Let Us Assess, Q1–Q6 – Solutions വിലയിരുത്താം
The first six end-of-chapter questions: which source turns chemical energy into electrical energy, what the two terminals of a cell need, the ammeter and voltmeter of Fig 7.34, the match-the-columns Table 7.10, the work of 50 J on 5 C, and the bulbs of Fig 7.35. Try each one in your science diary സയൻസ് ഡയറി first, then step through the solution.
- A continuous current needs a potential difference പൊട്ടൻഷ്യൽ വ്യത്യാസം kept up by a source of electricity വൈദ്യുതസ്രോതസ്സ്. V = W/Q, unit J/C = volt (V) (page 5).
- Sources of emf and their energy changes; cells in series and parallel (page 6).
- I = Q/t, unit ampere (A). The ammeter അമ്മീറ്റർ goes in series, the voltmeter വോൾട്ട്മീറ്റർ in parallel; each meter's + terminal faces the cell's + side (page 7).
- R = V/I, unit ohm (Ω) (page 8). Series and parallel connections (page 10).
Question 1 · Chemical energy → electrical energy
1. താഴെ കൊടുത്തവയിൽ രാസോർജത്തെ വൈദ്യുതോർജമാക്കുന്നത് ഏത്? Which of the following changes chemical energy രാസോർജം into electrical energy വൈദ്യുതോർജം?
Ask of each one: what goes in? A dynamo needs something to turn it; a solar cell needs light; a dry cell needs nothing but the chemicals inside it.
(b) a dynamo changes mechanical energy (the turning wheel) into electrical energy; (c) a solar cell changes light energy into electrical energy.
All three give electricity, so all three “look right”. The question asks about the input energy. Only the cell starts from chemical energy – that is why a dry cell is used up when its chemicals are spent.
Question 2 · What must the two terminals have?
2. ഒരു അടഞ്ഞ സെർക്കീട്ടിൽ വൈദ്യുതപ്രവാഹം ഉണ്ടാകണമെങ്കിൽ അതിലെ സെല്ലിന്റെ രണ്ട് ടെർമിനലുകൾ For an electric current to flow in a closed circuit, the two terminals ടെർമിനലുകൾ of the cell in it must
Remember the two water pots of Fig 7.3 and 7.4: water flowed only when the levels were different. What plays the role of the “level” for electric charges?
Model: the bulb is treated as a fixed resistance of 6 Ω, so I = (difference in potential) ÷ 6 Ω. Put both terminals at 9 V – both are at a “high potential”, yet nothing flows. That rules out option (a).
(a) is wrong: two terminals at the same high potential have zero difference – no current. (c) a temperature difference makes heat flow; (d) a height difference makes water flow (Table 7.1) – not electric charge.
Question 3 · Fig 7.34 – which circuit has both meters right?
3. താഴെക്കൊടുത്തവയിൽ അമ്മീറ്ററും വോൾട്ട്മീറ്ററും ശരിയായ രീതിയിൽ ബന്ധിപ്പിച്ച സെർക്കീട്ട് ഏത്? In which of the circuits below (Fig 7.34) are the ammeter അമ്മീറ്റർ and the voltmeter വോൾട്ട്മീറ്റർ connected correctly?
Check four things in each circuit: (1) is the ammeter in series? (2) is the voltmeter in parallel with the bulb? (3) and (4) does the current enter each meter at its + terminal? First find which side of the battery is + (the long line).
Model (Fig 7.34 gives no values): battery of two 1.5 V cells = 3 V; the bulb is a fixed 6 Ω; ideal ammeter (almost 0 Ω); voltmeter with a very large resistance.
| Circuit | Ammeter position | Ammeter + terminal | Voltmeter position | Voltmeter + terminal | Verdict |
|---|---|---|---|---|---|
| (a) | in series ✔ | current enters at + ✔ | in series ✘ | (enters at +) | ✘ |
| (b) | across the bulb ✘ | – | in series ✘ | – | ✘ |
| (c) | in series ✔ | + towards the cell's + ✔ | across the bulb ✔ | + on the higher-potential end ✔ | ✔ correct |
| (d) | in series ✔ | reversed ✘ | across the bulb ✔ | reversed ✘ | ✘ |
In (c) the ammeter is in series and the current enters it at its + terminal (the + is on the side joined to the cell's +); the voltmeter is in parallel with the bulb, with its + on the right-hand end of the bulb, which is at the higher potential. Both are correct.
(a) the voltmeter is in series ✘. (b) the ammeter is across the bulb and the voltmeter is in series ✘. (d) both meters are in the right places, but both are connected the wrong way round (+ and − reversed) ✘.
Looking only at where the meters are and ignoring the + and − marks – that makes (d) look correct. An analogue meter connected the wrong way round tries to swing backwards and hits its stop (it can be damaged); a digital meter shows a minus sign.
An ammeter has almost no resistance – put it across a component (as in (b)) and it becomes a short circuit. A voltmeter has a very large resistance – put it in series (as in (a)) and it almost stops the current, so the bulb goes out. This is the most common practical-exam mistake.
Question 4 · Match the columns (Table 7.10)
4. A കോളത്തിലുള്ളവയ്ക്ക് ചേരുന്നവ B, C എന്നീ കോളങ്ങളിൽ നിന്ന് എടുത്ത് എഴുതുക. For each item in column A, pick the matching items from columns B and C and write them.
| A | B | C |
|---|---|---|
| potential difference (V) പൊട്ടെൻഷ്യൽ വ്യത്യാസം | Q/t | ohm (Ω) ഓം |
| current (I) കറന്റ് | W/Q | volt (V) വോൾട്ട് |
| resistance (R) പ്രതിരോധം | V/I | ampere (A) ആമ്പിയർ |
Table 7.10
Each quantity is defined by a fraction: p.d. = work per unit charge; current = charge per unit time; resistance = p.d. per unit current. The unit is named after a scientist.
Matched: 0 / 3
| A | B | C |
|---|---|---|
| potential difference (V) | W/Q | volt (V) |
| current (I) | Q/t | ampere (A) |
| resistance (R) | V/I | ohm (Ω) |
Mixing up the letter V (the quantity, potential difference) with the unit V (volt), and the letter A for ampere with “area”. In “V = 6 V” the first V is the quantity, the second is the unit.
Question 5 · 50 J to carry 5 C from M to N
5. ഒരു വൈദ്യുത സെർക്കീട്ടിലെ M എന്ന ബിന്ദുവിൽ നിന്ന് 5 C വൈദ്യുത ചാർജ് N എന്ന ബിന്ദുവിൽ എത്തിക്കാൻ 50 J പ്രവൃത്തി ചെയ്തുവെങ്കിൽ M, N എന്നീ ബിന്ദുക്കൾക്കിടയിലെ പൊട്ടെൻഷ്യൽ വ്യത്യാസം എത്ര? If 50 J of work is done to carry an electric charge of 5 C from a point M to a point N in an electric circuit, what is the potential difference between M and N?
Potential difference = work done per unit charge. Which number goes on top?
Formula: V = W/Q.
Substitution: V = 50 J ÷ 5 C.
Answer: V = 10 V (10 J of work for every coulomb).
Writing Q/W = 5/50 = 0.1. Check the unit: the volt is joule per coulomb (J/C), so work goes on top. 1 V means 1 J for each 1 C.
Question 6 · Fig 7.35 – which bulbs are in series?
6. താഴെക്കൊടുത്തവയിൽ ബൾബുകളെ ശ്രേണീരീതിയിൽ ബന്ധിപ്പിച്ച സെർക്കീട്ടുകൾ ഏവ? In which of the circuits below (Fig 7.35) are the bulbs connected in series ശ്രേണീരീതി?
Put your finger on the battery's + terminal and follow the wire. If you must pass through both bulbs, one after the other, before getting back – series. If the path splits and each bulb has its own route – parallel സമാന്തരരീതി.
Score: 0 / 0
In (a) and (d) the two bulbs are on separate branches joined to the same two points, so each bulb gets the battery's full potential difference – they are in parallel. (In (c) the second bulb is drawn lower and the wire bends, but it is still one single loop – the shape of the drawing does not matter, only the paths.)
Imagine unscrewing one bulb. In a series circuit the other bulb goes out too (the only path is broken). In a parallel circuit the other bulb stays lit – that is why the lights and fans at home are in parallel.