Chapter 11 · Electricity

NCERT Class 10 Science Chapter 11 Intext Questions (Page 188)

All 5 questions of NCERT Class 10 Science Chapter 11 Intext Questions (Page 188) — Electricity — in the book's order, each with its answer and a step-by-step solution.

1. Q1
   Judge the equivalent resistance when ​the following are connected ​in ​parallel ​–

   1. (a)$1\Omega$ and $10^6\Omega$, ​
   2. (b)$1\Omega$ and $10^3\Omega$, and $10^6\Omega$.
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   The equivalent resistance is about $1\Omega$ in (a) and $0.999\Omega$ in (b), just below the smallest resistance of each set.

   (a) $\frac{1}{R_p}=\frac{1}{1}+\frac{1}{10^6}=1.000001{\Omega }^{-1}$, so $R_p=0.999999\Omega \approx 1\Omega$; (b) $\frac{1}{R_p}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^6}=1.001001{\Omega }^{-1}$, so $R_p\approx 0.999\Omega$. In both cases the equivalent resistance is a little smaller than $1\Omega$, the smallest resistance in the combination: a very large resistance placed in parallel carries almost no current and so hardly changes the result.

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2. Q2
   An electric lamp ​of $100\Omega$, ​a toaster ​of resistance $50\Omega$, and a water filter of resistance $500\Omega$ ​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?
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   $R=31.25\Omega$ and $I=7.04\text{A}$: the iron needs the equivalent resistance of the three appliances in parallel.

   The iron must have the equivalent resistance of the three appliances in parallel: $\frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}=\frac{16}{500}$, so $R=31.25\Omega$. The current through it is $I=\frac{V}{R}=\frac{220}{31.25}=7.04$ A.

   Step-by-step solution →
3. Q3
   What are ​the advantages ​of ​connecting ​electrical devices in parallel with the battery instead of connecting ​them in series?
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   In parallel each device gets the full voltage, draws its own current, keeps working if another fails, and the total resistance falls.

   Connecting devices in parallel with the battery has four advantages. (i) Each device gets the full battery voltage, because every branch is joined straight across the terminals; in series the voltage is divided among the devices, so none of them works at its rated voltage. (ii) Each device draws its own current $I=V/R$, decided by its own resistance, so appliances needing very different currents — a bulb and a heater, for instance — can run on the same supply; a series circuit forces one common current through all of them. (iii) If one device fails, only its own branch is broken and the rest keep working, whereas a break anywhere in a series circuit stops every device. Each branch can also have its own switch, so devices can be operated independently. (iv) The total resistance of a parallel combination is less than the smallest branch resistance, so the circuit can draw the larger total current that several appliances together need.

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4. Q4
   How ​can three resistors of resistances $2\Omega$, $3\Omega$, ​and $6\Omega$ be connected to give a total ​resistance ​of ​

   1. (a)$4\Omega$, ​
   2. (b)$1\Omega$?
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   (a) $3\Omega$ and $6\Omega$ in parallel, in series with $2\Omega$, give $4\Omega$; (b) all three in parallel give $1\Omega$.

   (a) Connect the $3\Omega$ and $6\Omega$ resistors in parallel and put this combination in series with the $2\Omega$ resistor. The parallel pair gives $\frac{1}{R^{\prime }}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$, so $R^{\prime }=2\Omega$, and the total is $R=2+2=4\Omega$. (b) Connect all three resistors in parallel: $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, so $R=1\Omega$.

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5. Q5
   What is

   1. (a)​the highest, ​
   2. (b)the lowest ​total ​resistance ​that can be ​secured by combinations of four coils of resistance $4\Omega$, $8\Omega$, $12\Omega$, $24\Omega$?
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   (a) Highest: $48\Omega$ with all four in series; (b) lowest: $2\Omega$ with all four in parallel.

   (a) The highest total resistance is obtained with all four coils in series: $R=4+8+12+24=48\Omega$. (b) The lowest is obtained with all four in parallel: $\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{12}{24}=\frac{1}{2}$, so $R=2\Omega$.

   Step-by-step solution →