Chapter 7 · Let us assessTextbook pages 155–156

Let Us Assess, Q7–Q11 & Extended Activities വിലയിരുത്താം · തുടർപ്രവർത്തനങ്ങൾ

The last five end-of-chapter questions – the 1 Ω and 2 Ω circuit of Fig 7.36, the fan and its regulator, the four voltmeters of Fig 7.37, a wire stretched to double its length and the graphs of Graph 7.2 – followed by the three extended activities തുടർപ്രവർത്തനങ്ങൾ: the resistance from Graph 7.3, three LEDs on a 9 V battery, and an old and a new torch cell on a multimeter.

📖 What you need (pages 8–10 of this guide)

Question 7 · Fig 7.36 – 1 Ω and 2 Ω on 6 V

7. താഴെക്കൊടുത്ത സെർക്കീട്ട് വിശകലനം ചെയ്ത് ചോദ്യങ്ങൾക്ക് ഉത്തരം എഴുതുക. Analyse the circuit below (Fig 7.36: a 1 Ω and a 2 Ω resistor, a 6 V battery and a switch) and answer the questions.

a) സ്വിച്ച് ഓണാക്കിയാൽ സെർക്കീട്ടിലെ സഫലപ്രതിരോധം കണക്കാക്കുക.
b) സെർക്കീട്ടിലൂടെയുള്ള കറന്റ് എത്ര?
c) 2 Ω പ്രതിരോധകത്തിന് ലഭിക്കുന്ന പൊട്ടെൻഷ്യൽ വ്യത്യാസം എത്ര?
d) 1 Ω പ്രതിരോധകത്തിലൂടെ പ്രവഹിക്കുന്ന കറന്റ് എത്ര?
  1. When the switch is turned on, calculate the effective resistance സഫലപ്രതിരോധം of the circuit.
  2. What is the current through the circuit?
  3. What is the potential difference across the 2 Ω resistor?
  4. What is the current flowing through the 1 Ω resistor?

The two resistors are on one single path – series. Add them, then use I = V/R. In series the current is the same everywhere; the p.d. across one resistor is I × (its own R).

🎬 Solution in exam layout – Fig 7.36 with meters added
🧪 Try other numbers – two resistors in series

Ideal meters. The bar shows how the battery's p.d. is shared: the bigger resistor always gets the bigger share, and the two shares add up to the battery's p.d.

Given: R₁ = 1 Ω and R₂ = 2 Ω in series; V = 6 V.
(a) Formula R = R₁ + R₂; substitution R = 1 Ω + 2 Ω; answer R = 3 Ω.
(b) Formula I = V/R; substitution I = 6 V ÷ 3 Ω; answer I = 2 A.
(c) Formula V₂ = I R₂; substitution V₂ = 2 A × 2 Ω; answer V₂ = 4 V.
(d) In series the same current flows through every part: current through 1 Ω = 2 A.
Check: V₁ = 2 A × 1 Ω = 2 V, and 2 V + 4 V = 6 V ✓.
⚠️ Common trap

In (c), using the whole 6 V: “I = 6 ÷ 2 = 3 A”. The 2 Ω resistor does not get the full 6 V – it shares it with the 1 Ω resistor. Always find the current first, then V = IR for that resistor alone.

Question 8 · The fan and its regulator

8. ഒരു വൈദ്യുത സെർക്കീട്ടിൽ ഫാനും ഫാനിന്റെ റെഗുലേറ്ററും തമ്മിൽ ബന്ധിപ്പിക്കുന്നത് ഏത് രീതിയിലാണ്? (സമാന്തരം / ശ്രേണി) In an electric circuit, how are a fan and its regulator connected to each other – in parallel സമാന്തരം or in series ശ്രേണി?

The regulator's job is to change the current through the fan. Which connection lets one component control the current in another?

🎬 Solution animation – regulator in series with the fan
🧪 Turn the regulator

Illustration only (no real values): setting 5 = least resistance in the regulator, so the most current flows and the fan runs fastest. Modern electronic regulators work differently inside, but they are still connected in series with the fan.

In series (ശ്രേണി). The regulator works like a rheostat – a variable resistance. Connected in series, the same current flows through the regulator and the fan, so changing the regulator's resistance changes the current through the fan and therefore its speed: more resistance → less current → slower fan.
If the regulator were in parallel, the fan would always get the full supply voltage (230 V) and its current would not change – the regulator could not control it.
⚠ Never experiment with the mains supply (230 V AC) at home – it can kill. Never open a regulator or switchboard; that is an electrician's job. Use only torch cells or a 9 V battery for experiments.

Question 9 · Fig 7.37 – which voltmeters read 10 V?

9. താഴെക്കൊടുത്ത സെർക്കീട്ടുകളിൽ 10 V പൊട്ടെൻഷ്യൽ വ്യത്യാസം കാണിക്കുന്ന വോൾട്ട്മീറ്റർ ഏതെല്ലാം? In which of the circuits below (Fig 7.37) does the voltmeter show a potential difference of 10 V?

Each circuit has two 10 Ω resistors in series (20 Ω). Find the current from the battery's p.d., then the p.d. across whatever the voltmeter is connected to. Equal resistors in series share the p.d. equally.

🎬 Solution animation – Fig 7.37 redrawn with live voltmeter readings
🧪 Move the voltmeter
(b) and (d). In each circuit R = 10 Ω + 10 Ω = 20 Ω.
(a) I = 10 V ÷ 20 Ω = 0.5 A; voltmeter across one 10 Ω: V = 0.5 × 10 = 5 V.
(b) I = 0.5 A; voltmeter across both: V = 0.5 × 20 = 10 V ✔.
(c) I = 20 V ÷ 20 Ω = 1 A; voltmeter across both: V = 1 × 20 = 20 V.
(d) I = 1 A; voltmeter across one 10 Ω: V = 1 × 10 = 10 V ✔.
💡 Shortcut

Two equal resistors in series share the battery's p.d. equally: each gets half. Across one resistor: half the battery; across both: the whole battery.

Question 10 · A 5 Ω wire stretched to double its length

10. 50 cm നീളമുള്ള ഒരു കമ്പിയുടെ പ്രതിരോധം 5 Ω ആണ്. ഇതിനെ വലിച്ചു നീട്ടി നീളം ഇരട്ടിയാക്കുന്നു എങ്കിൽ The resistance of a wire 50 cm long is 5 Ω. If it is stretched so that its length is doubled,

a) ചാലകത്തിന്റെ ഛേദതല പരപ്പളവിന് (വണ്ണത്തിന്) എന്ത് മാറ്റം സംഭവിക്കും?
b) ചാലകത്തിന്റെ പ്രതിരോധം എത്രയാകും?
  1. what change happens to the area of cross-section (thickness) ഛേദതല പരപ്പളവ് (വണ്ണം) of the conductor?
  2. what will the resistance of the conductor become?

Stretching does not add or remove metal – the volume (area × length) stays the same. Resistance goes up with length and up when the wire gets thinner.

🎬 Solution in exam layout – stretching the wire
🧪 Stretch it yourself
Given: L = 50 cm, R = 5 Ω; new length = 2 × 50 = 100 cm.
(a) The volume of metal stays the same: area × length = constant. The length doubles, so the area of cross-section becomes half – the wire becomes thinner.
(b) Formula: R ∝ length and R ∝ 1/area. Substitution: length × 2 → R × 2; area ÷ 2 → R × 2 again; R = 5 Ω × 2 × 2. Answer: R = 20 Ω (four times the original).
⚠️ Common trap

Answering 10 Ω – doubling for the length but forgetting that the stretched wire is also thinner. Stretching to n times the length makes the resistance n² times.

Question 11 · Graph 7.2 – which graph shows Ohm's law?

11. താഴെക്കൊടുത്തവയിൽ ഓം നിയമത്തെ സൂചിപ്പിക്കുന്ന ഗ്രാഫ് ഏതാണ്? Which of the graphs below (Graph 7.2, V on the Y-axis, I on the X-axis) represents Ohm's law?

Ohm's law: V ∝ I (at constant temperature). What does the graph of a direct proportion look like? Where must it start?

🎬 Solution animation – Graph 7.2 (a)–(d) redrawn
🧪 Ohm's law graph for any resistor
Graph (c) – a straight line through the origin O. Ohm's law says that, at constant temperature, the current is directly proportional to the potential difference: V ∝ I, so V/I = R is constant. A direct proportion is a straight line passing through the origin; its slope is the resistance.
(a) V stays the same whatever the current; (b) the current stays the same whatever the p.d.; (d) V decreases as I increases – none of these is a direct proportion.

Extended activity 1 · Resistance from Graph 7.3

1. ഓം നിയമത്തെ സൂചിപ്പിക്കുന്ന ഗ്രാഫ് ആണ് താഴെ കൊടുത്തത്. ഗ്രാഫ് വിശകലനം ചെയ്ത് ഗ്രാഫിൽ നിന്ന് പ്രതിരോധം കണ്ടുപിടിക്കുക. The graph below (Graph 7.3: voltage വോൾട്ടത (V) against current കറന്റ് (A)) represents Ohm's law. Analyse the graph and find the resistance from it.

Read any point on the line, for example where I = 1.5 A. R = V/I. Then check with another point – you should get the same value.

🎬 Solution in exam layout – reading Graph 7.3
🧪 Pick any point on the line
Given (from the graph): (0.5 A, 1.5 V), (1.0 A, 3.0 V), (1.5 A, 4.5 V); the line passes through O.
Formula: R = V/I (= slope of the V–I line).
Substitution: R = 4.5 V ÷ 1.5 A (also 3.0 ÷ 1.0 and 1.5 ÷ 0.5).
Answer: R = 3 Ω.

Extended activity 2 · Three 3 V LEDs on a 9 V battery

2. 3 V ന്റെ മൂന്ന് LED കൾ, 9 V ബാറ്ററി ഉപയോഗിച്ച് പ്രകാശിപ്പിക്കുന്നതിന്റെ സെർക്കീട്ട് ഡയഗ്രം വരച്ച്, സെർക്കീട്ട് നിർമ്മിച്ച് പ്രവർത്തിപ്പിക്കുക. Draw the circuit diagram സെർക്കീട്ട് ഡയഗ്രം for lighting three 3 V LEDs using a 9 V battery, then build the circuit and make it work.

Each LED needs 3 V. How can three of them share 9 V? And remember: an LED lets current through only one way – its long leg is +.

🎬 Solution animation – drawing the circuit
🧪 Test your circuit
Connect the three LEDs in series. In series the battery's p.d. is shared: 3 V + 3 V + 3 V = 9 V, so each LED gets the 3 V it needs.
Polarity: the long leg (+, anode) of the first LED goes towards the battery's +; the short leg (−) of each LED goes to the long leg of the next; the short leg of the last LED goes (through the switch) to the battery's −.
In parallel each LED would get the full 9 V – far too much – and would burn out. If one LED is turned the wrong way, it blocks the current and none of them glows.
🔧 Build card – LED chain (science fair circuit)

You need: a 9 V battery with a snap connector, three 3 V LEDs (white or blue LEDs need about 3 V), a small switch, connecting wires, a piece of cardboard or a breadboard, insulation tape. Optional but safer: one small resistor (for example 22–47 Ω, approx.).

  1. Look at each LED: the longer leg is + (anode), the shorter leg is − (cathode); the − side of the rim is often flat.
  2. Twist or plug the short leg of LED 1 to the long leg of LED 2, and the short leg of LED 2 to the long leg of LED 3.
  3. Join the long leg of LED 1 to the red (+) wire of the battery snap (through the small resistor, if you use one).
  4. Join the short leg of LED 3 to one terminal of the switch, and the other switch terminal to the black (−) wire.
  5. Check every joint, then clip on the battery and switch on. All three LEDs should glow equally.
  6. Nothing glows? One LED is probably reversed – turn it round. Draw the final circuit diagram in your science diary.

Why the resistor? A fresh 9 V battery can give a little more than 9 V; a small series resistor limits the current so the LEDs are not overdriven (they glow slightly dimmer). This is exactly why the circuit at the school science fair had a resistor next to the LED.

⚠ Never short-circuit a cell or battery – never join the + and − of the 9 V battery with a wire; never put cells in your mouth; stop if a battery gets hot; keep water and fire away; dispose of batteries (nickel, cadmium, lead) at e-waste collection points.
⚠ Never experiment with the mains supply (230 V AC) at home – it can kill. Use only torch cells or a 9 V battery.
🌐 Explore online needs internet · opens in a new tab

Extended activity 3 · A used torch cell and a new one on a multimeter

3. ഉപയോഗശൂന്യമായ ഒരു ടോർച്ച് സെല്ലിന്റെയും പുതിയ ടോർച്ച് സെല്ലിന്റെയും emf ഒരു മൾട്ടിമീറ്റർ ഉപയോഗിച്ച് അളക്കുക. അവയുടെ മൂല്യത്തിൽ എന്തെങ്കിലും വ്യത്യാസം കാണുന്നുണ്ടോ? എന്തായിരിക്കും കാരണം? Measure the emf of a used-up torch cell and of a new torch cell with a multimeter മൾട്ടിമീറ്റർ. Is there any difference in the values? What could be the reason?

emf is measured with no current drawn: set the multimeter to DC volts, touch the red probe to + and the black probe to −. The emf comes from a chemical reaction inside the cell – what happens to the chemicals as the cell is used?

🧪 Multimeter card – measure each cell

Sample readings – your real readings will differ: a new 1.5 V torch cell usually reads about 1.5–1.6 V; a used-up one perhaps about 1.2 V (approx.).

Yes – the used cell gives a smaller reading (for example about 1.2 V against about 1.5–1.6 V for the new cell; approx.).
Reason: the emf of a cell comes from the chemical reaction inside it (chemical energy → electrical energy). As the cell is used, the chemicals that take part in the reaction are used up, so it can no longer maintain the same potential difference between its terminals. A dry cell is a primary cell – it cannot be recharged, so it must be replaced.
🔭 Go further – internal resistance

An old cell also has a larger internal resistance. With no current drawn (the multimeter on volts draws almost none) it may still read close to 1.4 V, but as soon as it drives a torch bulb, part of its p.d. is lost inside the cell and the voltage across the bulb drops sharply – so the torch is dim.

⚠ Never put cells in your mouth or try to open them; dispose of used cells and batteries (nickel, cadmium, lead) at e-waste collection points – not in the dustbin or fire.

Check yourself

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