NCERT Solutions for Class 10 Science Chapter 12 Electricity
NCERT Solutions for Class 10 Science Chapter 12 Electricity provides simple and reliable answers to all textbook questions. Electricity is one of the most important and high-scoring chapters in the CBSE and SEBA Class 10 Board Exams. However, many students find its numerical problems, circuit diagrams, and formulas tricky to apply during exams.
In this article, we solve every in-text question, chapter-end exercise of Class 10 Science Chapter 12 Electricity in simple, step-by-step manner so that you can understand the concept easily and able to score full marks in SEBA/CBSE board exam.
Before solving this chapter’s in-text questions and chapter-end exercises, I am going to request you to first solve the chapter examples’ solutions.
In text 12.1 Page No: 200
1. What does an electric circuit mean?
Ans: A continuous and closed path through which an electric current flows is called an electric circuit. It consists of various electrical components, such as a cell, conductors, an electric bulb, an ammeter, a plug key, etc.
2. Define the unit of current.
Ans: The SI unit of electric current is the ampere (A). If one coulomb of electric charge flows through a conductor in one second, the current is said to be one ampere.
3. Calculate the number of electrons constituting one coulomb of charge.
Answer:
Given, total charge, \(Q = 1\,\text{C}\)
We know the charge of one electron, \(e = 1.6 \times 10^{-19}\,\text{C}\)
According to charge quantization, \( Q = ne \)
The number of electrons is given by, \( n=\frac{Q}{e} \)
∴ $ n=\frac{1}{1.6 \times 10^{-19}}$
$ n=6.25 \times 10^{18} $
Therefore, the number of electrons constituting one coulomb of charge is\(6.25 \times 10^{18}\)
In text 12.2 Page No: 202
1. Name a device that helps to maintain a potential difference across a conductor.
Answer: An electric cell or a battery helps to maintain a potential difference across a conductor.
2. What is meant by saying that the potential difference between two points is 1 V?
Answer: The potential difference between two points is said to be 1 volt (V) if 1 joule (J) of work is done to move 1 coulomb (C) of electric charge from one point to the other.
Thus, \(1\,\text{V} = \dfrac{1\,\text{J}}{1\,\text{C}} = 1\,\text{J C}^{-1}\).
3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer: Given,
- Potential difference, \(V = 6\,\text{V}\)
- Charge, \(Q = 1\,\text{C}\)
We know that,
$V = \frac{W}{Q}$
$W = V \times Q$
$W = 6 \times 1 = 6\text{ J}$
Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V battery.
In text 12.25 Page No: 209
1. On what factors does the resistance of a conductor depend?
Answer: The resistance of a conductor depends:
- on its length (l), \(R \propto \ell\)
- on its area of cross-section (A), \(R \propto \frac{1}{A}\)
- on the nature of its material, \(R \propto \rho \)
- on its temperature; the resistance increases with an increase in temperature.
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?
Answer:
The resistance of a conductor is given by the equation,
\[ R=\rho\frac{l}{A} \]
Thus, the resistance of a wire is inversely proportional to its cross-sectional area (R ∝ 1/A). This means that as the cross-sectional area increases, the resistance decreases, and vice versa. Therefore, a thick wire has has a larger cross-sectional area and hence less resistance than a thin wire of the same material and length. According to Ohm’s law, $ I = \frac{V}{R}$ smaller resistance allows a larger current to flow from the same source. Hence, current will flows more easily through a thick wire when both wires are connected to the same source.
3. Let the resistance of an electrical component remain 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?
Answer: The change in current flowing through the electrical component can be determined by using Ohm’s law.
According to Ohm’s law, \(I = \dfrac{V}{R}\)
where \(I\) is current, \(V\) is potential difference, and \(R\) is resistance.
When the potential difference decreases to half while the resistance remains constant, the new potential difference is \(V’ = \dfrac{V}{2}\).
Hence, the new current is \(I’ = \dfrac{V’}{R} = \dfrac{V/2}{R} = \dfrac{V}{2R} = \dfrac{I}{2}\).
Therefore, the current flowing through the electrical component is reduced by half.
4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because
(i) The resistivity of an alloy is generally higher than that of its constituent metals, so an alloy produces more heat.
(ii) Alloys do not oxidise (burn) easily at high temperatures. This helps the coil last longer.
5. Use the data in the table given below and answer the following questions.
(a) Which among iron and mercury is a better conductor?
Given, resistivity of iron= 10.0 × 10⁻⁸ Ω m and resistivity of mercury = 94.0 × 10⁻⁸ Ω m.
(b) Which material is the best conductor?
Answer:
(a) Iron is a better conductor than mercury because the resistivity of iron (\(10.0 \times 10^{-8}\ \Omega\,\text{m}\)) is lower than that of mercury (\(94.0 \times 10^{-8}\ \Omega\,\text{m}\)). A material with lower resistivity conducts electricity more easily.
(b) Silver is the best conductor among the materials listed in the table because it has the lowest resistivity, \(1.60 \times 10^{-8}\ \Omega\,\text{m}\).
In text 12.26 Page No: 213
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:
The required circuit diagram is shown below :
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:
Since all three resistors are connected in series, the total resistance will be
\(R = 5 + 8 + 12 = 25\,\Omega\).
The total potential difference is \(V = 2 + 2 + 2 = 6\,\text{V}\).
According to Ohm’s law,
\(I = \dfrac{V}{R} = \dfrac{6}{25} = 0.24\,\text{A}\).
Therefore, the ammeter reading is 0.24 A.
Now, for 12 Ω resistor \(V_1\) is:
\(V_1 = I \times R_1 = 0.24 \times 12 = 2.88\,\text{V}\).
Therefore, the voltmeter reading is 2.88 V.
In text 12.6 Page No: 216
1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and \(10^6\,\Omega\), (b) \(1\,\Omega\), \(10^3\,\Omega\), and \(10^6\,\Omega\).
Answer:
When resistors are connected in parallel, the equivalent resistance is given by \(\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots\).
(a) When \(1\,\Omega\) and \(10^6\,\Omega\) are connected in parallel.
Let the equivalent resistance be \(R\).
\(\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{10^6}\)
\(\dfrac{1}{R} = 1 + 0.000001 = 1.000001\)
\(R = \dfrac{1}{1.000001} \approx 0.999999\,\Omega\)
Therefore, the equivalent resistance is approximately \(1\,\Omega\).
(b) When \(1\,\Omega\), \(10^3\,\Omega\), and \(10^6\,\Omega\) are connected in parallel.
Let the equivalent resistance be \(R\).
\(\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{10^3} + \dfrac{1}{10^6}\)
\(\dfrac{1}{R} = 1 + 0.001 + 0.000001 = 1.001001\)
\(R = \dfrac{1}{1.001001} \approx 0.998\,\Omega\)
Therefore, the equivalent resistance is approximately \(0.998\,\Omega\).
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?
Answer
Given: \(R_1 = 100\,\Omega\), \(R_2 = 50\,\Omega\), \(R_3 = 500\,\Omega\), and \(V = 220\,\text{V}\).
Let R be the equivalent resistance of the circuit.
\(\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\).
\(\dfrac{1}{R} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500}\)
\(\dfrac{1}{R} = \dfrac{5+10+1}{500} = \dfrac{16}{500} = \dfrac{4}{125}\)
\(\therefore \dfrac{1}{R} = \dfrac{4}{125}\)
\(\therefore R = \dfrac{125}{4} = 31.25\,\Omega\)
Now using Ohm’s law, \(I = \dfrac{V}{R} = \dfrac{220}{31.25} = 7.04\,\text{A}\).
Therefore, the resistance of the electric iron is \(31.25\,\Omega\), and the current through it is \(7.04\,\text{A}\).
3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer: The following are the advantages of connecting electrical devices in parallel:
- In a parallel circuit, the voltage does not get divided among the appliances. The potential difference across each device is equal to the supply voltage, so every device works properly.
- If one device gets damaged or is switched off, the other devices keep working as usual, because the rest of the circuit stays complete.
- Also, the total resistance of the circuit decreases, so the battery can supply a larger current to the devices.
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) 1Ω?
(a) Here \(3\,\Omega\) and \(6\,\Omega\) resistors are connected in parallel.
So the equivalent resistance is given by,
\(\dfrac{1}{R} = \dfrac{1}{3} + \dfrac{1}{6}\)
\(\dfrac{1}{R} = \dfrac{2+1}{6} = \dfrac{3}{6} = \dfrac{1}{2}\)
\(\therefore R = 2\,\Omega\)
This combination is connected to a 2 Ω resistor in series.
Therefore, the total resistance is \(R_{\text{total}} = 2 + 2 = 4\,\Omega\).
Hence, the total resistance is \(4\,\Omega\).
(b) The following circuit diagram shows the connection of the three
resistors.
All the resistors are connected in parallel.
Therefore, their equivalent resistance will be given as,
\(\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}\)
\(\dfrac{1}{R} = \dfrac{3+2+1}{6} = \dfrac{6}{6} = 1\)
\(\therefore \dfrac{1}{R} = 1\)
\(\therefore R = 1\,\Omega\)
Therefore, the total resistance of the circuit is \(1\,\Omega\).
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
(a) The highest resistance is obtained when all four coils are connected in series.
The total resistance is given by:
R = 4Ω + 8Ω + 12 Ω + 24 Ω = 48 Ω.
Therefore, the highest total resistance is \(48\,\Omega\).
(b) The lowest resistance is obtained when all four coils are connected in parallel.
\(\dfrac{1}{R} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24}\)
\(\dfrac{1}{R} = \dfrac{6}{24} + \dfrac{3}{24} + \dfrac{2}{24} + \dfrac{1}{24}\)
\(\dfrac{1}{R} = \dfrac{6+3+2+1}{24} = \dfrac{12}{24} = \dfrac{1}{2}\)
\(\therefore \dfrac{1}{R} = \dfrac{1}{2}\)
\(\therefore R = 2\,\Omega\)
Therefore, the lowest total resistance is \(2\,\Omega\).
In text 12.7 Page No: 218
1. Why does the cord of an electric heater not glow while the heating element does?
Answer: The heat produced in a wire depends on its resistance (H = I²Rt). The heating element is made of an alloy like nichrome, which has very high resistance. When current passes through it, a large amount of heat is produced, and it becomes too hot and glows red.
On the other hand, the cord is made of copper or aluminum, which has very low resistance. The same current flows through it, but it produces only a negligible amount of heat. So the cord does not glow.
2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
Given:
Charge, \(Q = 96000\,\text{C}\)
Potential difference, \(V = 50\,\text{V}\)
Time, \(t = 1\,\text{hour} = 3600\,\text{s}\)
We know that, \(H = VIt\)
Again we know that \(I = \frac{Q}{t}\).
Therefore, \(I = \frac{96000}{3600}\,\text{A}\).
Now,
\(H = 50 \times \frac{96000}{3600}\times 3600\)
\(H = 48,00,000\,\text{J}\)
Hence, the heat generated is \(4.8 \times 10^6\,\text{J}\).
2. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s
Answer:
Given:
Resistance, \(R = 20\,\Omega\)
Current, \(I = 5\,\text{A}\)
Time, \(t = 30\,\text{s}\)
According to Joule’s law of heating, \(H = I^2Rt\).
\(H = 5^2 \times 20 \times 30\)
\(H = 25 \times 20 \times 30\)
\(H = 15000\,\text{J}\)
Hence, the heat developed is \(15,000\,\text{J}\).
In text 12.8 Page No: 220
1. What determines the rate at which energy is delivered by a current?
Answer: The rate at which energy is delivered by a current is determined by the electric power of the circuit. Electric power is the rate at which electrical energy is consumed or delivered. We can write it as: P = VI
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.
Answer:
Given:
Current, \(I = 5\,\text{A}\)
Potential difference, \(V = 220\,\text{V}\)
Time, \(t = 2\,\text{h} = 2 \times 3600= 7200\,\text{s} \)
(i) Power of the motor,\(P = VI\)
\(P = 220 \times 5\)
\(\therefore P = 1100\,\text{W}\)
(ii) Energy consumed by the motor, \(E = Pt\)
\(E = 1100 \times 7200\)
\(E = 7,920,000\,\text{J}\)
Therefore, the energy consumed is \(7.92 \times 10^6\,\text{J}\)
Hence, the power of the motor is \(1100\,\text{W}\), and the energy consumed in 2 hours is \(7.92 \times 10^6\,\text{J}\)
Exercises Page No: 221
1. A piece of wire of resistance \(R\) is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is \(R’\), then the ratio \(R/R’\) is _____.
(A) \(\frac{1}{25}\)
(B) \(\frac{1}{5}\)
(C) \(5\)
(D) \(25\)
Answer: (D) 25
Explanation:
When a wire is cut into five equal parts, the resistance of each part becomes \(\frac{R}{5}\).
Since the five equal parts are connected in parallel, their equivalent resistance is given by
\(\frac{1}{R’}=\frac{1}{R/5}+\frac{1}{R/5}+\frac{1}{R/5}+\frac{1}{R/5}+\frac{1}{R/5}\)
\(\frac{1}{R’}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}\)
\(\frac{1}{R’}=\frac{5 + 5 + 5 + 5 + 5}{R}\)
\(\frac{1}{R’}=\frac{25}{R}\)
Therefore,\(R’=\frac{R}{25}\).
Hence, \(\frac{R}{R’}=\frac{R}{R/25}=25\).
Therefore, the correct answer is (D) 25.
2. Which of the following does not represent electrical power in a circuit?
(A) \(I^2R\)
(B) \(IR^2\)
(C) \(VI\)
(D) \(\frac{V^2}{R}\)
Answer: (B) \(IR^2\)
Explanation:
Electrical power is given by the expression \(P = VI\).
According to Ohm’s law, \(V = IR\).
∴ \(P = (IR)I = I^2R\).
Similarly, using \(I = \frac{V}{R}\),
we get \(P = V \times \frac{V}{R} = \frac{V^2}{R}\).
Thus, \(I^2R\), \(VI\), and \(\frac{V^2}{R}\) represent electrical power. However, \(IR^2\) does not.
Therefore, the correct answer is (B) \(IR^2\).
3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be _____.
(A) 100 W
(B) 75 W
(C) 50 W
(D) 25 W
Answer: (D) 25 W
Explanation:
Given:
Rated voltage, \(V_1 = 220\,\text{V}\)
Rated power, \(P_1 = 100\,\text{W}\)
Operating voltage, \(V_2 = 110\,\text{V}\)
The energy consumed by the appliance is \(P = VI = \frac{V^2}{R}\)
So, we get \(R = \frac{V_1^2}{P_1}\).
\(R = \frac{220^2}{100} = 484\,\Omega\).
Now if the bulb is operated at 110 V.
Even if the supply voltage is reduced, the resistance remains the same.
Hence, the power consumed by the appliance is,
\(P_2 = \frac{V_2^2}{R}\)
\(P_2 = \frac{110^2}{484}\).
\(\therefore P_2 = 25\,\text{W}\)
Hence, the correct answer is (D) 25 W.
4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be _____.
(A) 1:2
(B) 2:1
(C) 1:4
(D) 4:1
Answer: (C) 1:4
Explanation:
Let the resistance of each wire be \(R\), and let the potential difference be \(V\) and the time be \(t\).
The equivalent resistance in series is \(R_s = R+R = 2R\).
Heat produced in series combination, \(H_s = \frac{V^2t}{R_s}\).
Therefore, \(H_s = \frac{V^2t}{2R}\).
The equivalent resistance in parallel is \(R_p = \frac{R \times R}{R+R} = \frac{R}{2}\).
Therefore, \(H_p = \frac{V^2t}{R_p} = \frac{V^2t}{R/2} = \frac{2V^2t}{R}\).
The ratio of heat produced in series and parallel combinations would be,
\(\frac{H_s}{H_p} = \frac{V^2t}{2R} \div \frac{2V^2t}{R} = \frac{1}{4}\).
Hence, the ratio of heat produced in series and parallel combinations is \(1:4\).
The correct answer is (C).
5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured.
6. A copper wire has diameter 0.5 mm and resistivity of \(1.6 \times 10^{-8}\,\Omega\,\text{m}\). What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Given:
Diameter, \(d = 0.5\,\text{mm} = 0.5 \times 10^{-3}\,\text{m}\)
Radius, \(r = \frac{d}{2} = 0.25 \times 10^{-3}\,\text{m}\)
Resistivity, \(\rho = 1.6 \times 10^{-8}\,\Omega\,\text{m}\)
Resistance, \(R = 10\,\Omega\)
\(\pi = 3.14\)
We know that, \(R = \frac{\rho \ell}{A}\)
Therefore, \(\ell = \frac{RA}{\rho}\)
The cross-sectional area is \(A = \pi r^2\).
\(A = 3.14 \times (0.25 \times 10^{-3})^2\)
\(A = 3.14 \times 0.0625 \times 10^{-6}\)
\(A = 1.9625 \times 10^{-7}\,\text{m}^2\)
Now, \(\ell = \frac{10 \times 1.9625 \times 10^{-7}}{1.6 \times 10^{-8}}\)
\(\ell = 122.65625\,\text{m}\)
Hence, the length of the wire is approximately 122.66 m.
Now,
For the same material and length, resistivity \(\rho\) and length \(\ell\) remain constant.
The original resistance is \(R = \frac{\rho \ell}{\pi r^2} = 10\,\Omega\).
The new resistance is \(R’ = \frac{\rho \ell}{\pi(2r)^2}\). [Since if the diameter is doubled, then the radius is also doubled]
\(R’ = \frac{\rho \ell}{4\pi r^2}\).
Therefore, \(R’ = \frac{R}{4} = \frac{10}{4} = 2.5\,\Omega\).
Change in resistance \(= \text{Original resistance} – \text{New resistance}\).
\(= 10 – 2.5 = 7.5\,\Omega\).
So, when the diameter is doubled, the resistance decreases from \(10\,\Omega\) to \(2.5\,\Omega\). Hence, the resistance decreases by \(7.5\,\Omega\).
7. The potential difference (\(V\)) across a resistor and the current (\(I\)) flowing through it are given in the following table:
| Current, \(I\) (A) | Potential Difference, \(V\) (V) |
|---|---|
| 0.5 | 1.6 |
| 1.0 | 3.4 |
| 2.0 | 6.7 |
| 3.0 | 10.2 |
| 4.0 | 13.2 |
Plot a graph between \(V\) and \(I\), and calculate the resistance of the resistor.
Ans: The voltage is plotted on the x-axis and current is plotted on the y-axis.
The VI graph is plotted in the following figure.
The slope of the line gives the value of resistance.
Now,
Point A: $I_1 = 1.0\text{ A}$, $V_1 = 3.4\text{ V}$
Point B: $I_2 = 3.0\text{ A}$, $V_2 = 10.2\text{ V}$
The change in potential difference is:
The change in current is:
$$\Delta I = 3.0\text{ A} – 1.0\text{ A} = 2.0\text{ A}$$
Using Ohm’s Law:
$$R = \frac{\Delta V}{\Delta I} = \frac{6.8\text{ V}}{2.0\text{ A}} = 3.4\ \Omega$$
Hence, the resistance of the resistor is \(3.3\,\Omega\).
8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Given:
Potential difference, \(V = 12\,\text{V}\)
Current, \(I = 2.5\,\text{mA} = 2.5 \times 10^{-3}\,\text{A} = 0.0025\,\text{A}\)
According to Ohm’s law, \(R = \frac{V}{I}\).
So, \(R = \frac{12}{0.0025} = 4800\,\Omega\).
\(\therefore R = 4800\,\Omega = 4.8\,\text{k}\Omega\)
The resistance of the resistor is \(4800\,\Omega\) or \(4.8\,\text{k}\Omega\).
9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer:
Given:
Potential difference, \(V = 9\,\text{V}\)
Resistances in series: \(0.2\,\Omega\), \(0.3\,\Omega\), \(0.4\,\Omega\), \(0.5\,\Omega\) and \(12\,\Omega\).
In a series circuit, the total resistance is the sum of all the resistances.
\(\therefore R = 0.2 + 0.3 + 0.4 + 0.5 + 12\)
\(R = 13.4\,\Omega\)
Now, using Ohm’s law.
\(I = \frac{V}{R} = \frac{9}{13.4}\)
\(I \approx 0.671\,\text{A}\)
Therefore, the current flowing through the \(12\,\Omega\) is \(0.67\,\text{A}\).
10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Given:
Resistance of each resistor, \(R = 176\,\Omega\)
Potential difference, \(V = 220\,\text{V}\)
Total current, \(I = 5\,\text{A}\)
Let $R_p$ be the total (equivalent) resistance of the parallel combination.
According to Ohm’s law,
$R_p = \frac{V}{I}$
$R_p = \frac{220}{5} = 44\,\Omega$
Now,
Let the number of resistors required be \(n\).
For \(n\) identical resistors connected in parallel, we have
\(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \dots + n\text{ times} = \frac{n}{R}\)
\(R_{\text{p}} = \frac{R}{n}\).
Therefore, \(n = \frac{R}{R_{\text{p}}} = \frac{176}{44} = 4\).
Therefore, 4 resistors of \(176\,\Omega\) are required.
11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Answer:
(i) To get 9 Ω
Connect two resistors in parallel, then connect the third resistor in series with this pair, as shown in the figure.
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
\(\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}\)
\(\therefore R_p = \frac{6}{2} = 3\,\Omega\)
Now, the third resistor 6Ω is connected in series with 3Ω
∴ \(R = 3 + 6 = 9\,\Omega\)
(ii) To get 4 Ω
In the above figure, two 6 Ω resistors are connected in series.
∴ \(R_s = 6 + 6 = 12\,\Omega\)
Now, the third 6 Ω resistor is in parallel with 12 Ω
∴ \(\frac{1}{R} = \frac{1}{12} + \frac{1}{6}\)
\(\frac{1}{R} = \frac{1+2}{12} = \frac{3}{12}\)
\(\therefore R = \frac{12}{3} = 4\,\Omega\)
12. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of a 220 V line if the maximum allowable current is 5 A?
Answer:
Given:
Power of each bulb, \(P = 10\,\text{W}\)
Potential difference, \(V = 220\,\text{V}\)
Maximum allowable current, \(I = 5\,\text{A}\)
Let $R_p$ be the combined resistance of all the bulbs connected in parallel.
According to Ohm’s law,
$R_p = \frac{V}{I}$
$R_p = \frac{220}{5} = 44\,\Omega$
Now, we know that the power of each bulb is given by,
$P = \frac{V^2}{R}$
$R = \frac{V^2}{P}$
$R = \frac{220 \times 220}{10} = 4840\,\Omega$
Now,
Let \(n\) be the number of bulbs required.
For \(n\) identical bulbs connected in parallel, we have
\(\frac{1}{R_{\text{p}}} = \frac{1}{R} + \frac{1}{R} + \dots + n\text{ times} = \frac{n}{R}\)
\(R_{\text{p}} = \frac{R}{n}\)
$44 = \frac{4840}{n}$
$n = \frac{4840}{44} = 110$
Therefore, 110 bulbs can be connected in parallel across the 220 V supply line.
13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
Given:
Potential difference, \(V = 220\,\text{V}\)
Resistance of coil A, \(R_A = 24\,\Omega\)
Resistance of coil B, \(R_B = 24\,\Omega\)
(i) When the coils are used separately:
The resistance is \(24\,\Omega\).
According to Ohm’s law, \(I = \frac{V}{R}\)
Therefore,
\(I = \frac{220}{24} = 9.17\,\text{A}\)
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) When the coils are connected in series:
The equivalent resistance of two resistors connected in series is
\(R_{\text{eq}} = R_A + R_B\)
\(R_{\text{eq}} = 24 + 24 = 48\,\Omega\)
Therefore, the current is
\(I = \frac{V}{R_{\text{eq}}} = \frac{220}{48}\)
\(I \approx 4.58\,\text{A}\)
Therefore, a current of 4.58 A flows through the series circuit.
(iii) When the coils are connected in parallel:
The equivalent (total) resistance connected in parallel is
\(\frac{1}{R_{\text{eq}}} = \frac{1}{R_A} + \frac{1}{R_B}\)
\(\frac{1}{R_{\text{eq}}} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24}\)
\(R_{\text{eq}} = \frac{24}{2} = 12\,\Omega\)
Therefore, the current is
\(I = \frac{V}{R_{\text{eq}}} = \frac{220}{12}\)
\(I \approx 18.33\,\text{A}\)
The current in the parallel circuit is 18.33 A.
14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
(i) When 1 Ω and 2 Ω resistors are connected in series:
Given:
\(V = 6\,\text{V}\)
\(R_1 = 1\,\Omega\)
\(R_2 = 2\,\Omega\)
In a series circuit, the equivalent resistance is
\(\therefore R_{\text{eq}} = R_1 + R_2\)
\(R_{\text{eq}} = 1 + 2 = 3\,\Omega\)
Now, according to Ohm’s law,
\(I = \frac{V}{R_{\text{eq}}}\)
\(I = \frac{6}{3} = 2\,\text{A}\)
Therefore, the power used in the \(2\,\Omega\) resistor is
\(P = I^2R\)
\(P = 2^2 \times 2\)
\(P = 4 \times 2 = 8\,\text{W}\)
Hence, the power used in the \(2\,\Omega\) resistor is 8 W.
(ii) When 12 Ω and 2 Ω resistors are connected in parallel:
Given:
\(V = 4\,\text{V}\)
\(R_1 = 12\,\Omega\)
\(R_2 = 2\,\Omega\)
In a parallel circuit, the potential difference across each resistor remains the same.
Therefore, the power used in the \(2\,\Omega\) resistor is
\(P = \frac{V^2}{R}\)
\(P = \frac{4^2}{2}\)
\(P = \frac{16}{2} = 8\,\text{W}\)
Hence, the power used in the \(2\,\Omega\) resistor is 8 W.
Therefore, the power used in the \(2\,\Omega\) resistor is the same (8 W) in both circuits.
15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Given:
Power of first lamp, \(P_1 = 100\,\text{W}\)
Power of second lamp, \(P_2 = 60\,\text{W}\)
Potential difference, \(V = 220\,\text{V}\)
We know that, \(P = VI\)
Therefore,
\(I = \frac{P}{V}\)
For the first lamp,
\(I_1 = \frac{P_1}{V}\)
\(I_1 = \frac{100}{220} \,\text{A}\)
For the second lamp,
\(I_2 = \frac{P_2}{V}\)
\(I_2 = \frac{60}{220} \,\text{A}\)
Now, the total current drawn from the line is
\(I = I_1 + I_2\)
\(I = \frac{100}{220} + \frac{60}{220}\)
\(I = \frac{100+60}{220}\)
\(I = \frac{160}{220}\)
\(I = 0.727\,\text{A}\)
Therefore, the total current drawn from the line is 0.727 A.
16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer: Given:
For the TV set:
Power, \(P_1 = 250\,\text{W}\)
Time, \(t_1 = 1\,\text{hr} = 3600\,\text{s}\)
For the toaster:
Power, \(P_2 = 1200\,\text{W}\)
Time, \(t_2 = 10\,\text{min} = 600\,\text{s}\)
(i) Energy consumed by the TV set:
\(E_1 = P_1 \times t_1\)
\(E_1 = 250 \times 3600\)
\(E_1 = 900000\,\text{J} = 9 \times 10^5\,\text{J}\)
(ii) Energy consumed by the toaster:
\(E_2 = P_2 \times t_2\)
\(E_2 = 1200 \times 600\)
\(E_2 = 720000\,\text{J} = 7.2 \times 10^5\,\text{J}\)
Therefore, the TV set uses more energy than the toaster.
17. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer: Given:
- Resistance, \(R = 8\,\Omega\)
- Current, \(I = 15\,\text{A}\)
- Time, \(t = 2\,\text{hr}\)
We know that the rate at which heat is developed in an electric heater is equal to its electrical power.
According to Joule’s law of heating,
\(P = I^2R\)
\(⇒P = 15^2 \times 8\)
\(⇒P = 225 \times 8\)
∴ \(P = 1800\,\text{W}\)
Therefore, the electric heater produces heat at the rate of 1800 W or 1.8 kW.
18. Explain the following.
(a) Why is tungsten used almost exclusively for the filament of electric lamps?
Answer:
Tungsten has a very high melting point (about 3380 °C) and a high resistivity. Due to its high melting point, it can withstand very high temperatures without melting. Its high resistivity allows it to produce a large amount of heat and light when electric current passes through it. That is why tungsten is used for the filament of electric lamps.
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Answer:
The conductors of electric heating devices are made of an alloy for the following reasons.
(i) An alloy has a higher resistivity than a pure metal. So a large amount of heat is produced when current passes through it.
(ii) An alloy has a high melting point and does not oxidise (burn) easily, even at high temperatures. So, the heating element lasts longer.
c) Why is the series arrangement not used for domestic circuits?
The series arrangement is not used for domestic circuits for the following reasons:
(i) In a series circuit, same current flows through every component. But different household appliances require different currents to work properly.
(ii) If one appliance fails or is switched off, the circuit is broken and all other appliances stop working.
(d) How does the resistance of a wire vary with its area of cross-section?
Answer:
The resistance (\(R\)) of a wire is inversely proportional to its cross-sectional area (\(A\)). This means that when the cross-sectional area of a wire increases, its resistance decreases, and vice versa.
Mathematically, \(R \propto \frac{1}{A}\).
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
Copper and aluminium are good conductors of electricity. They also have very low resistivity, so they offer little resistance to the flow of current. Due to this, only a small amount of electrical energy is lost as heat when electricity passes through them. That is why they are usually used for electricity transmission.