Chapter 7 · Solved paper40 marks

Model Test – Solved ഉത്തരങ്ങൾ

Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated circuit or working for every numerical and explanation question, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.

📐 How the figures are drawn

Circuits use the symbols of Table 7.5: in a cell the long thin line is + and the short thick line is −. The amber arrowheads show the conventional current (കറന്റ്), from + to − in the outer circuit; more current = faster arrows. The part of the circuit being worked on is drawn in amber. Meters are ideal, and every number is worked out with V = W/Q, I = Q/t, V = IR and the series and parallel rules.

SECTION AAnswer all · 8 × 1 = 8 marks

The SI unit of electric charge is .

(1 mark)

💡 Named after Charles-Augustin de Coulomb.

✅ Answer: coulomb (C) (കൂളോം (C))

Electric charge (വൈദ്യുതചാർജ്) is measured in coulomb (C). The charge of one electron is negative and very small.

Marking: 1 mark.

📖 Review this on Page 1 · Charges and the coulomb

Which of the following changes light energy into electrical energy?

(a) dry cell(b) dynamo(c) solar cell(d) car battery

(1 mark)

💡 Which one works only in light?

✅ Answer: (c) solar cell

A solar cell: light → electrical. A dry cell and a car battery: chemical → electrical; a dynamo: mechanical → electrical (Table 7.2).

Marking: 1 mark for (c).

📖 Review this on Page 6 · Table 7.2, sources of emf

The potential difference between the terminals of a cell when no current is drawn from it is called its .

(1 mark)

💡 The open-circuit potential difference of a source.

✅ Answer: emf (electromotive force) (വിദ്യുത്ചാലകബലം (emf))

The emf is the potential difference between the terminals of a source in an open circuit – the largest potential difference it can give (key box 5). Its unit is the volt.

Marking: 1 mark.

📖 Review this on Page 5 · Key box 5: emf

Find the relation and fill in: potential difference : volt :: electric current :

(1 mark)

💡 Named after André-Marie Ampère.

✅ Answer: ampere (A) (ആമ്പിയർ (A))

Potential difference is measured in volt; electric current in ampere (A) = coulomb per second (key box 10). Resistance is measured in ohm (Ω).

Marking: 1 mark.

📖 Review this on Page 7 · Key box 10: the ampere

Three identical 1.5 V cells are connected in parallel (+ to +, − to −). The potential difference they give is

(a) 0.5 V(b) 1.5 V(c) 3 V(d) 4.5 V

(1 mark)

💡 Parallel cells do not add their emfs.

✅ Answer: (b) 1.5 V

Identical cells in parallel (സമാന്തരരീതി) give the emf of one cell, 1.5 V – they only make it last longer (key box 9). In series they would give 3 × 1.5 = 4.5 V.

Marking: 1 mark for (b).

📖 Review this on Page 6 · Combination of cells, key box 9

Name the component whose circuit symbol is shown in the figure.

(1 mark)

💡 Look at the two little arrows – it gives out light.

✅ Answer: LED (light emitting diode)

A triangle with a bar, and two small arrows pointing outwards for the light it gives out: an LED (Table 7.5). The triangle points in the direction of the current – an LED lights only when connected the right way round.

Marking: 1 mark.

📖 Review this on Page 7 · Table 7.5, circuit symbols · Page 4 · Symbols

An electric current in a metal wire is a flow of

(a) protons(b) free electrons(c) positive ions(d) whole atoms

(1 mark)

💡 Which particles are free to move in a metal?

✅ Answer: (b) free electrons

A metal has many free electrons (സ്വതന്ത്ര ഇലക്ട്രോണുകൾ); a source makes them drift from − to +. The conventional current is taken from + to − (key box 11).

Marking: 1 mark for (b).

📖 Review this on Page 7 · Current in a conductor (Fig 7.17) · Page 1 · Conductors

The length of a wire is doubled, while its material and thickness stay the same. Its resistance

(a) becomes half(b) stays the same(c) doubles(d) becomes four times

(1 mark)

💡 Longer path → more collisions.

✅ Answer: (c) doubles

Resistance (പ്രതിരോധം) is proportional to length: a longer conductor has more resistance (Fig 7.25, Table 7.9). Double the length → double the resistance.

Marking: 1 mark for (c).

📖 Review this on Page 9 · Factors affecting resistance

SECTION BAnswer all · 5 × 2 = 10 marks

12 C of charge flows through a bulb in 4 s. (a) Find the current through the bulb. (b) How much charge flows through it in one minute?

(2 marks)

💡 Current = charge ÷ time; change 1 minute into 60 s.

🎬 Step by step

✅ Answer: (a) 3 A (b) 180 C

Given: Q = 12 C, t = 4 s

Formula: I = Q ÷ t

Substitution: 12 ÷ 4

Answer: (a) I = 3 A

Given: I = 3 A, t = 1 min = 60 s

Formula: Q = I × t

Substitution: 3 × 60

Answer: (b) Q = 180 C

Marking: (a) 1 mark (formula ½, answer with unit ½); (b) 1 mark.

📖 Review this on Page 7 · I = Q/t · Page 11 · 10 C in 5 s, 1.5 A for 3 s

10.

24 J of work is done in moving 4 C of charge from a point A to a point B of a circuit. (a) Find the potential difference between A and B. (b) How much work is needed to move 10 C of charge between the same two points?

(2 marks)

💡 1 V means 1 J of work for every coulomb.

🎬 Step by step

✅ Answer: (a) 6 V (b) 60 J

Given: W = 24 J, Q = 4 C

Formula: V = W ÷ Q

Substitution: 24 ÷ 4

Answer: (a) V = 6 V

Given: V = 6 V, Q = 10 C

Formula: W = V × Q

Substitution: 6 × 10

Answer: (b) W = 60 J

Marking: (a) 1 mark; (b) 1 mark.

📖 Review this on Page 5 · V = W/Q, the volt · Page 12 · Let us assess Q5

11.

Draw the circuit diagram of a circuit with a cell, a switch, a bulb, an ammeter to measure the current through the bulb and a voltmeter to measure the potential difference across the bulb. Mark the + and − terminals of both meters.

(2 marks)

💡 Ammeter: in the same loop. Voltmeter: across the bulb.

🎬 Step by step

✅ Answer: ammeter in series, voltmeter in parallel with the bulb, + terminals towards the + of the cell

Draw the cell, switch, bulb and ammeter in series in one loop, and the voltmeter in parallel (across) the bulb (Table 7.6, Fig 7.21). The + terminal of each meter goes on the side connected to the + terminal of the cell (the long line); current flows from + to − in the outer circuit.

Marking: ½ ammeter in series, ½ voltmeter in parallel, ½ correct symbols, ½ correct + and − marks.

📖 Review this on Page 7 · Table 7.6, Fig 7.21 · Page 12 · Let us assess Q3

12.

The figure shows two 1.5 V cells joined in three ways, (a), (b) and (c), each with a voltmeter. Write the reading of the voltmeter in each case. Which arrangement must never be used?

(2 marks)

💡 Series adds, parallel keeps one cell’s emf, opposing cancels.

🎬 Step by step

✅ Answer: (a) 3 V (b) 1.5 V (c) 0 V; never use (c)

(a) Series (+ of one cell to − of the other): the emfs add, 1.5 + 1.5 = 3 V.

(b) Parallel (+ to +, − to −): the emf of one cell, 1.5 V.

(c) Joined + to +: the two cells oppose each other, 0 V. Such an arrangement must never be used (Fig 7.13 (c), Table 7.4).

Marking: ½ mark for each reading and ½ for naming (c).

📖 Review this on Page 6 · Table 7.4, Fig 7.13 · Page 14 · Battery lab

13.

A fan and its regulator are connected in series, but the fan and the light of a room are connected in parallel. Give a reason for each.

(2 marks)

💡 What is the same in series? What is the same in parallel?

🎬 Step by step

✅ Answer: regulator in series to control the fan’s current; fan and light in parallel so each gets 230 V and works independently

Fan and regulator in series (ശ്രേണീരീതി): the same current flows through both, so changing the regulator’s resistance changes the current through the fan – and its speed (Let us assess Q8).

Fan and light in parallel (സമാന്തരരീതി): each gets the full 230 V, each can be switched on and off by itself, and if one fails the other keeps working.

Marking: 1 mark for each reason.

📖 Review this on Page 13 · Let us assess Q8 · Page 10 · Household circuits · Page 14 · Wiring and regulator labs

SECTION CAnswer all · 4 × 3 = 12 marks

14.

The table shows the readings of an experiment with a resistor, at constant temperature.

Potential difference V (volt)	1.5	3.0	4.5	6.0
Current I (ampere)	0.25	0.50	0.75	?

(a) Find the resistance of the resistor. (b) Find the missing current. (c) State Ohm’s law.

(3 marks)

💡 V ÷ I is the same in every column.

🎬 Step by step

✅ Answer: (a) 6 Ω (b) 1.0 A (c) I ∝ V at constant temperature

Given: V = 1.5 V, I = 0.25 A (any pair of readings)

Formula: R = V ÷ I

Substitution: 1.5 ÷ 0.25 = 3.0 ÷ 0.50 = 4.5 ÷ 0.75

Answer: (a) R = 6 Ω (V/I is the same for every pair)

Given: V = 6.0 V, R = 6 Ω

Formula: I = V ÷ R

Substitution: 6.0 ÷ 6

Answer: (b) I = 1.0 A

(c) Ohm’s law (ഓം നിയമം): at constant temperature, the current through a conductor is directly proportional to the potential difference between its ends (V/I = constant = R).

Marking: 1 mark for each part.

📖 Review this on Page 8 · Ohm’s law, Table 7.7

15.

A 5 Ω and a 10 Ω resistor are connected in series with a 4.5 V battery and a switch (figure). (a) Find the effective resistance. (b) Find the current in the circuit. (c) Find the potential difference across each resistor.

(3 marks)

💡 In series, add the resistances; the current is the same everywhere.

🎬 Step by step

✅ Answer: (a) 15 Ω (b) 0.3 A (c) 1.5 V and 3 V

Given: R₁ = 5 Ω, R₂ = 10 Ω in series, V = 4.5 V

Formula: R = R₁ + R₂

Substitution: 5 + 10

Answer: (a) R = 15 Ω

Given: V = 4.5 V, R = 15 Ω

Formula: I = V ÷ R

Substitution: 4.5 ÷ 15

Answer: (b) I = 0.3 A

(c) The same 0.3 A flows through both: V₁ = 0.3 × 5 = 1.5 V; V₂ = 0.3 × 10 = 3 V. Check: 1.5 + 3 = 4.5 V. The bigger resistor takes the bigger share.

Marking: 1 mark for each part ((c): ½ + ½).

📖 Review this on Page 10 · Series combination (Fig 7.29) · Page 13 · Let us assess Q7

16.

A 6 Ω and a 12 Ω resistor are connected in parallel to a 6 V battery, with an ammeter in the main wire (figure). (a) Find the effective resistance. (b) Find the current through each resistor. (c) What does the ammeter read?

(3 marks)

💡 In parallel the potential difference is the same; the currents add.

🎬 Step by step

✅ Answer: (a) 4 Ω (b) 1 A and 0.5 A (c) 1.5 A

Given: R₁ = 6 Ω, R₂ = 12 Ω in parallel, V = 6 V

Formula: R = R₁R₂ ÷ (R₁ + R₂)

Substitution: (6 × 12) ÷ (6 + 12) = 72 ÷ 18

Answer: (a) R = 4 Ω – less than the smaller resistor

(b) Each resistor has the full 6 V across it: I₁ = 6 ÷ 6 = 1 A; I₂ = 6 ÷ 12 = 0.5 A.

Marking: 1 mark for each part.

📖 Review this on Page 10 · Parallel combination (Fig 7.30) · Page 11 · 6 Ω ∥ 3 Ω question

17.

An LED works at 2 V with a current of 20 mA. It is to be lit with a 9 V battery. (a) What potential difference must a resistor in series with it take? (b) Find the resistance needed. (c) What would happen if the LED were connected to the battery without the resistor?

(3 marks)

💡 The resistor takes what the LED does not need; 20 mA = 0.02 A.

🎬 Step by step

✅ Answer: (a) 7 V (b) 350 Ω (c) too much current – the LED burns out

(a) In series the potential differences add: 9 = 2 + V_R, so V_R = 9 − 2 = 7 V.

Given: V_R = 7 V, I = 20 mA = 0.02 A

Formula: R = V ÷ I

Substitution: 7 ÷ 0.02

Answer: (b) R = 350 Ω

(c) The LED would get the full 9 V – far more than it needs – so a very large current would flow and the LED would burn out. That is why the science-fair LED had a resistor in series with it (Fig 7.33).

Marking: 1 mark for each part (convert mA to A!).

📖 Review this on Page 10 · Fig 7.33 and the LED resistor · Page 14 · LED lab

SECTION DAnswer all · 2 × 5 = 10 marks

18.

A uniform nichrome wire 1 m long has a resistance of 8 Ω. (a) What is the resistance of a 50 cm piece of this wire? (b) Another nichrome wire is also 1 m long but has double the area of cross-section. What is its resistance? (c) The 1 m wire (8 Ω) is stretched evenly to 2 m. Find its new resistance. (d) Why is nichrome, and not copper, used for the heating coil of an iron box? (e) How does a rheostat change the current in a circuit?

(5 marks)

💡 R ∝ length and R ∝ 1/area; stretching changes both.

🎬 Step by step

✅ Answer: (a) 4 Ω (b) 4 Ω (c) 32 Ω (d) high resistance (e) changes the length of wire in the circuit

(a) Resistance ∝ length: half the length → 8 ÷ 2 = 4 Ω.

(b) Resistance ∝ 1/area: double the area → 8 ÷ 2 = 4 Ω.

(c) Stretched to double the length, the volume stays the same, so the area becomes half. Length × 2 → R × 2; area ÷ 2 → R × 2 again: 8 × 2 × 2 = 32 Ω (Let us assess Q10).

(d) Nichrome (നിക്രോം) has a high resistance, so it gets very hot when a current flows; copper’s resistance is very low, so a copper coil would hardly heat and would draw a dangerously large current. (Nichrome also does not burn away easily when red-hot.)

(e) A rheostat (റിയോസ്റ്റാറ്റ്) has a long resistance wire wound on a tube and a sliding contact (സമ്പർക്കകം). Moving the slider changes the length of wire in the circuit, so the resistance changes – more length, more resistance, less current (I = V/R).

Marking: 1 mark for each part.

📖 Review this on Page 9 · Factors, rheostat (Fig 7.25–7.27) · Page 13 · Let us assess Q10 · Page 14 · Heater lab

19.

In the circuit in the figure, a 2 Ω resistor (R₁) is in series with a parallel combination of 6 Ω (R₂) and 3 Ω (R₃). The battery gives 3 V. (a) Find the effective resistance of the parallel part. (b) Find the total resistance of the circuit. (c) What does the ammeter read? (d) What does the voltmeter read? (e) Find the current through R₂ and through R₃.

(5 marks)

💡 First replace the parallel pair by one resistor.

🎬 Step by step

✅ Answer: (a) 2 Ω (b) 4 Ω (c) 0.75 A (d) 1.5 V (e) 0.25 A, 0.5 A

Given: R₂ = 6 Ω ∥ R₃ = 3 Ω

Formula: R_p = R₂R₃ ÷ (R₂ + R₃)

Substitution: (6 × 3) ÷ (6 + 3) = 18 ÷ 9

Answer: (a) R_p = 2 Ω

(b) R₁ is in series with the parallel part: R = 2 + 2 = 4 Ω.

Given: V = 3 V, R = 4 Ω

Formula: I = V ÷ R

Substitution: 3 ÷ 4

Answer: (c) I = 0.75 A – the ammeter reading

(d) The voltmeter is across the parallel part: V_p = I × R_p = 0.75 × 2 = 1.5 V. (The other 1.5 V is across R₁: 0.75 × 2 = 1.5 V.)

(e) Both have 1.5 V across them: I₂ = 1.5 ÷ 6 = 0.25 A; I₃ = 1.5 ÷ 3 = 0.5 A. Check: 0.25 + 0.5 = 0.75 A ✔.

Marking: 1 mark for each part.

📖 Review this on Page 10 · Series, parallel and Fig 7.32 · Page 11 · Fig 7.32 question

Marks summary

Section	Questions	Marks each	Total	Topics
A	1–8	1	8	coulomb, solar cell, emf, ampere, cells in parallel, LED symbol, free electrons, resistance and length
B	9–13	2	10	I = Q/t, V = W/Q, drawing a circuit with meters, cell combinations, regulator in series and appliances in parallel
C	14–17	3	12	Ohm’s law from a table, series circuit, parallel circuit, the LED resistor
D	18–19	5	10	factors of resistance, stretched wire, nichrome and the rheostat; a series–parallel circuit with meters
Total			40

⏱️ Time management (15 min cool-off + 90 minutes)

Cool-off (15 min) – read the whole paper. Mark every mA, minute and kΩ that must be converted, and decide which circuits are series and which parallel.
Section A – about 10 minutes. One word or one option; write the unit where one is asked.
Section B – about 20 minutes. Given, Formula, Substitution, Answer for Q9 and Q10; draw Q11 neatly with a ruler and mark + and − on both meters.
Section C – about 25 minutes. Check V/I with every column in Q14; in series add, in parallel use product ÷ sum; 20 mA = 0.02 A.
Section D – about 25 minutes. Replace the parallel pair by one resistor first; answer every part.
Checking – keep 10 minutes: units V, A, Ω, C, J on every answer? parallel resistance smaller than the smallest? branch currents adding up to the main current?

⚠ Never experiment with the mains supply (230 V AC) at home – it can kill. Use only torch cells or a 9 V battery.

🎉 Finished the chapter?

Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5 to 10 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.