Chapter 12 · Electricity

NCERT Exemplar Class 10 Science Chapter 12 Short Answer Questions

All 10 questions of NCERT Exemplar Class 10 Science Chapter 12 Short Answer Questions (Electricity), in the book's order, each with its answer and a step-by-step solution.

1. Q19
   A child has drawn the electric circuit to study Ohm’s law as ​shown in Figure 12.6. His teacher told that the ​circuit diagram needs correction. Study the circuit diagram and redraw it after ​making all ​corrections.

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   The ammeter must be in series, the voltmeter across $R$, meter terminals correct and cells aiding; redraw the circuit with these corrections.

   The diagram has four faults. (1) The ammeter is connected in parallel with the resistor $R$; it must be connected in series with $R$. (2) The voltmeter is connected in series in the main loop; it must be connected in parallel across $R$. (3) The ammeter terminals are reversed — the current arrives at its $-$ terminal; the $+$ terminal of each meter must face the $+$ terminal of the battery, so that the current enters each meter at its $+$ terminal. (4) The cells of the battery are not all the same way round — the two middle plates are both long, so two cells oppose the other two and the net emf would be zero; every cell must be joined with its $+$ plate to the $-$ plate of the next. The corrected circuit is a single series loop of battery (all cells aiding), plug key, ammeter and $R$, with the voltmeter joined across $R$ alone; it then gives $R=V/I$.

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2. Q20
   Three 2 ​Ω resistors, ​A, B and C, are connected as ​shown ​in ​Figure 12.7. Each ​of them dissipates energy and can withstand a maximum ​power of 18W without melting. Find the maximum current that can flow through the three resistors?

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   The maximum current is 3 A, limited by resistor A, which carries the whole current and dissipates its rated 18 W.

   The maximum current that can flow is $I=3$ A. At that current resistor A carries the whole $3$ A and dissipates its full rated $18$ W, while B and C each carry $1.5$ A and dissipate only $4.5$ W, so A is the resistor that sets the limit.

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3. Q21
   Should ​the resistance ​of an ammeter be ​low or ​high? ​Give ​reason.
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   An ammeter should have low resistance, ideally zero, so that connecting it in series does not change the current.

   The resistance of an ammeter should be **low**, and ideally zero. An ammeter is connected in series, so its resistance adds to the resistance of the circuit and the current becomes $I=\frac{V}{R+r}$ instead of $\frac{V}{R}$. Only if $r$ is very small compared with $R$ does the meter leave the current essentially unchanged, so that what it reads is the current that was flowing before it was put in.

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4. Q22
   Draw a circuit ​diagram of an electric circuit ​containing a cell, a key, an ammeter, a ​resistor of 2 Ω in series with a combination of two resistors ​(4 Ω each) in parallel and a voltmeter across the ​parallel combination. Will the potential difference across the ​2 Ω resistor be the same as ​that across the parallel combination of 4Ω resistors? Give reason.
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   Yes, the potential differences are equal: the two $4\Omega$ resistors in parallel make $2\Omega$, equal to the series resistor carrying the same current.

   Yes, the two potential differences are the same. The two $4\Omega$ resistors in parallel are equivalent to $R_p=2\Omega$, which is equal to the series resistor. Since the two sections are in series they carry the same current $I$, so by $V=IR$ each has a potential difference of $2I$ — that is, each takes half the potential difference of the cell. With a $6$ V cell, for instance, $I=1.5$ A and each section drops $3$ V.

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5. Q23
   How ​does use ​of ​a fuse ​wire protect ​electrical appliances?
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   A fuse, a high-resistance low-melting-point wire in series, melts when the current is excessive and breaks the circuit, saving the appliance.

   A fuse is a short piece of wire of high resistance and low melting point, connected in series in the live wire, with a current rating slightly above the circuit’s normal working current. It therefore carries the whole current the appliance draws. If a short circuit or an overload makes the current exceed the rating, the heat developed in the fuse, $H=I^{2}Rt$, rises as the square of the current; the fuse wire quickly reaches its melting point and melts, breaking the circuit. The supply is cut off before the large current can overheat the appliance or its wiring, so the fuse is destroyed instead of the appliance.

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6. Q24
   What is electrical resistivity? In a series electrical circuit comprising a resistor ​made up of a metallic wire, ​the ammeter reads 5 ​A. The reading of the ammeter ​decreases ​to half when the length ​of ​the wire is doubled. Why?
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   Resistivity is the resistance of unit length and unit cross-section; doubling the length doubles $R$, halving the current to 2.5 A.

   Electrical resistivity is the resistance of a conductor of that material of unit length and unit area of cross-section: from $R=\frac{\rho l}{A}$ we get $\rho =\frac{RA}{l}$, and its SI unit is the ohm metre ($\Omega$ m). It is a property of the material and its temperature, not of the size or shape of the conductor. When the length of the wire is doubled, $\rho$ and $A$ stay the same, so $R$ doubles. The same potential difference then drives half the current, $I=\frac{V}{2R}$, and the ammeter reading falls from $5$ A to $2.5$ A.

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7. Q25
   What ​is ​the commercial unit ​of electrical energy? Represent it ​in ​terms of ​joules.
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   The commercial unit is the kilowatt hour, and $1\text{kW h}=3.6\times 10^6$ J.

   The commercial unit of electrical energy is the kilowatt hour (kW h), called one "unit" of electricity on an electricity bill. Since $1$ kW $=1000$ W and $1$ h $=3600$ s, $1$ kW h $=1000\times 3600=3.6\times 10^6$ J.

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8. Q26
   A current of ​1 ampere flows in a series circuit containing an electric ​lamp ​and a ​conductor ​of 5 Ω when connected to a 10 V battery. Calculate ​the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and ​potential difference across the lamp will take place? Give reason.
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   The lamp is $5\Omega$; adding the parallel $10\Omega$ resistor changes neither the current through the conductor nor the potential difference across the lamp.

   Resistance of the lamp: the total resistance is $R=V/I=10/1=10\Omega$, so $R_{\text{lamp}}=10-5=5\Omega$. After the $10\Omega$ resistor is joined in parallel with the series combination there is **no** change: the $5\Omega$ conductor still carries $1$ A and the lamp still has $5$ V across it. The reason is that the new resistor is connected across the same two points as the series combination, so the potential difference driving that combination is still $10$ V and its resistance is still $10\Omega$. Only the current drawn from the battery changes, from $1$ A to $2$ A, the extra ampere going through the new branch.

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9. Q27
   Why ​is parallel arrangement ​used in ​domestic wiring?
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   In parallel each appliance gets the full 220 V and works independently, and the supply can deliver the large total current.

   Domestic wiring is in parallel for four reasons. (1) Every appliance is connected directly across the live and neutral wires, so each receives the full supply voltage of $220$ V that it is designed for; in series that voltage would be shared out. (2) Each appliance has its own branch and its own switch, so it can be operated independently and a break in one branch does not stop the others. (3) Each appliance draws the current its own resistance demands, $I=V/R$, so a $100$ W lamp and a $1000$ W heater can work together; in series both would be forced to carry the same current. (4) The equivalent resistance of a parallel combination is smaller than any single branch, so the supply can deliver the large total current the whole household load requires.

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10. Q28
    $B_1$, $B_2$ and $B_3$ are three identical bulbs ​connected as shown in Figure 12.8. ​When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

    1. (i)What happens to the glow of the other two bulbs when the bulb $B_1$ gets fused?
    2. (ii)What happens to the reading of $A_1$, $A_2$, $A_3$ and A ​when the bulb $B_2$ ​gets fused?
    3. (iii)How much power is dissipated in ​the circuit when all the three bulbs glow ​together?

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    The other bulbs keep glowing; when $B_2$ fuses, $A_2$ reads zero and $A$ reads 2 A; the circuit dissipates 13.5 W.

    (i) The other two bulbs, $B_2$ and $B_3$, go on glowing with exactly the same brightness as before. They are connected in parallel directly across the battery, so each still has the full $4.5$ V across it and still carries $1$ A; only the reading of A falls, from $3$ A to $2$ A. (ii) $A_2$ reads zero, because its branch is broken; $A_1$ and $A_3$ are unaffected and each still reads $1$ A; and A reads $1+1=2$ A instead of $3$ A. (iii) $P=VI=4.5\times 3=13.5$ W.

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