NCERT Solutions Class 10 Science Chapter 11 In-text Questions — Electricity
NCERT Class 10 Science — Electricity, the seven in-text QUESTIONS boxes on pages 172, 174, 181, 185, 188, 190 and 192. All 23 questions solved.
Also here. Solutions to the end-of-chapter questions for this chapter and the activities are on their own pages.
Chapter 11 is the most heavily interrupted chapter in the book: seven boxes, one after each idea is introduced, and together they are a complete rehearsal for the end-of-chapter Exercises. The definitions come first, then Ohm’s law, then combinations of resistors, then heating and power. Five relations carry all twenty-three questions:
- Current and charge — $I = \dfrac{Q}{t}$, with $1\ \text{A} = 1\ \text{C s}^{-1}$.
- Potential difference — $V = \dfrac{W}{Q}$, with $1\ \text{V} = 1\ \text{J C}^{-1}$.
- Ohm’s law and resistivity — $V = IR$, and $R = \rho\dfrac{l}{A}$.
- Combinations — $R_s = R_1 + R_2 + \dots$ in series, and $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ in parallel.
- Heat and power — $H = VIt = I^2Rt$, and $P = VI = I^2R = \dfrac{V^2}{R}$.
Key insight. Ask what is shared before you calculate. In series the current is the same everywhere, because there is only one path; in parallel the potential difference is the same across every branch, because they are joined between the same two points. Every combination question in these boxes, and the reason a heater’s element glows while its cord does not, comes straight out of that one question.
Questions, page 172. After Section 11.1, on electric current and circuits.
Page 172, Question 1
What does an electric circuit mean?
Solution. Charge will not flow unless there is somewhere for it to go and come back from. A circuit is that closed path: a source of potential difference such as a cell or battery, connecting wires, the component the current is meant to pass through, and usually a switch to open and close the path.
The word that matters is closed. Open the plug key and the path is broken, no charge flows, and the current is zero everywhere at once — not just at the break.
An electric circuit is a continuous, closed path along which an electric current flows — a cell or battery, connecting wires, the electrical components, and a switch or plug key to complete or break the path.
Page 172, Question 2
Define the unit of current.
Solution. Current is the rate at which charge crosses a section of the conductor, $I = Q/t$, so its unit follows from the units of charge and time. Setting $Q = 1$ C and $t = 1$ s gives the unit of current.
The SI unit of electric current is the ampere (A). One ampere is the current when one coulomb of charge flows through any cross-section of a conductor in one second:
$$1\ \text{A} = \frac{1\ \text{C}}{1\ \text{s}}$$
Page 172, Question 3
Calculate the number of electrons constituting one coulomb of charge.
Solution. A coulomb is made up of the charges of a whole number of electrons, and each electron carries $1.6 \times 10^{-19}$ C. So the number needed is the total charge divided by the charge on one electron:
$$n = \frac{Q}{e} = \frac{1\ \text{C}}{1.6 \times 10^{-19}\ \text{C}} = 6.25 \times 10^{18}$$
The chapter itself says “nearly $6 \times 10^{18}$”, which is this number rounded.
$$n = \frac{Q}{e} = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}\ \text{electrons}$$
Questions, page 174. After Section 11.2, on electric potential and potential difference.
Page 174, Question 1
Name a device that helps to maintain a potential difference across a conductor.
Solution. A potential difference does not maintain itself: as charge flows, it would die away unless something keeps pushing charge back. A cell does that work, using the energy of a chemical reaction inside it, and a battery is several cells joined together.
A cell, or a battery of several cells. (Any source of emf, such as a generator, does the same job.)
Page 174, Question 2
What is meant by saying that the potential difference between two points is 1 V?
Solution. Potential difference is defined as work done per unit charge, $V = W/Q$, so a value of 1 V is a statement about how much work moving one coulomb takes.
$$1\ \text{V} = \frac{1\ \text{J}}{1\ \text{C}}$$
It means 1 joule of work is done in moving a charge of 1 coulomb from one point to the other: $1\ \text{V} = 1\ \text{J C}^{-1}$.
Page 174, Question 3
How much energy is given to each coulomb of charge passing through a 6 V battery?
Solution. This is the previous definition used the other way round. If $V = W/Q$, then $W = VQ$, and here the charge is one coulomb.
$$W = VQ = 6\ \text{V} \times 1\ \text{C} = 6\ \text{J}$$
$$W = VQ = 6\ \text{V} \times 1\ \text{C} = 6\ \text{J}$$
Questions, page 181. After Section 11.5, on the factors that decide resistance, and Table 11.2.
Page 181, Question 1
On what factors does the resistance of a conductor depend?
Solution. The chapter arrives at this by experiment, and the result is collected into one relation:
$$R = \rho\,\frac{l}{A}$$
Reading it off: resistance grows in proportion to the length $l$, because the charge has further to push through; it falls as the area of cross-section $A$ grows, because there is more room to flow through; and it depends on the material through the resistivity $\rho$, which is why a copper wire and a nichrome wire of the same shape behave quite differently. Resistance also changes with temperature, as the chapter notes under the table.
On its length (directly proportional), its area of cross-section (inversely proportional), the material it is made of (through its resistivity $\rho$), and its temperature:
$$R = \rho\,\frac{l}{A}$$
Page 181, Question 2
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Solution. Same material and same source, so the only difference is the area of cross-section. Since $R = \rho l / A$, a larger area means a smaller resistance — the thick wire has the lower resistance of the two.
The source fixes $V$, so by Ohm’s law $I = V/R$ the smaller resistance carries the larger current. Doubling the radius quarters the resistance, because the area goes as the square of the radius.
Through the thick wire. Its larger area of cross-section gives it a smaller resistance ($R \propto 1/A$), and with the same potential difference across it, $I = V/R$ is therefore larger.
Page 181, Question 3
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Solution. With $R$ held constant, Ohm’s law makes the current directly proportional to the potential difference:
$$I = \frac{V}{R} \quad\Longrightarrow\quad I’ = \frac{V/2}{R} = \frac{1}{2}\cdot\frac{V}{R} = \frac{I}{2}$$
The current is also halved. With $R$ constant, $I = V/R$ is directly proportional to $V$, so halving $V$ halves $I$.
Page 181, Question 4
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Solution. Two properties matter, and Table 11.2 shows the first directly: the resistivity of an alloy is much higher than that of its constituent metals. Nichrome is $100 \times 10^{-6}\ \Omega\,\text{m}$ against iron’s $10.0 \times 10^{-8}\ \Omega\,\text{m}$ — about a thousand times larger. A higher resistance produces more heat for the same current, since $H = I^2Rt$, which is the whole point of a heating element.
The second is durability. Alloys do not oxidise — burn — readily at high temperatures, so the element survives being made red hot again and again. A pure metal coil would oxidise and fail quickly.
Because an alloy has a much higher resistivity than its constituent pure metals, so it produces far more heat for the same current ($H = I^2Rt$), and because alloys do not oxidise readily at high temperature, so the element is not destroyed by being heated red hot repeatedly.
Page 181, Question 5
Use the data in Table 11.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Solution. A better conductor is one that opposes the current less, so the comparison is between resistivities, and the smaller value wins.
From Table 11.2, iron has resistivity $10.0 \times 10^{-8}\ \Omega\,\text{m}$ and mercury $94.0 \times 10^{-8}\ \Omega\,\text{m}$. Iron’s is smaller by a factor of about nine, so iron conducts better.
Looking down the whole conductors column, the smallest entry is silver at $1.60 \times 10^{-8}\ \Omega\,\text{m}$ — just below copper’s $1.62 \times 10^{-8}$.
(a) Iron, with $\rho = 10.0 \times 10^{-8}\ \Omega\,\text{m}$ against mercury’s $94.0 \times 10^{-8}\ \Omega\,\text{m}$.
(b) Silver, with the lowest resistivity in the table, $\rho = 1.60 \times 10^{-8}\ \Omega\,\text{m}$.
Questions, page 185. After Section 11.6.1, on resistors in series.
Page 185, Question 1
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Solution. “All connected in series” means one single loop with no branch anywhere: the charge leaving the battery must pass through the 5 Ω, then the 8 Ω, then the 12 Ω, then the key, and back. The order the components are placed in around the loop makes no difference to the circuit — only that there is one path.
Each cell is drawn as a long line for the positive terminal and a short thick line for the negative, and three of them in a row make the battery, of $3 \times 2 = 6$ V.
A single closed loop, drawn above: the three cells in series make a 6 V battery, and the 5 Ω, 8 Ω and 12 Ω resistors and the plug key all lie on that one loop, so the same current passes through every one of them.
Page 185, Question 2
Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Solution. The two meters go in opposite ways, and for a reason. The whole current has to pass through an ammeter for it to be measured, so the ammeter goes in series, anywhere on the loop. A potential difference is between two points, so the voltmeter is joined to those two points — one terminal to each end of the 12 Ω resistor — that is, in parallel with it.
Now the readings. In series the resistances add:
$$R_s = 5\ \Omega + 8\ \Omega + 12\ \Omega = 25\ \Omega$$
The battery of three 2 V cells gives $V = 6$ V, so the ammeter reads
$$I = \frac{V}{R_s} = \frac{6\ \text{V}}{25\ \Omega} = 0.24\ \text{A}$$
The same current passes through the 12 Ω resistor, so the voltmeter across it reads
$$V_{12} = IR = 0.24\ \text{A} \times 12\ \Omega = 2.88\ \text{V}$$
The ammeter reads $0.24\ \text{A}$ and the voltmeter reads $2.88\ \text{V}$.
$$R_s = 25\ \Omega,\qquad I = \frac{6}{25} = 0.24\ \text{A},\qquad V_{12} = 0.24 \times 12 = 2.88\ \text{V}$$
Questions, page 188. After Section 11.6.2, on resistors in parallel.
Page 188, Question 1
Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω, and 10⁶ Ω.
Solution. The word is “judge”, not “calculate”: the point is to see what parallel combination does without grinding through the arithmetic. In parallel the conductances add, so the equivalent resistance is always smaller than the smallest resistance in the combination. Adding a very large resistance in parallel adds almost no conductance, so it barely changes anything.
(a) $\dfrac{1}{R_p} = \dfrac{1}{1} + \dfrac{1}{10^6} = 1.000001$, so $R_p \approx 0.999999\ \Omega$ — just under 1 Ω.
(b) $\dfrac{1}{R_p} = \dfrac{1}{1} + \dfrac{1}{10^3} + \dfrac{1}{10^6} = 1.001001$, so $R_p \approx 0.999\ \Omega$ — again just under 1 Ω.
The book’s wording of part (b) is its own; read as printed it lists three resistances, 1 Ω, 10³ Ω and 10⁶ Ω, in parallel, which is how it is taken here.
(a) $R_p \approx 0.999999\ \Omega$, that is, just under $1\ \Omega$.
(b) $R_p \approx 0.999\ \Omega$, again just under $1\ \Omega$.
In both, the equivalent resistance is less than the smallest resistance in the combination, and the very large resistances make almost no difference.
Page 188, Question 2
An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Solution. An iron that draws as much current as the three together, from the same source, must have the same resistance as their parallel combination — that is what “equivalent resistance” means. So find $R_p$ first:
$$\frac{1}{R_p} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} = \frac{5 + 10 + 1}{500} = \frac{16}{500}$$
$$R_p = \frac{500}{16} = 31.25\ \Omega$$
The iron therefore has a resistance of 31.25 Ω, and the current through it is
$$I = \frac{V}{R} = \frac{220\ \text{V}}{31.25\ \Omega} = 7.04\ \text{A}$$
$$R = 31.25\ \Omega, \qquad I = \frac{220}{31.25} = 7.04\ \text{A}$$
Page 188, Question 3
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Solution. Think about what goes wrong in series. The same current must pass through everything, so the devices cannot be switched on and off separately, and if one fails the path is broken and all of them stop. Worse, the battery’s voltage is shared out between them, so no device gets the voltage it was designed for.
In parallel each device is joined across the same two points, so each gets the full potential difference of the battery and works at its rated value; each branch can have its own switch and be used independently; and one device failing leaves the other branches intact. The total resistance also falls below the smallest branch resistance, so the circuit can draw the larger current that several appliances together need.
In parallel, every device gets the full battery voltage and so works at its rated value; each can be switched on and off independently; a failure in one does not stop the others; and the total resistance is reduced, so the circuit can supply the larger total current the devices need. This is why house wiring is in parallel.
Page 188, Question 4
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Solution. Series raises the total above the largest resistance; parallel lowers it below the smallest. Both targets here — 4 Ω and 1 Ω — are below the largest resistance of 6 Ω, so parallel must be involved in each.
(a) 4 Ω. Notice that 4 = 2 + 2, and that 3 Ω and 6 Ω in parallel come to exactly 2 Ω:
$$\frac{1}{R_p} = \frac13 + \frac16 = \frac{2+1}{6} = \frac12 \quad\Longrightarrow\quad R_p = 2\ \Omega$$
Put that combination in series with the 2 Ω resistor:
$$R = 2\ \Omega + 2\ \Omega = 4\ \Omega$$
(b) 1 Ω. Connect all three in parallel:
$$\frac{1}{R_p} = \frac12 + \frac13 + \frac16 = \frac{3+2+1}{6} = 1 \quad\Longrightarrow\quad R_p = 1\ \Omega$$
(a) Join the 3 Ω and 6 Ω in parallel (giving 2 Ω) and put that in series with the 2 Ω: $2 + 2 = 4\ \Omega$.
(b) Join all three in parallel: $\frac12 + \frac13 + \frac16 = 1$, so $R_p = 1\ \Omega$.
Page 188, Question 5
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Solution. The two extremes are the two pure arrangements, and no mixture can beat either. Series gives the highest possible total, because every resistance adds; parallel gives the lowest, because every added branch gives the current one more route.
(a) Highest — all four in series:
$$R_s = 4 + 8 + 12 + 24 = 48\ \Omega$$
(b) Lowest — all four in parallel:
$$\frac{1}{R_p} = \frac14 + \frac18 + \frac{1}{12} + \frac{1}{24} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac12$$
$$R_p = 2\ \Omega$$
(a) Highest: $48\ \Omega$, with all four coils in series.
(b) Lowest: $2\ \Omega$, with all four coils in parallel.
Questions, page 190. After Section 11.7, on the heating effect of electric current.
Page 190, Question 1
Why does the cord of an electric heater not glow while the heating element does?
Solution. The cord and the element are in series, so exactly the same current passes through both. Since $H = I^2Rt$ and $I$ and $t$ are shared, the heat produced in each depends only on its resistance.
The element is a long, thin coil of nichrome, an alloy of high resistivity, so its resistance is large and it gets hot enough to glow. The cord is thick copper, whose resistivity is about a thousand times smaller, and it is short and wide as well — so its resistance is tiny and the heat it produces is negligible. It stays barely warm.
Because the two carry the same current but have very different resistances, and $H = I^2Rt$. The element is a long thin coil of nichrome, of high resistivity and large resistance, so it becomes hot enough to glow; the cord is short and thick and made of copper, of very low resistivity, so its resistance and the heat produced in it are negligible.
Page 190, Question 2
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Solution. The work done in moving a charge $Q$ through a potential difference $V$ is $W = VQ$, and in a purely resistive path all of it appears as heat:
$$H = VQ = 50\ \text{V} \times 96000\ \text{C} = 4.8 \times 10^6\ \text{J}$$
The “one hour” is not needed for the heat itself — the charge already carries the whole account. It would be needed only if the question asked for the current or the power ($I = 96000/3600 = 26.7$ A).
$$H = VQ = 50 \times 96000 = 4.8 \times 10^6\ \text{J}$$
Page 190, Question 3
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Solution. Here the current and the resistance are given rather than the voltage, so the convenient form of Joule’s law is $H = I^2Rt$:
$$H = I^2Rt = (5\ \text{A})^2 \times 20\ \Omega \times 30\ \text{s}$$
$$H = 25 \times 20 \times 30 = 15000\ \text{J} = 1.5 \times 10^4\ \text{J}$$
$$H = I^2Rt = 25 \times 20 \times 30 = 1.5 \times 10^4\ \text{J}$$
Questions, page 192. The last box, after Section 11.8 on electric power.
Page 192, Question 1
What determines the rate at which energy is delivered by a current?
Solution. The rate of doing work is power, and for an electric circuit the energy delivered in time $t$ is $W = VIt$. Dividing by $t$:
$$P = \frac{W}{t} = VI$$
So the rate is fixed by two things together — the potential difference across the device and the current through it. Where Ohm’s law applies this can be rewritten as $I^2R$ or $V^2/R$, but the quantity being named is the same: electric power.
Electric power, $P = VI$ — the rate is determined by the potential difference across the device and the current through it (equivalently $I^2R$, or $V^2/R$).
Page 192, Question 2
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Solution. The power follows straight from $P = VI$:
$$P = VI = 220\ \text{V} \times 5\ \text{A} = 1100\ \text{W} = 1.1\ \text{kW}$$
The energy is the power multiplied by the time. In joules, the time must be in seconds — two hours is $2 \times 3600 = 7200$ s:
$$W = Pt = 1100\ \text{W} \times 7200\ \text{s} = 7.92 \times 10^6\ \text{J}$$
In the commercial unit, keep the power in kilowatts and the time in hours:
$$W = 1.1\ \text{kW} \times 2\ \text{h} = 2.2\ \text{kW h}$$
$$P = VI = 1100\ \text{W} = 1.1\ \text{kW}$$
$$W = Pt = 1100 \times 7200 = 7.92 \times 10^6\ \text{J} = 2.2\ \text{kW h}$$
Common mistakes
- Adding resistances in parallel as though they were in series in the page-188 questions. The parallel formula gives $1/R_p$, not $R_p$ — forgetting the final reciprocal is the single commonest slip in this chapter. A quick check costs nothing: $R_p$ must come out smaller than the smallest branch.
- Using the time in Question 2 of the page-190 box. The charge and the potential difference already fix the energy, $H = VQ$. Dividing by the hour first and then multiplying back is extra work that invites an error.
- Leaving the time in hours in $W = Pt$. $1100 \times 2$ gives 2200, which is neither joules nor kilowatt hours. Seconds go with joules; hours go with kilowatt hours and kilowatts.
- Putting the voltmeter in series in the page-185 question. A voltmeter has a very high resistance; in series it would nearly stop the current and read almost the whole battery voltage. It belongs across the component, and the ammeter belongs in the loop.
- Answering the page-181 alloy question with resistivity alone. Half the answer is that alloys do not oxidise readily when red hot. An element that produced plenty of heat but burnt out in a week would be no use.
- Reading “better conductor” as “larger resistivity” in the table question. The better conductor is the one with the smaller resistivity, which is why silver, with the smallest number in the table, is the best conductor in it.
Practise next
- The end-of-chapter Exercises for this chapter, where Questions 4, 14 and 15 use the same series-and-parallel reasoning on harder numbers, and Question 18 asks why domestic wiring is parallel.
- Chapter 12, Magnetic Effects of Electric Current — its in-text questions take the current set up here and ask what it does to a compass needle.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.