NCERT Exemplar Solutions Class 10 Science Chapter 11

Read it like a circuit, not a picture. The trap here is that the three diagrams look different, so students feel one must give more current. But electric current cares only about two numbers: the driving voltage and the total resistance of the loop.

Concept used. For a single series loop, I = V/R from Ohm's law. The ammeter is an in-series device, so the reading equals the loop current irrespective of its position. Charge is conserved, so the same current passes through the cell, the resistor, the key and the ammeter.

1. List what is fixed. Same cell ⇒ same V. Same resistor ⇒ same R. Same single loop in each diagram. There is nothing left to make I differ.
2. Apply Ohm's law. I = V/R gives one value of I. Three identical (V,R) pairs give three identical currents.
3. Kill the misconception. Moving the ammeter from the top wire to the side wire does not add or remove resistance. An ideal ammeter has near-zero resistance, so it never throttles the current.
4. Cross-check with conservation of charge. In a series loop the same number of electrons per second crosses every cross-section, so every meter on the loop must read the same.

Option (d): the ammeter reads the same current in all three circuits.

Pick the right power formula first. The battery voltage is fixed at 12 V, while the current through each network changes with the arrangement. So the safe formula is P = V²/R_eq, not P = I² R. Voltage is the constant you can lean on.

Concept used. Heat dissipated in time t is H = P t with P = V²/R_eq when the terminal voltage is constant. Equal voltage across equal equivalent resistance forces equal heating, no matter how the individual resistors are wired internally.

1. Reduce each network. Replace every series/parallel block by one R_eq. The three given networks are designed to collapse to the same R_eq.
2. Heat in equal time. H = V²R_eq t. With V = 12 V common and R_eq equal, H is the same for all.
3. Why not use P = I²R? You could, but the current is different in each case, so you would have to find I separately every time. Using the constant voltage is faster and avoids errors.

Option (a): heat produced is the same in all the cases.

Spot the word "resistivity". The trap is to confuse resistivity (ρ) with resistance (R). Resistance follows the shape of the wire; resistivity is baked into the material itself.

Concept used. Resistivity is defined as the resistance of a unit cube of the material, ρ = R A / l. Rearranged from R = ρ l/A, it is a per-material number measured in Ωm. It is an intensive property, so cutting, stretching or reshaping the same metal leaves ρ unchanged.

1. Test each option with the cube definition. A unit cube has fixed l and A by definition, so length, thickness and shape are already removed from ρ.
2. What does change ρ? Only two things: switching to a different material, and changing the temperature (heating a metal raises ρ). Neither is listed except material.
3. Eliminate (a), (b), (c). Each describes how R varies with geometry, which is the wrong quantity for this question.

Option (d): resistivity is a material property.

Two-step plan. First convert current and time into charge, then convert charge into a count of electrons. Both steps are one-line formulas, so this is a pure plug-in numerical.

Concept used. Current is charge per second (I = Q/t), and charge is quantised in units of the electronic charge e = 1.6× 10⁻¹⁹ C, so Q = ne. Combining, n = It/e.

1. Charge first. Q = I t = 1 × 16 = 16 C.
2. Now the electron count. n = Qe = 161.6× 10⁻¹⁹.
3. Do the arithmetic carefully. n = 161.6× 10¹⁹ = 10 × 10¹⁹ = 10²⁰.
4. Sanity check the power of ten. A 1 A current is ordinary, 16 s is short, yet 10²⁰ electrons flow. That huge number is why we never count electrons one by one in circuits; we track charge in coulombs.

Option (a): n = 10²⁰ electrons.

Two checkboxes per circuit. Run a quick two-point check on every option: (1) Is the ammeter in series? (2) Is the voltmeter in parallel? A circuit passes only if both boxes are ticked.

Concept used. An ideal ammeter has near-zero resistance, so it must sit in the main line (series) where it does not disturb the current. An ideal voltmeter has near-infinite resistance, so it must sit across (in parallel with) the element whose voltage is wanted, drawing almost no current.

1. Series test for the ammeter. If the ammeter has a wire bypassing it, it is in parallel: wrong. It must be in the single current path.
2. Parallel test for the voltmeter. The voltmeter's two leads must touch the two ends of the resistor only, not break the main loop.
3. Polarity test for the cell. Current should leave the positive terminal, pass through the resistor, and return. A reversed cell or open key disqualifies the circuit.
4. Survivor. Only circuit (ii) satisfies all three checks, so it is the correctly connected circuit.

Option (b): circuit (ii).

Maximum means series. Remember the pair: series stacks resistance up (maximum), parallel splits it down (minimum). For "maximum", reach straight for series.

Concept used. In series the same current passes through each resistor and the voltages add, so the resistances simply add: R_s = R_i. With n equal resistors of value r, R_s = nr.

1. Identify the goal. Maximum total resistance ⇒ series connection.
2. Use R_s = nr. R_s = 5 × 15 = 1 Ω.
3. Compare with the options. 1 Ω is larger than a single 15 Ω resistor, as series should give. Option (b) 10 Ω is impossible from these values, and (a) equals one resistor alone.

Option (d): 1 Ω.

Minimum means parallel. The partner rule to Q6: for the smallest resistance, put everything in parallel. Equal resistors in parallel just divide by how many you have.

Concept used. In parallel the voltage is common and currents add, so conductances (1/R) add. For n equal resistors, R_p = r/n, which is smaller than a single resistor.

1. Goal to connection. Minimum total ⇒ parallel.
2. Use R_p = r/n. R_p = 155 = 125 Ω.
3. Reality check. 125 Ω is smaller than one 15 Ω resistor, exactly as parallel should be. Option (d) 25 Ω is the reciprocal trap; do not forget the final inversion.

Option (b): 125 Ω.

Follow the terminals. Trace the chain of cells and watch the + and − marks. A cell helps only if its + feeds into the next cell's −. Any cell facing backwards fights the others.

Concept used. The emf of a series battery is the algebraic sum of cell emfs. Aligned cells add (+V); a reversed cell subtracts (−V). Maximum total emf needs every cell aligned in the same sense.

1. Mark the polarity. For each cell write down which end is +. Walk along the series line.
2. Check the junctions. Every junction should join a + to the next cell's −. If two like terminals meet, a cell is reversed and the total drops.
3. Select the all-aligned diagram. Only diagram (i) keeps every cell pointing the same way, so its emfs add to the maximum. The other diagrams have one or more cells flipped.

Option (a): diagram (i).

Start from the definition, then substitute. Voltage has a one-line definition, V = W/Q. The only twist here is to express Q through current and time, which gives the answer instantly.

Concept used. Potential difference is energy per unit charge, measured in volts where 1 V = 1 J/C. Charge relates to current by Q = It. So V = W/(It).

1. Write the definition. V = WQ, the work per coulomb.
2. Replace Q. Use Q = It to get V = WIt.
3. Unit check kills the wrong options. Volt must be joule/coulomb. Option (a) gives J/(A·s) = J/C, correct. Options (b) and (d) multiply by charge (units J·C), which is wrong, and (c) has time in the wrong place.

Option (a): V = WCurrent×Time.

Keep R constant, balance the ratio. Since R ∝ l/A for a fixed material, holding R fixed means l/A is fixed. Double the top (l), and you must double the bottom (A) to keep the ratio unchanged.

Concept used. R = ρ l/A with ρ a material constant. Equal R and equal ρ force l/A to be equal for both wires, so A ∝ l when R is held constant.

1. Set the ratio constant. l₁A₁ = l₂A₂ because R and ρ are the same.
2. Plug in l₂ = 2l₁. lA = 2lA₂ ⇒ A₂ = 2A.
3. Physical reading. A longer wire would normally have more resistance; making it twice as thick (double area) halves the resistance back to the original, so R stays R.

Option (c): 2A.

Slope is the whole story. For a V (y-axis) versus I (x-axis) plot, resistance is just the slope. Rank the lines by steepness and you have ranked the resistances.

Concept used. From V = IR, plotting V against I gives a straight line through the origin with gradient R. The steeper the line, the larger the resistance. (If I were on the vertical axis instead, the order would flip, so always note which quantity is on which axis.)

1. Confirm the axes. Here V is vertical and I is horizontal, so slope = V/I = R directly.
2. Compare slopes. The line rising most steeply belongs to the highest resistance. In the plot that is R₃, then R₂, then R₁.
3. Write the order. R₃ > R₂ > R₁, which is option (c).

Option (c): R₃ > R₂ > R₁.

Square the factor, not the percentage. The classic trap is to say "current up 100 %, so power up 100 % or 200 %". Power goes as I², so doubling current multiplies power by 2² = 4, a 300 % increase (not 400 %).

Concept used. For a fixed resistance, P = I²R. Scaling current by a factor k scales power by k². The percentage increase is (k² − 1)× 100.

1. Find the factor k. A 100 % rise means I → 2I, so k = 2.
2. Scale the power. P → k² P = 4P.
3. Percentage increase, not the multiple. (4 − 1)× 100 = 300 %. The power becomes four times as large, which is an increase of three times the original.

Option (c): 300 %.

Ask: did the material or temperature change? Resistivity reacts only to those two. If an option leaves both untouched, ρ is safe.

Concept used. Resistivity is an intensive material property; it depends on the type of material and its temperature, never on geometry. Reshaping a conductor alters R through l and A, but the per-cube resistance ρ stays the same.

1. Scan the options for material or temperature. (a), (b) and (d) all change one or both, so each alters ρ.
2. Find the geometry-only change. Option (c) changes only the shape, leaving material and temperature fixed.
3. Conclude. With material and temperature held constant, ρ is unchanged, so (c) is correct.

Option (c): shape change does not affect resistivity.

Parallel: rating equals brightness. On the mains (constant voltage), a bulb's printed wattage is its actual power, so the higher-watt bulb is simply brighter. Rank by rating and test the options.

Concept used. Parallel bulbs share the same voltage V. Power at fixed voltage is P = V²/R, and the rated wattage already assumes this rated voltage. So a 100 W bulb genuinely outshines a 60 W bulb, which outshines a 40 W bulb, when all three sit across the same supply.

1. Same V for all. Parallel guarantees equal voltage, so the rated wattages apply directly.
2. Rank by power. P_C(100) > P_B(60) > P_A(40), hence C brightest, A dimmest.
3. Match the only true statement. Among the options, "B brighter than A" agrees with P_B > P_A. The others contradict the ranking.

Option (c): brightness of B > brightness of A.

Series first, then heat. Two clean steps: get the common current from the total resistance, then apply H = I²Rt only to the 4 Ω resistor, since that is what the question asks about.

Concept used. Series resistors carry the same current I = V/R_series. Joule's heating law H = I²Rt gives the heat in any single resistor using its own R and the shared I.

1. Add the resistances. R_series = 2 + 4 = 6 Ω.
2. Find the shared current. I = 66 = 1 A.
3. Heat in the chosen resistor. H = I² R t = (1)²(4)(5) = 20 J.
4. Quick check via V₄. The 4 Ω resistor drops V₄ = IR = 1×4 = 4 V, so H = V₄ I t = 4×1×5 = 20 J, the same answer.

Option (c): 20 J.

Compute the current, then round up. Two parts: get the working current from I = P/V, then pick a fuse rated just above it. A fuse rated below the working current would melt during normal use.

Concept used. Power P = VI gives the steady current. A fuse is a thin wire that melts (breaking the circuit) once current exceeds its rating, so its rating must sit just above the appliance's normal current, never below.

1. Working current. I = PV = 1000220 ≈ 4.5 A.
2. Rule out too-small fuses. 1 A, 2 A and 4 A are all below 4.5 A, so each would blow during ordinary operation.
3. Pick the next value up. 5 A is just above 4.5 A, so it carries normal current yet still protects against overloads.

Option (d): 5 A.

Two slogans answer everything. "Series shares current, parallel shares voltage." Hold those two and every option falls into place.

Concept used. A series path has a single current common to all elements; a parallel network has a single voltage common to all branches. These are direct consequences of charge conservation (series) and the definition of potential difference between two common nodes (parallel).

1. Series current. One loop ⇒ one current, so the 2 Ω and 4 Ω resistors carry the same I. Option (b) is true.
2. Series voltage. V = IR with common I but different R gives different drops, so option (c) ("same p.d. in series") is false.
3. Parallel. Equal voltage across both branches makes (d) ("different p.d. in parallel") false, and unequal currents make (a) ("same current in parallel") false.

Option (b): same current flows in the series connection.

Power versus energy. The trap mixes power units with energy units. Power is energy per second (watt); multiply power by time and you get energy (kilowatt hour, watt second). Only one option is a power unit.

Concept used. P = VI gives power the unit volt·ampere, which equals the watt. Energy is E = Pt, with units like watt second (joule) and kilowatt hour. Spotting the hidden time factor separates the two families.

1. Build the unit from P = VI. Volt times ampere is the natural unit of power, and it equals one watt.
2. Flag the energy units. "Kilowatt hour" and "watt second" both carry a time unit, so they measure energy, not power.
3. Reject the nonsense unit. "Joule second" is neither power (J/s) nor a standard energy unit; it is the inverted combination. Only (a) survives.

Option (a): volt ampere.

Fix the two meters. In almost every "correct this Ohm's-law circuit" question, the error is the same: ammeter and voltmeter are swapped. Put the ammeter back into the main line and hang the voltmeter across the resistor.

Concept used. An ideal ammeter has very low resistance, so it must sit in series where it does not disturb the current. An ideal voltmeter has very high resistance, so it must sit in parallel across the element whose voltage is to be measured, drawing almost no current.

1. Build the main loop. Cell, key and resistor in one series path, cell terminals connected correctly so current can flow.
2. Ammeter in series. Break the main wire once and insert the ammeter so the full loop current passes through it.
3. Voltmeter in parallel. Connect the voltmeter's two leads only to the two ends of the resistor, without breaking the main loop.
4. Verify with readings. Now a change in R changes the ammeter reading I and the voltmeter reading V together, and V/I stays constant, confirming Ohm's law.

Corrected circuit: cell, key, resistor and ammeter in series; voltmeter in parallel across the resistor.

Find the weakest link. Each resistor has the same melting power (18 W) and the same resistance (2 Ω), but A carries the whole current while B and C split it. So A hits its power limit first; A sets the maximum.

Concept used. Safe current per resistor is I = √P/R. In a series-parallel network the series resistor (A) carries the full current, and the parallel branches (B, C) each carry a share. The overall limit is the smallest of the individual safe currents referred to the line.

1. Safe current of A. I_A = √182 = √9 = 3 A.
2. Where A sits. A is in series with the B∥C block, so the line current equals I_A. Maximum line current = 3 A.
3. Branch currents. B and C are equal, so each takes half: I_B = I_C = 1.5 A. Their power, (1.5)²(2) = 4.5 W, is far below 18 W, confirming A is the bottleneck.

Maximum (line) current = 3 A; B and C each carry 1.5 A.

A good meter does not disturb what it measures. The whole point of an ammeter is to report the true current. If it added noticeable resistance, it would itself lower that current, defeating the purpose.

Concept used. For an in-series ammeter, the loop current is I = V/(R_circuit + R_ammeter). To keep I as close as possible to the true value V/R_circuit, the term R_ammeter must be negligible.

1. Write the loaded current. With the ammeter in series, I = VR_circuit + R_ammeter.
2. Demand minimal disturbance. For I to equal the true current, we need R_ammeter ≪ R_circuit, i.e. as low as possible.
3. State the ideal. An ideal ammeter has zero resistance, so it reads the exact circuit current without altering it.

Low resistance, so the ammeter does not reduce the current it is meant to measure.

Reduce, then compare. The key move is to collapse the two 4 Ω resistors into one equivalent. It turns out to be exactly 2 Ω, the same as the lone series resistor, so the voltages match.

Concept used. Two equal resistors r in parallel give r/2. Here 4/2 = 2 Ω. Series elements share the current, so equal resistances carrying the same current drop equal voltages, V = IR.

1. Collapse the parallel pair. R_p = 42 = 2 Ω.
2. See the series chain. The circuit is now a 2 Ω resistor in series with a 2 Ω block, sharing one current I.
3. Compare the drops. V_2Ω = I× 2 and V_block = I× 2, so the two potential differences are equal.

Yes, the potential differences are equal because both stages have 2 Ω and share the same current.

A deliberate weak point. A fuse is designed to be the first thing to fail. It is the cheapest, most easily replaced part, so the circuit sacrifices the fuse to save the expensive appliance.

Concept used. Heat produced in the fuse is H = I²Rt, so heat rises sharply with current (square law). The fuse wire's material and thickness fix a melting current; above it, the wire melts in a fraction of a second and opens the circuit.

1. Series placement. The fuse carries the same current as the appliance, so it senses any overload directly.
2. Square-law heating. A doubled current quadruples the heating in the fuse, so even a moderate overload melts it quickly.
3. Circuit breaks. Once melted, the gap stops all current. The appliance, now carrying no current, is protected from overheating, insulation damage and fire.

The fuse wire melts on excess current, breaking the circuit and protecting the appliance.

Define, then track the proportionality. First give the clean definition of resistivity. Then for the numerical part, note that R is proportional to length, so doubling length doubles R and, at fixed voltage, halves the current.

Concept used. ρ is an intensive material property defined by R = ρ l/A. Holding V, ρ and A fixed, I = V/R ∝ 1/l. So current is inversely proportional to length.

1. State the definition. Resistivity is the resistance of a conductor of unit length and unit cross-sectional area, measured in Ωm.
2. Length doubles ⇒ R doubles. From R ∝ l, the new resistance is 2R.
3. Current halves. With the same battery voltage, I_new = V2R = I_old2 = 52 = 2.5 A.

ρ = resistance of a unit-length, unit-area conductor; the current halves to 2.5 A because doubling the length doubles the resistance.

Energy, not power. The "unit" on an electricity bill is a unit of energy, the kilowatt hour, not a unit of power. Convert kilowatts to watts and hours to seconds, then multiply.

Concept used. Electrical energy E = Pt. The joule (W·s) is tiny for household use, so suppliers bill in kilowatt hours. One kWh is simply P = 1000 W sustained for t = 3600 s.

1. Name the unit. The commercial unit is the kilowatt hour (kWh), one "unit" of electricity.
2. Convert to SI. 1 kW = 1000 W and 1 h = 3600 s.
3. Compute joules. 1 kWh = 1000 × 3600 = 3.6×10⁶ J.

kilowatt hour (kWh); 1 kWh = 3.6×10⁶ J.

Two parts, two ideas. Part one is plain Ohm's law subtraction to get the lamp resistance. Part two tests whether you know that a parallel branch does not disturb the voltage across an existing branch.

Concept used. Series: R_total = V/I, and resistances add, so R_lamp = R_total − R_conductor. Parallel: branches between the same two nodes share the same voltage, so adding a new branch leaves the original branch's voltage (and hence its current) unchanged.

1. Lamp resistance. R_total = 101 = 10 Ω,    R_lamp = 10 − 5 = 5 Ω.
2. Voltage across the original branch is fixed. The series branch is still connected straight across the 10 V battery, so its terminal voltage stays 10 V.
3. No change in that branch. With the same 10 V across the same 10 Ω branch, its current is still 1 A, so the current in the 5 Ω conductor and the voltage across the lamp are both unchanged. The new 10 Ω simply draws its own extra current straight from the battery.

Lamp = 5 Ω; the parallel 10 Ω causes no change in the 5 Ω branch current or the lamp voltage, because the branch voltage stays at 10 V.

Picture your home. A fan, a fridge and a bulb all need the same 220 V, all must switch independently, and one failing must not switch off the rest. Only parallel wiring does all three.

Concept used. Parallel branches share the same voltage and have independent current paths. Series wiring would force one current through everything, drop the voltage across each device and switch them all off together if one broke. Parallel removes every one of those problems.

1. Same full voltage. Each branch sits across the supply, so every appliance gets its rated 220 V regardless of the others.
2. Independent operation. A separate switch in each branch turns that appliance on or off without disturbing the rest.
3. Fault tolerance and own current. If one appliance fails (its branch opens), the others still have complete paths and keep running, and each draws only the current it needs.

Because parallel wiring gives every appliance the full voltage and an independent path, so devices work at rated voltage and switch on or off separately.

Each branch is independent. The power of parallel wiring is that each bulb lives on its own branch. Removing one branch (a fused bulb) leaves the others exactly as they were, because their voltage never changes.

Concept used. Identical parallel bulbs each see the full supply voltage and carry equal currents, I_branch = I_total/n. The main current is the sum of branch currents, and the total power is P = VI_total.

1. Branch currents. 3 A among three identical bulbs gives 1 A each.
2. (i) B₁ fused. B₂ and B₃ still sit across the same 4.5 V, so each still carries 1 A and glows just as brightly as before. Brightness unchanged.
3. (ii) B₂ fused. A₂ = 0 (open branch). A₁ and A₃ stay at 1 A each. Main ammeter A = 1 + 1 = 2 A.
4. (iii) Total power, all glowing. P = VI = 4.5× 3 = 13.5 W.

(i) other two unchanged in brightness; (ii) A₁=1 A, A₂=0, A₃=1 A, A=2 A; (iii) P = 13.5 W.

Compare resistance, then test the broken bulb. Part (a) is decided by total resistance for the same supply; part (b) is decided by whether breaking one bulb breaks the only path or just one of several.

Concept used. For a common source voltage V, the set's power is P = V²/R_eq. Series gives R_eq = 3R (low power, dim); parallel gives R_eq = R/3 (high power, bright). A series loop has one path; a parallel network has independent paths.

1. Equivalent resistances. Series: 3R. Parallel: R/3. The parallel value is nine times smaller.
2. Brightness. P = V²/R_eq, so the parallel set dissipates about nine times the total power of the series set. Parallel bulbs are clearly brighter, so brightness is not the same.
3. (b) Series break. One fused bulb opens the single loop; current drops to zero and every bulb goes dark.
4. (b) Parallel break. One fused bulb removes only its own branch; the remaining two still sit across the full V and glow unchanged.

(a) No, parallel bulbs are brighter than series bulbs. (b) Series: all go out; Parallel: the rest keep glowing.

State, verify, then qualify. A full-mark answer has three parts: the precise statement (with the constant-temperature condition), a clear experiment with a graph, and an honest note that the law has limits.

Concept used. Ohm's law V = IR is an empirical relation valid for metallic (ohmic) conductors at constant temperature. The straight-line V-I graph is its signature, and a curved graph signals a non-ohmic device.

1. Statement with the key condition. At constant temperature, V ∝ I, so V = IR with R constant. The condition "constant temperature" must be stated.
2. Circuit. Resistor, battery, key and ammeter in series; voltmeter across the resistor. A rheostat or extra cells let you change V.
3. Take readings and plot. Record several (V, I) pairs. The constant ratio V/I = R and a straight line through the origin verify the law; the slope gives R.
4. Limits. The law breaks when temperature changes (a wire heated by large current has higher R) and for non-ohmic elements like filament lamps, diodes and semiconductors, whose V-I graphs are curved.

V = IR at constant temperature; verified by a straight-line V-I graph; not valid under all conditions (fails with temperature change and for non-ohmic devices).

Change one factor at a time. A clean experiment isolates each variable: vary length alone, then area alone, then material alone, while keeping the others fixed. That is how you prove what R depends on.

Concept used. R = ρ l/A. To test the role of each symbol, hold the other two constant and watch how R responds. Resistivity ρ (Ωm) is the material constant left over once geometry is fixed.

1. Define and unit. ρ is the resistance of a unit-cube of the material; unit Ωm, depending only on material and temperature.
2. Measure R. Use ammeter (series) and voltmeter (across the wire); R = V/I.
3. Length test. Same material and area, different lengths ⇒ R rises with l (R ∝ l).
4. Area test. Same material and length, different areas ⇒ R falls as A rises (R ∝ 1/A).
5. Material test. Same length and area, different metals ⇒ different R, so R depends on ρ. Together: R = ρ l/A.

ρ = resistance of a unit-length, unit-area conductor (Ωm); the experiment shows R ∝ l, R ∝ 1/A, and material dependence, i.e. R = ρ l/A.

One meter, several positions. The experiment is simply to move the ammeter around the series loop and show its reading never changes. Identical readings everywhere is the proof.

Concept used. A series circuit has a single conducting path. By conservation of charge, the charge per second (current) crossing any cross-section must be the same. An ammeter reads exactly that current wherever it is inserted in the loop.

1. Build the series loop. Battery, key, R₁, R₂, R₃ all in one line.
2. Insert and read. Put the ammeter before R₁ and record I. Then shift it to the gaps after R₁, after R₂, and after R₃, recording each time.
3. Compare. All readings are equal. Equal current at every point is exactly what "same current flows through every part" means, confirming the series property.

Move one ammeter to different points of the series circuit; equal readings everywhere prove the current is the same through every part.

One voltmeter, three branches. Measure the voltage across each parallel resistor in turn. If all three readings agree (and match the battery), the equal-voltage property is proved.

Concept used. Parallel branches share the same pair of nodes, so the potential difference between those nodes is common to all branches. A voltmeter, connected across each branch, reports this shared voltage.

1. Wire three resistors in parallel across the battery, key in the main line.
2. Sweep the voltmeter. Read the voltage across R₁, then R₂, then R₃.
3. Compare. All three readings are equal and equal the battery voltage. Equal voltage across each branch is the defining property of a parallel arrangement, so the conclusion follows.

Connect a voltmeter across each parallel resistor; equal readings (equal to the supply voltage) confirm the same potential difference across each.

Law, demo, then everyday devices. A complete answer states H = I²Rt, describes a water-heating experiment that proves it, and lists four common heating appliances.

Concept used. Joule's law H = I²Rt quantifies the electrical energy dissipated as heat in a resistor. The square dependence on current is why heating appliances use thick, high-resistance nichrome elements and draw large currents.

1. State the law. H = I²Rt: heat rises with the square of current, with resistance, and with time.
2. Calorimeter demonstration. A nichrome coil in water, fed by a measured current for a measured time, raises the water temperature. The measured heat gained equals I²Rt, proving the law.
3. Why nichrome. It has high resistivity and a high melting point, so it produces a lot of heat (I²R) without melting, ideal for heating elements.
4. Four applications. Electric heater, electric iron, geyser, and bulb filament (also toasters and ovens) all rely on this effect.

Heating effect: H = I²Rt; show by heating water with a current-carrying coil; applications include the electric heater, iron, geyser and bulb filament.

Reduce, then march through (a) to (e). Collapse the parallel pair first; everything else is one chain of Ohm's-law and power steps, since the 4 Ω resistor carries the whole line current.

Concept used. Two equal resistors r in parallel give r/2, so the 8 Ω pair becomes 4 Ω. In the resulting series circuit, I = V/R_total, V = IR and P = I²R, and series elements share one common current.

1. (a) Parallel block. R_p = 8 × 816 = 4 Ω.
2. Total and current. R_total = 4 + 4 = 8 Ω, so I = 88 = 1 A. (b) the 4 Ω resistor carries this 1 A.
3. (c) and (d). V₄ = IR = 1×4 = 4 V; P = I²R = 1×4 = 4 W.
4. (e) Series means equal current. The 4 Ω resistor is in series with the parallel block, so an ammeter before and after the block reads the same 1 A. No difference.

(a) 4 Ω; (b) 1 A; (c) 4 V; (d) 4 W; (e) no difference (series, same current).