Chapter 12 · Electricity

NCERT Exemplar Class 10 Science Chapter 12 Long Answer Questions

All 7 questions of NCERT Exemplar Class 10 Science Chapter 12 Long Answer Questions (Electricity), in the book's order, each with its answer and a step-by-step solution.

1. Q29
   Three incandescent bulbs of ​100 W each are connected in series in an ​electric circuit. In another ​circuit another set of three ​bulbs of the same ​wattage ​are connected in parallel to the same source.

   1. (a)Will the bulb in the two circuits glow with the same brightness? Justify your answer.
   2. (b)Now let ​one bulb in both the circuits get ​fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
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   No: parallel bulbs are nine times as bright; when one fuses, series bulbs go out but parallel bulbs keep glowing.

   (a) No. Each bulb has resistance $R=V^2/100$. In the parallel circuit every bulb has the full source voltage $V$ across it, so each dissipates its rated $100$ W. In the series circuit the three equal resistances share the voltage, so each bulb has only $V/3$ across it and dissipates $P=\frac{(V/3{)}^2}{R}=\frac{100}{9}\approx 11.1$ W. A parallel bulb is therefore nine times as bright as a series bulb, and the two circuits take $300$ W and $33.3$ W in all. (b) In the series circuit the remaining bulbs do **not** glow: a fused bulb breaks the only path, so no current flows anywhere and all three go out. In the parallel circuit the remaining bulbs **do** go on glowing, with unchanged brightness, because each has its own path across the source and only the fused branch is broken.

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2. Q30
   State ​Ohm’s law? How can it be ​verified experimentally? Does ​it ​hold good ​under ​all ​conditions? ​Comment.
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   Ohm's law is $V=IR$ at constant temperature; a straight $V$–$I$ graph verifies it, but it fails for non-ohmic devices and heated filaments.

   Ohm’s law: the potential difference across the ends of a given metallic conductor is directly proportional to the current through it, provided its temperature and physical state remain unchanged; that is, $V\propto I$, so $V=IR$, where the constant $R$ is the resistance. Verification: connect a nichrome wire in series with a battery, a plug key, a rheostat and an ammeter, and a voltmeter in parallel across the wire. Vary the current with the rheostat and record four or five pairs of readings $(I,V)$. The ratio $V/I$ comes out the same for every pair, and a graph of $V$ against $I$ is a straight line through the origin whose slope is $R$ — which verifies the law. Comment: it does **not** hold under all conditions. It applies only to ohmic conductors whose temperature and physical state are constant. It fails when the conductor heats up (the filament of a lamp, whose V–I graph curves upward) and for non-ohmic devices such as semiconductor diodes, transistors, LEDs, thermistors, electrolytes and gases in a discharge tube, whose V–I graphs are not straight lines.

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3. Q31
   What is electrical ​resistivity of a ​material? What is its unit? ​Describe ​an experiment to ​study the factors on ​which the resistance of conducting ​wire depends.
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   Resistivity is the resistance of unit length and unit area, in $\Omega$ m; varying wire length, area and material shows $R=\frac{\rho l}{A}$.

   The electrical resistivity of a material 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 $\rho =\frac{RA}{l}$. Its SI unit is the ohm metre ($\Omega$ m). It depends only on the material and its temperature, not on the size or shape of the conductor. Experiment: fix a test wire between terminals X and Y, connect X and Y in series with a cell, a plug key and an ammeter, and connect a voltmeter across X and Y; then $R=V/I$. Repeat with different test wires, changing one factor at a time. (i) With the same material and thickness, a wire of double the length gives double the resistance, so $R\propto l$. (ii) With the same material and length, a wire of double the area of cross-section gives half the resistance, so $R\propto 1/A$. (iii) With the same length and thickness but a different metal, the resistance is different, so it depends on the nature of the material. Combining these, $R=\frac{\rho l}{A}$. Heating the wire also raises its resistance, so resistance depends on temperature as well.

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4. Q32
   How will you infer ​with the help of an experiment that ​the same current flows ​through every ​part of the circuit containing three resistances in series ​connected to ​a ​battery?
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   An ammeter gives the same reading at every point of the series circuit, so the same current flows through every part.

   Connect three unequal resistors — say $R_1=1\Omega$, $R_2=2\Omega$ and $R_3=3\Omega$ — in series with a battery and a plug key, leaving four gaps in the loop: P before $R_1$, Q between $R_1$ and $R_2$, S between $R_2$ and $R_3$, and T after $R_3$. Insert the ammeter at P, close the key and note the reading; then open the key, move the same ammeter to Q, then to S, then to T, changing nothing else, and note the reading each time. All four readings are found to be equal — $1$ A each in this example. The inference is that the same current flows through every part of a series circuit. The reason is that there is only one path and no junction at which charge can be added or removed, and charge is conserved, so the same charge passes every cross-section each second. A voltmeter across each resistor, by contrast, reads $1$ V, $2$ V and $3$ V, adding to the $6$ V of the battery: in a series circuit it is the potential difference that is divided, not the current.

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5. Q33
   How will ​you conclude that the same potential difference (voltage) exists ​across three ​resistors connected in ​a ​parallel arrangement to ​a ​battery?
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   Equal voltmeter readings (6 V) across each resistor and across X and Y show the same potential difference across parallel resistors.

   Join three unequal resistors — say $R_1=2\Omega$, $R_2=3\Omega$ and $R_3=6\Omega$ — between the same two points X and Y, and connect X and Y through a plug key and an ammeter to a battery of, say, $6$ V. Close the key and connect one voltmeter across $R_1$; record the reading. Then move the same voltmeter across $R_2$, then across $R_3$, and finally across X and Y, changing nothing else. All four readings are found to be equal, each $6$ V, and equal to the potential difference of the battery. Hence the same potential difference exists across every resistor of a parallel arrangement. The reason is that each resistor has one end at X and the other at Y, and the potential difference between two given points has one definite value. Measuring the branch currents confirms what does differ: $3$ A, $2$ A and $1$ A, adding to the $6$ A read by the ammeter in the main line — in parallel it is the current that divides, not the voltage.

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6. Q34
   What is Joule’s heating effect? How can ​it ​be ​demonstrated ​experimentally? List ​its four applications in ​daily ​life.
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   Electrical energy turns into heat in a resistance ($H=I^{2}Rt$), shown by a coil heating water, and used in irons, bulbs, fuses and geysers.

   Joule’s heating effect is the conversion of electrical energy into heat when a current is passed through a conductor having resistance. In a time $t$ a current $I$ carries a charge $Q=It$ through a potential difference $V$, so the work done is $W=VIt$, and in a resistor all of it appears as heat: $H=VIt=I^{2}Rt=\frac{V^{2}t}{R}$. Hence Joule’s law: $H\propto I^2$, $H\propto R$ and $H\propto t$. Demonstration: immerse a nichrome coil in a known mass of water in a calorimeter with a thermometer and a stirrer, and connect it in series with a battery, a key, a rheostat and an ammeter, with a voltmeter across the coil. Pass a steady current for a measured time: the temperature rises, and the heat gained, $mc\Delta \theta$, equals $VIt$. Doubling the current for the same time makes the rise about four times as large ($H\propto I^2$); using a coil of larger resistance at the same current increases the rise in proportion to $R$; doubling the time doubles the rise. Simply put: a thin nichrome wire glows while the thick copper leads carrying the same current stay cool. Four applications: (1) electric heating appliances — electric iron, room heater, toaster and electric kettle, with a nichrome heating element; (2) the filament of an incandescent electric bulb, a tungsten filament heated to about $2500{}^{\circ }$C so that it glows; (3) the electric fuse, which melts and breaks the circuit when the current is excessive; (4) the electric geyser or immersion heater for heating water.

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7. Q35
   Find out the following in the electric circuit given in Figure 12.9

   1. (a)Effective resistance ​of two 8 Ω resistors in the combination
   2. (b)​Current flowing through ​4 Ω resistor
   3. (c)Potential difference across ​4 Ω resistance
   4. (d)Power ​dissipated in 4 Ω ​resistor
   5. (e)Difference in ammeter readings, if ​any.

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   (a) $4\Omega$; (b) $1$ A; (c) $4$ V; (d) $4$ W; (e) the two ammeters read the same, so the difference is zero.

   (a) The two $8\Omega$ resistors are in parallel between C and D, so $\frac{1}{R_p}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$, giving $R_p=4\Omega$. (b) The total resistance is $4+4=8\Omega$, so the current in the main line, and therefore through the $4\Omega$ resistor, is $I=8/8=1$ A. (c) The potential difference across the $4\Omega$ resistance is $V=IR=1\times 4=4$ V. (d) The power dissipated in it is $P=I^{2}R=(1{)}^2\times 4=4$ W. (e) There is **no** difference: $A_1$ and $A_2$ both lie in the main line outside the parallel section, so each reads the full current of $1$ A and $A_1-A_2=0$.

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