NCERT Solutions Class 10 Science Chapter 11 Electricity

Picture-first reading. Think of the circuit as a closed loop that water could run around. Just as water needs an unbroken pipe loop and a pump to keep flowing, charge needs an unbroken wire loop and a cell to keep moving. The cell is the pump; the wires are the pipes; the bulb is the place where the moving charge does useful work.

The key word in the definition is closed. A single battery and a bulb sitting on a table are not a circuit until you join them with wires into a loop. The moment the loop is complete, charges drift steadily and the bulb glows. Break the loop at any point and the bulb goes dark instantly, even though the cell is still full of energy. In a board answer always name the four parts (source, connecting wires, switch and device) and use the words continuous and closed; those two adjectives carry the marks.

Common board phrasing. CBSE often pairs this with a circuit diagram. If asked, draw a cell, an ammeter in series, a bulb and a plug key, all joined into one loop, with the conventional current shown leaving the positive terminal.

Answer: A closed, continuous conducting loop (source + wires + switch + device) along which current can flow.

Strategic angle. A unit is always defined by setting every other quantity in its defining equation to one. Here the defining equation is I = Q/t, so put Q = 1 C and t = 1 s and read off I = 1 A. This same trick defines the volt later (set W = 1 J, Q = 1 C) and the ohm (set V = 1 V, I = 1 A).

The ampere is one of the seven base SI units, named after the French physicist André-Marie Ampère. It is worth remembering the size of a coulomb in everyday terms: one coulomb is the charge of about 6.25 × 10¹⁸ electrons. So a one-ampere current means roughly six billion billion electrons sweeping past a point each second, which is why even a small torch bulb carries an enormous number of charge carriers. A board answer that gives the equation, the substitution and the symbol form 1 A =1 C s^-1 is complete.

Answer: 1 ampere = 1 coulomb per second (1 A=1 C s^-1).

Strategic angle. This is a one-line application of charge quantisation Q = ne. Solve for the count n = Q/e and substitute. The whole problem is the single division 1 ÷ (1.6 × 10^-19).

The result, 6.25 × 10¹⁸, is worth committing to memory because it is the bridge between the everyday unit (coulomb) and the microscopic unit (electron charge). It tells you that the coulomb is a huge amount of charge on the atomic scale. To handle the power of ten cleanly, write 1/1.6 = 0.625 first, then move the negative exponent across: 0.625 × 10¹⁹ = 6.25 × 10¹⁸. Students who try to divide the powers of ten in their head usually slip by a factor of ten here, so always separate the mantissa from the exponent. The same relation Q = ne also lets you reverse the question: given a charge, the number of electrons is fixed because charge cannot take fractional values of e.

Answer: n = 6.25 × 10¹⁸ electrons per coulomb.

Quick reading. The single word the examiner wants is cell (or battery). But understanding why a cell is the right answer is what protects the mark in a tricky multiple-choice version.

A potential difference is like a height difference for water: water keeps flowing down a slope only if something keeps lifting it back up. In a circuit the cell is that lift. Its chemical reaction does work to separate charge, holding one terminal at a higher potential than the other even when no current is drawn. Connect a wire and that stored potential difference drives the charges around. A generator or a dynamo does the same job using mechanical energy instead of chemical energy, so in a board answer “a cell, a battery or a generator” is acceptable, but the cell is the standard textbook answer. The key idea is that the device must keep spending energy to hold the terminals apart in potential; a charged object with no energy source cannot maintain a steady current.

Answer: A cell or battery (also a generator) maintains the potential difference.

Strategic angle. Define the unit by setting both other quantities in V = W/Q to one. With W = 1 J and Q = 1 C you read off 1 V directly, exactly as the ampere came from I = Q/t.

The deeper meaning is that voltage measures energy per unit charge. A 1.5 V cell does 1.5 joules of work on every coulomb of charge that passes through the circuit. This is why a higher-voltage source can push the same charge harder and deliver more energy. Note the contrast with current: current counts how much charge moves per second, while voltage measures how much energy each unit of charge carries. Mixing the two is the most common error here, so anchor your answer on the phrase “work done per unit charge”. In symbols, 1 V =1 J C^-1, and that compact form is what a marker looks for.

Answer: 1 V is the potential difference when 1 J of work moves 1 C of charge (1 V = 1 J C^-1).

Quick reading. The number 6 in “6 V” already is the energy in joules given to each coulomb, because the volt is defined as joule per coulomb. So the answer is 6 J with almost no calculation.

The reason this works is the definition V = W/Q. Rearranged it gives W = VQ, and for one coulomb (Q = 1) the work equals the voltage in numerical value. This is a useful sanity check throughout the chapter: the voltage rating of any source tells you, at a glance, the energy per coulomb it supplies. For a 6 V battery driving a current, every coulomb that completes the loop receives 6 J from the chemical store and then delivers that 6 J to the devices in the circuit. Energy is conserved, so the energy the battery gives out equals the energy the resistors and bulbs use up. Tie your answer to W = VQ and state the unit (joule) clearly.

Answer: W = VQ = 6 × 1 = 6 J per coulomb.

Structural angle. Sort the factors into two groups: geometry (the size and shape of the wire, captured by l and A) and material (captured by the resistivity ρ, with temperature shifting ρ). The single equation R = ρ l / A holds all three of the first kind in one place.

The geometric dependence is easy to feel physically. A long, thin wire is like a long, narrow corridor: charges bump along it with difficulty, so resistance is high. A short, fat wire is a wide, short hallway, so charges pass easily and resistance is low. That is why thick copper cables are used to carry heavy household currents. The material factor is separate: even with identical length and area, a nichrome wire resists far more than a copper wire because nichrome's resistivity is about sixty times larger. Temperature acts through the material: as a metal heats up, its atoms vibrate more and scatter the moving electrons, raising the resistance. So the complete answer names l, A, ρ (material) and temperature, with the relation R = ρ l / A tying the geometric factors together.

Answer: R depends on l (directly), A (inversely), the material's resistivity ρ, and temperature; R = ρ l/A.

Quick reading. Same material, same length, so only the thickness changes. Thickness sets the area A, and R ∝ 1/A, so thicker means less resistance and therefore more current.

A helpful picture is water flowing through pipes. A wide pipe lets more water through per second than a narrow one for the same pump pressure, because there is simply more room for flow. A thick wire is the wide pipe for charge. Doubling the radius of a wire multiplies its area by four (since area depends on radius squared), so the resistance falls to one-quarter and the current rises four times. This is exactly why appliances that draw heavy currents, such as a geyser or an air conditioner, are wired with thick copper cables, while a low-current device like a phone charger uses thin wire. Always justify the answer with R = ρ l/A and then Ohm's law; stating the conclusion without the reason loses half the marks.

Answer: Thick wire: larger A gives smaller R, so a larger current flows.

Strategic angle. Treat resistance as the fixed conversion factor between voltage and current. With R held constant, I = V/R is a straight proportionality, so whatever you do to V happens to I.

The cleanest way to answer “what happens” questions is to take the ratio of the new case to the old case so the unknown R cancels:

I₂/I₁ = (V₂/R)/(V₁/R) = V₂/V₁ = 1/2.

This shows I₂ = I₁/2 without ever needing the numerical values of V or R. The ratio method is a powerful habit for the whole chapter: it works for any “if this is halved/doubled” problem and avoids carrying unknown constants. Physically, the result makes sense because the resistor obeys Ohm's law, so its V versus I graph is a straight line through the origin; moving to half the voltage simply slides you halfway down that line to half the current.

Answer: Current becomes half; I₂/I₁ = V₂/V₁ = 1/2.

Strategic angle. Anchor the answer on what a heating coil has to do: convert electrical energy into heat efficiently (H = I²Rt) and stay solid and unburnt while red hot. Then check which properties deliver that, and you find all of them point to an alloy.

A pure metal fails on several counts. Copper has very low resistivity, so a copper coil would barely warm up for the same current. Many pure metals also melt at lower temperatures or oxidise quickly when heated in air, so a pure-metal coil would either melt or corrode away in days. An alloy fixes each problem at once: its disordered mix of atoms scatters electrons strongly, giving high resistivity and so strong heating; its high melting point lets it glow without melting; and its resistance to oxidation lets it survive repeated red-hot use. Nichrome is the classic example. In a board answer, give the three reasons (high resistivity, high melting point, low oxidation) and link the first to H = I²Rt to show why more resistance means more heat.

Answer: Alloys give high resistivity (more heat), high melting point (no melting) and resist oxidation (long life), so they beat pure metals for heating coils.

Quick reading. Conductivity ranks inversely with resistivity, so the material with the smallest ρ in the table is the best conductor. Just read the numbers and pick the smallest.

It is worth noticing how wide the range is. Silver (1.60 × 10^-8) and copper (1.62 × 10^-8) sit at the very bottom and are almost equal, which is why copper, being far cheaper, is the workhorse of household wiring even though silver edges it slightly. Iron (10.0 × 10^-8) is a middling conductor, and mercury (94.0 × 10^-8) is a poor one for a metal, nearly sixty times worse than silver. So for part (a), iron clearly beats mercury, and for part (b) silver takes the top spot. The lesson behind the numbers is that resistivity, not resistance, is the right quantity for comparing materials, because it strips out the effect of length and thickness and leaves only the material's own nature.

Answer: (a) Iron. (b) Silver.

Picture-first reading. A series circuit is just one continuous loop with everything threaded on it like beads on a string. Start at the positive end of the battery, pass through each resistor in turn, through the plug key, and return to the negative end. There are no branches.

The detail students often miss is that three 2 V cells in series act as one 6 V source, so label the battery 6 V. Because there is only one path, the same current flows through the 5 Ω, 8 Ω and 12 Ω resistors, while the voltage of the battery splits up among them in proportion to their resistances. The total resistance is the simple sum 5 + 8 + 12 = 25 Ω, which sets the current at 6/25 = 0.24 A once the key is closed. When drawing for the board, use the standard symbols: long-and-short parallel lines for the cells, a rectangle for each resistor, and a small gap with a movable arm for the plug key. A neat single-loop sketch with correct symbols and the 6 V label earns full marks.

Answer: Series loop: 6 V battery, 5 Ω + 8 Ω + 12 Ω resistors and a plug key, all on one path.

Strategic angle. Solve the chain in order: total resistance gives the current (the ammeter reading), and the current times the chosen resistor gives that resistor's voltage (the voltmeter reading). No new circuit ideas are needed beyond Ohm's law applied twice.

The single most useful fact in a series circuit is that the current is the same everywhere, so the ammeter will read 0.24 A no matter where on the loop it sits. The voltmeter, by contrast, reads only the share of voltage that falls across the 12 Ω resistor, which is I R₃ = 0.24 × 12 = 2.88 V. As a check, the three voltage drops are 0.24 × 5 = 1.2 V, 0.24 × 8 = 1.92 V and 0.24 × 12 = 2.88 V, and these add to 1.2 + 1.92 + 2.88 = 6 V, exactly the battery voltage, which confirms energy conservation around the loop. Connecting the meters correctly matters in the diagram: the ammeter in the main line, the voltmeter bridging only the two ends of the 12 Ω resistor.

Answer: Ammeter = 0.24 A; voltmeter across 12 Ω = 2.88 V.

Strategic angle. Do not grind the full arithmetic; use the rule that parallel resistance is always less than the smallest branch. When the smallest branch is far smaller than the rest, the others barely contribute and R_p sits just under that smallest value.

Here the smallest resistor is 1 Ω in both parts. The reciprocal 1/1 = 1 swamps the tiny reciprocals 1/10³ = 0.001 and 1/10⁶ = 0.000001, so 1/R_p is barely above 1 and R_p is barely below 1 Ω. This “smallest wins” intuition is the whole point of the word judge in the question: the examiner wants you to estimate, not to compute to six decimal places. The same idea explains why adding more parallel branches can only lower the total resistance, never raise it, and why a single low-resistance fault (a short circuit) across a device drops the combined resistance almost to zero. State the estimate (≈ 1 Ω) and the reason (smallest branch dominates) for a complete answer.

Answer: Both combinations give R_p ≈ 1 Ω (just below the smallest, 1 Ω, resistor).

Strategic angle. The iron is just a stand-in for the three appliances: “same current at the same voltage” means “same resistance”. So the problem reduces to finding the parallel equivalent of 100, 50 and 500, then applying Ohm's law once.

The arithmetic is cleanest in fractions over 500: 1/100 = 5/500, 1/50 = 10/500 and 1/500 = 1/500, which sum to 16/500, giving R_p = 500/16 = 31.25 Ω. The current is then 220/31.25 = 7.04 A. As a quick cross-check, add the three branch currents directly: the lamp draws 220/100 = 2.2 A, the toaster 220/50 = 4.4 A and the filter 220/500 = 0.44 A, which total 2.2 + 4.4 + 0.44 = 7.04 A, exactly matching. This branch-current check is a reliable way to confirm any parallel-circuit answer, because in parallel the total current is always the sum of the branch currents.

Answer: Iron resistance = 31.25 Ω; current = 7.04 A.

Strategic angle. Compare the two arrangements on the things that matter to a user: does each device get its rated voltage, and does one device failing knock out the rest? Parallel wins on both, which is why it is the standard for domestic circuits.

The first advantage is voltage. In parallel, every branch sees the full source voltage, so a 220 V appliance actually runs on 220 V. In series, the supply voltage is shared, so each device gets only a fraction and none runs at its rated value. The second advantage is independence: parallel branches are separate paths, so switching off the toaster or having a bulb fuse leaves the rest untouched, whereas a single break anywhere in a series loop stops every device. A third practical gain is that the parallel combination has lower total resistance, so the mains can deliver enough current for many appliances at once, and each can draw exactly the current it needs. Together these reasons make parallel wiring the only sensible choice for a home, and a good board answer lists at least the voltage and the independence points explicitly.

Answer: Parallel gives each device full voltage, independent on/off operation, lower total resistance with more available current, and a suitable current per device.

Strategic angle. Work backwards from each target. A target below the smallest resistor (1 < 2) screams “all in parallel”; a target equal to a clean sum of parts suggests a parallel-then-series mix.

For part (b), the all-parallel sum is especially neat: 1/2 + 1/3 + 1/6 has the common denominator 6, giving (3 + 2 + 1)/6 = 6/6 = 1, so R_p = 1 Ω exactly. For part (a), notice that 3 Ω and 6 Ω in parallel give a tidy 2 Ω (since 1/3 + 1/6 = 1/2), and adding the spare 2 Ω resistor in series lands on 4 Ω. The general method for these “design a network” questions is to test whether the target is below the smallest branch (then go fully parallel) or whether some subset combines into a round number that the rest can be added to. Always show the reciprocal arithmetic so the marker can see the parallel step, then state the final series addition.

Answer: (a) (3 Ω ∥ 6 Ω) + 2 Ω = 4 Ω. (b) 2 Ω ∥ 3 Ω ∥ 6 Ω = 1 Ω.

Strategic angle. The extremes are fixed by the rules of combination: series only ever adds, so it gives the maximum; parallel only ever reduces below the smallest branch, so it gives the minimum. No clever mixed network can do better than these two limits.

For the highest, the all-series sum is immediate: 4 + 8 + 12 + 24 = 48 Ω. For the lowest, the all-parallel sum is cleanest over the common denominator 24: the reciprocals become 6 + 3 + 2 + 1 all over 24, which is 12/24 = 1/2, so R_p = 2 Ω. Notice that the lowest value, 2 Ω, is indeed below the smallest coil (4 Ω), exactly as the parallel rule predicts. This pair of extremes is worth internalising because it appears again in the exercises (for example the toaster-coil problem): whenever a question asks for the maximum or minimum obtainable resistance from a set of coils, the answer is the simple series sum and the all-parallel value, with no need to explore other configurations.

Answer: (a) 48 Ω (all series). (b) 2 Ω (all parallel).

Strategic angle. Pin the answer on the fact that the cord and the element carry the same current but have very different resistances. With I fixed, H = I²Rt says heat is proportional to resistance, so the high-resistance part heats up while the low-resistance part does not.

The cord is built to conduct, not to heat: thick copper with tiny resistance, so I²R is negligible and it stays cool. The element is built to heat: long, thin nichrome with large resistance, so I²R is large and it glows red. A common trap is to think the element “draws more current” than the cord, but in a series path the current is identical everywhere; it is the resistance, not the current, that differs. This is the same principle that makes a fuse wire (thin, of a low-melting alloy) the deliberately weak, hot spot of a circuit. For a full board answer, state that the current is equal, contrast the resistances of copper cord and nichrome element, and conclude with H = I²Rt.

Answer: Same current, but the element's high resistance makes I²Rt large, so it glows; the low-resistance copper cord stays cool.

Strategic angle. The problem hands you both the charge and the voltage, so the energy is simply their product, H = VQ. The mention of “one hour” is a distractor; it would only be needed if you had to compute the charge from a current and a time.

The physics is energy conservation: every coulomb gains energy V as it passes through the potential difference, and that energy turns into heat in the resistor. So total heat is voltage times total charge, 50 × 96000 = 4.8 × 10⁶ J. If you preferred, you could find the current first, I = Q/t = 96000/3600 = 26.67 A, and then use H = VIt = 50 × 26.67 × 3600, which gives the same 4.8 × 10⁶ J, but that is a longer path to the same answer. Recognising which formula needs the fewest steps, here H = VQ, is the mark of an efficient solver and saves time in an exam.

Answer: H = VQ = 50 × 96000 = 4.8 × 10⁶ J.

Strategic angle. All three quantities for Joule's law are given directly, so plug straight into H = I²Rt. The only care needed is to square the current before multiplying.

Taking it in order, I² = 5² = 25, then 25 × 20 = 500, then 500 × 30 = 15 000 J. A neat cross-check is to find the power first: P = I²R = 25 × 20 = 500 W, and over 30 s the energy is P t = 500 × 30 = 15 000 J, the same result. This two-route agreement is reassuring and shows that H = I²Rt is just power multiplied by time. The iron dissipates 500 watts, which is a realistic figure for a small electric iron, so the answer passes a common-sense check too. State the formula, show the squaring of the current, and keep the unit (joule) on the final line.

Answer: H = I²Rt = 25 × 20 × 30 = 15 000 J.

Strategic angle. The phrase “rate at which energy is delivered” is the dictionary definition of power, so the one-word answer is power, and the equation is P = VI. Naming the three equivalent forms shows full command of the idea.

Power tells you how fast a device converts electrical energy into other forms, such as heat or light. A 100 W bulb uses energy twice as fast as a 50 W bulb at the same voltage, which is why it glows brighter and also why it adds more to the electricity bill per hour. The three algebraic forms, P = VI, P = I²R and P = V²/R, are all the same relation rewritten with Ohm's law, and choosing the right one for the data given saves work. For example, if you know voltage and resistance but not current, P = V²/R is the direct route. The key teaching point for a board answer is to identify the quantity (power) and back it with the defining equation P = VI.

Answer: Electric power P = VI (also I²R or V²/R) sets the rate of energy delivery.

Strategic angle. Two short steps: power is the product VI, and energy is power times time. The neat part is the unit choice, keep power in kilowatts and time in hours to read the energy directly in kilowatt-hours, the unit on your electricity bill.

The power is 220 × 5 = 1100 W, or 1.1 kW. Over two hours the energy is 1.1 × 2 = 2.2 kWh, which is two and a fifth “units” of electricity. If the question wants joules instead, convert the time to seconds: 2 h = 7200 s, and W = 1100 × 7200 = 7.92 × 10⁶ J. Both numbers describe the same energy; the kWh form is for billing, the joule form is the SI unit. Note that 1 kWh equals 3.6 × 10⁶ J, so 2.2 kWh = 2.2 × 3.6 × 10⁶ = 7.92 × 10⁶ J, which confirms the two answers agree. Showing both forms, with the kWh as the headline, makes for a complete solution.

Answer: P = 1100 W; energy in 2 h = 2.2 kWh = 7.92 × 10⁶ J.

Strategic angle. Two independent factors of five combine into a factor of twenty-five. Cutting into five equal parts divides each part's resistance by five (length down by five), and putting five equal parts in parallel divides again by five. Multiply the two and you get R' = R/25.

The shortcut worth remembering is that for n equal pieces recombined in parallel, the resistance drops by n². The first power of n comes from the shorter length of each piece, the second from the parallel combination of n of them. So with n = 5 the resistance falls to 1/25 of the original, making R/R' = 25 and the answer option (d). This kind of “cut and reconnect” reasoning is a favourite in objective tests because it rewards understanding the two effects separately rather than memorising a single result. As a sanity check, note that parallel always lowers resistance, so R' < R and the ratio R/R' must be greater than one, which immediately rules out options (a) and (b).

Answer: R/R' = 25, option (d).

Strategic angle. Start from the one definition P = VI and use Ohm's law to generate the other valid forms, then spot the odd one out. Substituting V = IR gives I²R, and substituting I = V/R gives V²/R. The expression IR² cannot be produced this way.

A fast, reliable filter is dimensional analysis. Power must have units of watts, which are volt-amperes. Check each option: VI is volt-ampere (correct); I²R is ampere² × ohm, and since ohm = volt per ampere, this becomes ampere × volt (correct); V²/R is volt² per ohm = volt² × ampere per volt = volt-ampere (correct). But IR² is ampere × ohm², which is ampere × (volt per ampere)² = volt² per ampere, not watts. So IR² fails the unit check and is the answer, option (b). Carrying units alongside symbols is the surest way to catch a wrong formula in objective questions like this.

Answer: IR² is not power; option (b).

Strategic angle. The bulb's resistance does not change with the supply voltage, so use P = V²/R and note that power is proportional to the square of the voltage. Halving the voltage from 220 V to 110 V therefore divides the power by four.

The cleanest route avoids computing the resistance at all. Take the ratio of new to old power:

P'/P = (V'²/R)/(V²/R) = (V'/V)² = (110/220)² = (1/2)² = 1/4.

So P' = 100/4 = 25 W, option (d). If you do want the resistance, it is R = 220²/100 = 484 Ω, and P' = 110²/484 = 25 W, the same answer. The important conceptual point, and a frequent exam trap, is that the bulb does not keep delivering 100 W at any voltage: the 100 W rating only holds at the rated 220 V. At lower voltage it draws less current and dissipates less power, so it glows dimmer.

Answer: P' = (1/2)² × 100 = 25 W, option (d).

Strategic angle. The voltage is the same in both cases, so the right heat formula is H = V²t/R, not H = I²Rt. With V fixed, heat is inversely proportional to resistance, so the combination with the smaller resistance produces the larger heat.

Series gives R_s = 2R and parallel gives R_p = R/2, a factor of four apart. Since heat goes as 1/R, the series case (large R) produces less heat and the parallel case (small R) produces more, in the ratio H_s : H_p = R_p : R_s = (R/2) : (2R) = 1 : 4. The crucial decision here is choosing the voltage-based form of the heat law, because the problem fixes the potential difference; a student who reaches for I²Rt has to first work out two different currents and is far more likely to slip. Whenever the voltage is held constant across both arrangements, reach for H = V²t/R and the comparison collapses to a simple resistance ratio.

Answer: H_s : H_p = 1 : 4, option (c).

Strategic angle. Anchor the answer on the job of the instrument: a voltmeter compares the potential at two points, so it must touch both points at once, which means a parallel connection. Contrast this with an ammeter, which counts the current and so must sit in the current's path, in series.

The reason a voltmeter is connected in parallel is that potential difference is defined between two points, and to read it the meter must bridge exactly those two points. For this to be accurate the voltmeter must barely affect the circuit, so it is built with a very high internal resistance and draws almost no current. If it had low resistance, it would divert current through itself and lower the very voltage it is trying to measure. This pairing is worth memorising as a matched set: ammeter in series with low resistance, voltmeter in parallel with high resistance. Stating both the connection (parallel) and the reason (high resistance, negligible current) gives a complete board answer.

Answer: In parallel across the two points; it has very high resistance and draws negligible current.

Strategic angle. Rearrange R = ρ l/A for the length, compute the area from the diameter, and substitute. For the second part use the scaling R ∝ 1/d² rather than recomputing everything.

The area is the step where care is needed: convert 0.5 mm to 5 × 10^-4 m, halve it for the radius, then square. That gives A = π (2.5 × 10^-4)² = 1.96 × 10^-7 m². The length follows as l = RA/ρ = (10 × 1.96 × 10^-7)/(1.6 × 10^-8) ≈ 122.7 m, a surprisingly long wire, which makes sense because copper's resistivity is so low that a great length is needed to reach 10 Ω. For the diameter doubling, avoid redoing the arithmetic: since A ∝ d² and R ∝ 1/A, we get R ∝ 1/d², so doubling d cuts R to a quarter, giving 2.5 Ω. Recognising this inverse-square dependence on diameter is the efficient, exam-smart move.

Answer: l ≈ 122.7 m; doubling the diameter quarters the resistance to 2.5 Ω.

Strategic angle. Recognise that this is Ohm's law in graphical form. Because V = IR, the V-against-I plot is a straight line through the origin and its slope is the resistance, so the resistance is just V/I read off the line.

The neatest value to use is a point with round numbers, such as I = 2.0 A, V = 6.7 V, giving R = 6.7/2.0 ≈ 3.4 Ω. Averaging all five ratios (3.2, 3.4, 3.35, 3.4, 3.3) gives about 3.33 Ω, which is the best single estimate and matches the slope. The slight scatter in the ratios is normal experimental error, which is exactly why drawing a best-fit straight line and taking its slope is more reliable than using any one reading. When you plot, choose scales that spread the points across the graph, mark them clearly, and draw the straight line that passes through the origin and lies closest to all the points. State the slope as the resistance, around 3.4 Ω, for the final answer.

Answer: Straight-line graph through the origin; slope = R ≈ 3.4 Ω.

Strategic angle. This is a direct one-step use of R = V/I. The only real task is the unit conversion: the current is in milliamperes and must become amperes before dividing.

Writing the current as 2.5 × 10^-3 A and dividing, R = 12/(2.5 × 10^-3). Handle the powers of ten cleanly: 12/2.5 = 4.8, and dividing by 10^-3 multiplies by 10³, giving 4.8 × 10³ = 4800 Ω. The large value makes sense because a tiny current of only a few milliamperes flows from a fairly ordinary 12 V battery, which is only possible if the resistor is large, in the kilo-ohm range. A good habit in any Ohm's-law problem is to convert all quantities to base SI units (volts, amperes, ohms) before substituting, so milliamperes, kilo-ohms and millivolts never trip you up. Quote the answer as 4800 Ω or equivalently 4.8 kΩ.

Answer: R = V/I = 12/(2.5 × 10^-3) = 4800 Ω.

Strategic angle. The phrase “in series” is the whole hint: add all the resistances to get one total, find the one current, and that current is what flows through every resistor including the 12 Ω one. There is no need to treat the 12 Ω resistor separately.

The total resistance is 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω, where the four small resistors barely matter next to the 12 Ω. The current is I = 9/13.4 ≈ 0.67 A. Because the resistors share a single path, this same 0.67 A passes through the 12 Ω resistor. A useful sanity check is to notice that the 12 Ω resistor dominates the total, so the current is close to what a lone 12 Ω resistor would give, 9/12 = 0.75 A; the small extra resistances pull it down slightly to 0.67 A, which is consistent. The key conceptual takeaway is that “current through a particular series resistor” is always just the single circuit current.

Answer: I = 9/13.4 = 0.67 A through the 12 Ω resistor (and every other series resistor).

Strategic angle. Find the resistance the line needs to draw 5 A, then see how many 176 Ω resistors in parallel produce that value. For identical resistors in parallel the shortcut R_p = r/n turns the problem into one division.

The needed resistance is R_p = V/I = 220/5 = 44 Ω. Since n equal 176 Ω resistors in parallel give 176/n, setting 176/n = 44 yields n = 4. A quick check confirms it: four 176 Ω resistors in parallel give 176/4 = 44 Ω, and the current is 220/44 = 5 A, exactly as required. The general lesson is that adding more identical resistors in parallel steadily lowers the combined resistance and raises the current the line can deliver, which is why distribution networks add parallel paths to carry more load. State the target resistance, the relation R_p = 176/n, and the final count of four.

Answer: n = 176/44 = 4 resistors in parallel.

Strategic angle. Compare each target with a single 6 Ω resistor. A target above 6 (9 Ω) means you need a series addition; a target below 6 (4 Ω) means you need a parallel reduction. That choice fixes the design.

For 9 Ω, two 6 Ω in parallel give a tidy 3 Ω (half of 6, since identical pairs in parallel halve), and adding the third in series gives 3 + 6 = 9 Ω. For 4 Ω, two 6 Ω in series give 12 Ω, and bringing the third 6 Ω in parallel with that gives 1/(1/12 + 1/6) = 4 Ω. Both networks use all three resistors, as the question demands. The broad method for “design a resistance” problems is to first settle whether the target is above or below a single unit, then combine a sub-group and attach the remaining resistor in the opposite mode. Always show the reciprocal arithmetic for the parallel step so the marker can follow the design.

Answer: (i) (6∥ 6) + 6 = 9 Ω. (ii) (6 + 6)∥ 6 = 4 Ω.

Strategic angle. Each bulb is an independent parallel branch at the full 220 V, so it draws a fixed current set by its power rating. The total current is the number of bulbs times that per-bulb current, and the 5 A ceiling caps how many you may connect.

A single 10 W bulb at 220 V draws P/V = 10/220 ≈ 0.0455 A. Dividing the 5 A budget by this gives n = 5/(10/220) = 5 × 220/10 = 110 bulbs. The algebra is neatest if you rearrange to n = (I_total × V)/P = (5 × 220)/10 = 110 in one step, avoiding the small decimal. This is exactly how an electrician reasons about how many lights a circuit can safely carry before the fuse blows: total load current must stay under the rated limit. Connecting the bulbs in parallel is what keeps each at its rated 220 V and 10 W; in series they would share the voltage and none would light properly.

Answer: n = (5 × 220)/10 = 110 lamps.

Strategic angle. The voltage is fixed at 220 V, so the only thing that changes between the three cases is the resistance connected. Work out the resistance for each arrangement, then read off the current from I = V/R.

A single coil is 24 Ω, giving 220/24 ≈ 9.2 A. In series the resistance doubles to 48 Ω, halving the current to about 4.6 A. In parallel the resistance halves to 12 Ω, doubling the single-coil current to about 18.3 A. The pattern is worth noticing: relative to one coil, series gives half the current and parallel gives double, because series doubles the resistance and parallel halves it. This is exactly how a multi-setting heater or oven controls its heat output, low setting in series (small current, less heat), high setting in parallel (large current, more heat). Quote all three currents and, ideally, point out that parallel gives the most heat because it draws the most current at the same voltage.

Answer: ≈ 9.2 A (single), ≈ 4.6 A (series), ≈ 18.3 A (parallel).

Strategic angle. Choose the power formula that matches what each circuit hands you. The series circuit fixes the current through the 2 Ω resistor, so P = I²R is natural; the parallel circuit fixes the voltage across it (4 V), so P = V²/R is natural.

In circuit (i) the resistors are in series, so the current is 6/(1+2) = 2 A and flows through the 2 Ω resistor, giving P = 2² × 2 = 8 W. In circuit (ii) the resistors are in parallel, so each gets the full 4 V, and the 2 Ω resistor dissipates 4²/2 = 8 W. The two answers coincide at 8 W, which is a neat result the question is designed to reveal: a resistor's power depends on the actual current through it or voltage across it, not on whether the circuit happens to be series or parallel. Picking the power formula that uses the quantity you already know is the time-saver here; forcing the other formula would mean first finding the unknown current or voltage.

Answer: P_(i) = I²R = 8 W and P_(ii) = V²/R = 8 W; equal power.

Strategic angle. Both lamps are across the same 220 V, so the simplest route is to add their powers and divide once by the voltage: I = (P₁ + P₂)/V. This works only because they share the same voltage, which parallel connection guarantees.

The total power is 100 + 60 = 160 W, and the line current is 160/220 = 0.727 A. If you prefer, find the branch currents separately: 100/220 = 0.455 A and 60/220 = 0.273 A, which add to 0.727 A, the same result. This agreement illustrates the rule that in parallel the branch currents add to give the total line current. Notice that the brighter 100 W lamp draws the larger current, since at a fixed voltage current is proportional to power. A real mains connection would feed many such parallel loads, and the total current they draw decides the size of fuse and wiring needed, which is why this sum-of-powers calculation is a practical everyday one.

Answer: I = (100 + 60)/220 = 0.727 A.

Strategic angle. Energy is power times time, so a high-power device used briefly can still use less total energy than a lower-power device used for longer. Compute W = Pt for each in consistent units and compare the numbers.

Converting both times to seconds keeps the comparison honest. The TV runs for a full hour (3600 s) at 250 W, using 250 × 3600 = 9 × 10⁵ J. The toaster runs for only ten minutes (600 s) at 1200 W, using 1200 × 600 = 7.2 × 10⁵ J. So despite the toaster's much higher power rating, the TV uses more energy because it runs six times longer. This is the central insight the question tests: your electricity bill depends on energy (P × t), not on power alone. A short burst from a powerful appliance can cost less than a long stint from a modest one. If you prefer commercial units, the TV uses 0.25 kWh and the toaster 0.2 kWh, leading to the same conclusion.

Answer: TV set uses more energy (9 × 10⁵ J vs 7.2 × 10⁵ J).

Strategic angle. The word “rate” is the key: rate of heat production is power, so the answer is P = I²R and the 2-hour figure is a distractor. Substitute the current and resistance and you are done.

With I = 5 A and R = 44 Ω, P = 5² × 44 = 25 × 44 = 1100 W. That is 1100 joules of heat every second. If the question had instead asked for the total heat over 2 hours, you would multiply by the time in seconds: 1100 × 7200 = 7.92 × 10⁶ J. Telling the two apart is the whole point of including the time data: a rate is per second (watts), a total is over the whole duration (joules). The heater's 1100 W rating is realistic for a small room heater. State the answer as 1100 W and make clear it is a rate (power), not a total energy.

Answer: Rate of heat = P = I²R = 1100 W.

Strategic angle. Group the five parts by the principle each uses. Parts (a) and (b) are about the heating effect and material properties (high resistivity and high melting point); part (c) is about series versus parallel behaviour; parts (d) and (e) are about R = ρ l/A and resistivity.

For (a), a lamp filament must reach a white heat to emit light, so it needs a metal that survives extreme temperature without melting: tungsten, with its very high melting point and high resistivity, is almost the only suitable choice. For (b), a heating coil must produce much heat and endure repeated red-hot use in air, so an alloy is chosen for its high resistivity, high melting point and resistance to oxidation. For (c), series wiring at home would deny each appliance its rated voltage and would let a single failure kill the whole circuit, so parallel wiring is used instead. For (d), R = ρ l/A shows resistance is inversely proportional to area, so thicker wire conducts better. For (e), copper and aluminium are picked because their resistivity is very low, keeping transmission losses small, and they are cheap and ductile. Tying each answer back to either R = ρ l/A or H = I²Rt shows the examiner you see the unifying physics rather than five memorised facts.

Answer: (a) High melting point and resistivity. (b) High resistivity, high melting point, no oxidation. (c) Parallel gives full voltage and independent operation. (d) R ∝ 1/A. (e) Low resistivity, cheap, ductile.