During 180 s, E = Pt = 20 × 180 = the same 3,600 J.

Electric heaters, filament bulbs and protective fuses are examples that can be discussed in an age-appropriate classroom explanation. Do not ask a child to test heating by touching a live wire or household electrical fitting. Real mains electricity requires proper adult supervision and certified equipment; numerical questions can be practised safely using diagrams, calculators and a notebook.

Circuit diagrams: ammeter, voltmeter and connections

Many marks are lost in the drawing rather than in the arithmetic. In a basic school circuit used for measurement, an ammeter is placed in series to measure the current passing through the component of interest, while a voltmeter is connected in parallel across that component to measure potential difference. Use clear conventional symbols, joined wires, correct meter polarity for direct-current diagrams, and labels for the source and resistor.

A helpful study exercise is to draw a simple cell, switch, resistor and ammeter in one loop, then place the voltmeter across the resistor. Explain why the voltmeter is not simply inserted into the same series path. Students should also recognise that a disconnected switch opens the circuit, so an ordinary steady current does not flow through that loop. Clear drawings can be practised with a ruler and pencil without constructing the circuit physically.

If a question asks for the effect of adding a resistor in series to an ideal fixed-voltage source, the total resistance increases and total current decreases. In an ideal parallel addition, total resistance decreases and source current increases. Such predictions assume the stated circuit conditions; real batteries and electrical devices have limits. For a practical explanation, a teacher can use safe simulations or approved school apparatus rather than improvised mains wiring.

Common mistakes and a five-step answer check

The most common Class 10 Electricity numerical errors are predictable: putting milliamperes into a formula without converting to amperes; using minutes where seconds are expected; adding parallel resistances as if they were in series; calculating electricity consumption in kW without multiplying by hours; dropping a squared term in I²R or V²/R; and writing a physically impossible unit at the end.

Use this five-step routine each time: (1) copy the givens with symbols and units; (2) identify the exact unknown; (3) draw the circuit or state the formula with correct conditions; (4) substitute values, show the units and calculate; (5) check the number by reversing the calculation or comparing it with what should happen physically. The final answer should be clearly marked, not hidden between several lines of rough work.

For example, if two positive resistors of 6 Ω and 3 Ω are in parallel, an answer of 9 Ω fails the basic smaller-than-the-smallest check. If a 12 V supply is across 6 Ω, a current of 72 A fails V/R = 2 A and suggests multiplication was used instead of division. A student who learns to catch these errors independently is developing a more important skill than memorising the answers to a practice sheet.

A seven-day revision plan for the Electricity chapter

A practical week of revision can be divided into short, focused tasks. On Day 1, revise current, charge, potential difference and units; solve five Q = It and V = W/Q questions. On Day 2, practise Ohm’s law and explain each step aloud. On Day 3, label current–voltage graphs, analyse slopes and review the effect of wire length and area. On Day 4, solve series-resistor questions and check total voltage drops. On Day 5, work on parallel branches and total current.

On Day 6, combine electrical power, energy consumption and Joule heating. On Day 7, attempt a mixed exercise without looking at notes, mark it honestly, then redo only the types of mistakes you made. A short reflection at the end of each session is useful: which formula was hardest to select, and was the main problem understanding, units or arithmetic?

Parents can help by asking one simple question: “What was given and why did you choose that formula?” They do not need to provide the answer immediately. Students who are frequently anxious about Science numericals may benefit from a smaller number of well-explained questions, followed by independent practice. Our board examination study strategy describes how to combine revision with timed work and feedback.

Practice set: eight numericals with answers

Question 1: A current of 0.6 A flows for 40 s. Find charge. Answer: Q = It = 0.6 × 40 = 24 C. Question 2: 72 J of work moves 9 C of charge. Find potential difference. Answer: V = W/Q = 72/9 = 8 V.

Question 3: A 9 V supply sends 0.3 A through an ohmic resistor. Find resistance. Answer: R = V/I = 9/0.3 = 30 Ω. Question 4: Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series. Find equivalent resistance. Answer: 2 + 3 + 5 = 10 Ω.

Question 5: Two resistors of 4 Ω and 12 Ω are in parallel. Find equivalent resistance. Answer: (4 × 12)/(4 + 12) = 48/16 = 3 Ω. Question 6: A resistor carries 2 A with 7 V across it. Find power. Answer: P = VI = 7 × 2 = 14 W.

Question 7: A 0.8 kW appliance works for 3 hours. Find energy used in household units. Answer: E = 0.8 × 3 = 2.4 kWh. Question 8: Find heat produced by a 10 Ω resistor carrying 1.5 A for 20 s. Answer: H = I²Rt = 1.5² × 10 × 20 = 450 J.

Cover the answers and attempt all eight in one sitting. Then make a small correction table with three columns: your answer, the correct method and the reason for the difference. This trains students to notice when the problem is unit conversion, an incorrect connection rule, an arithmetic slip or selection of the wrong formula.

FAQs for students and parents

Which formulas should a Class 10 student learn first? Begin with I = Q/t, V = W/Q and V = IR. Once those are understood, add the equivalent resistance rules, P = VI, energy E = Pt and H = I²Rt. Learn the meaning and units of each quantity rather than memorising formula rearrangements without examples.

Why do I get different answers for resistance questions? Check whether components are in series or parallel, and whether the voltage and current refer to the whole circuit or to one component. Also inspect the scale used in the question. A reading of 500 mA is 0.5 A, not 500 A.

How many numericals should be practised each day? There is no single number for every learner. Six to ten well-explained problems may be more useful than thirty copied solutions. Increase quantity only when the student can choose the equation without hints and can check the units.

Is this a list of guaranteed CBSE exam questions? No. It is a learning guide built around standard Electricity concepts. For the applicable year, check the official CBSE 2026–27 curriculum and NCERT textbooks, along with the student’s school instructions. Textbook editions and assessment patterns can change.

Can one-to-one tuition help with Physics numericals? It can provide a place for the student to ask questions, practise a step-by-step method and receive feedback. The match depends on the tutor’s subject knowledge and teaching style. Parents may review our Class 9–10 tutor options and request a suitable teacher for their location and schedule.