NCERT Solutions For Class 10 Physics: Electricity

Intext Questions (Page 172-173)

Question 1: What does an electric circuit mean?

Answer-
An electric circuit means a continuous and closed path of an electric current. If the circuit is broken anywhere, the current stops flowing. A typical circuit comprises a cell (or battery), an electric bulb, an ammeter, and a plug key, with the current flowing from the positive terminal to the negative terminal.

Question 2: Define the unit of current.

Answer-
The SI unit of electric current is the ampere ($\text{A}$), named after Andre-Marie Ampere. One ampere ($\text{1 A}$) is defined as the flow of one coulomb ($\text{1 C}$) of charge per second ($\text{1 s}$).
$1 \text{ A} = 1 \text{ C}/1 \text{ s}$.

Question 3: Calculate the number of electrons constituting one coulomb of charge.

Answer-
The charge contained in nearly $\underline{6 \times 10^{18} \text{ electrons}}$ is equivalent to one coulomb ($\text{1 C}$). Since one electron possesses a negative charge of $\underline{1.6 \times 10^{-19} \text{ C}}$, the number of electrons constituting one coulomb is $\frac{1 \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} \approx \underline{6 \times 10^{18} \text{ electrons}}$.

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Intext Questions (Page 174)

Question 1: Name a device that helps to maintain a potential difference across a conductor.

Answer-
A device that helps to maintain a potential difference across a conductor is a battery, which consists of one or more electric cells. The chemical action within a cell generates the potential difference across its terminals.

Question 2: What is meant by saying that the potential difference between two points is 1 V?

Answer-
Saying that the potential difference between two points is 1 volt ($\text{1 V}$) means that 1 joule ($\text{1 J}$) of work is done to move a charge of 1 coulomb ($\text{1 C}$) from one point to the other in a current-carrying conductor.
$1 \text{ V} = 1 \text{ J}/1 \text{ C}$.

Question 3: How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer-
The potential difference ($\text{V}$) is defined as the work done ($\text{W}$) per unit charge ($\text{Q}$), $\text{V} = \text{W}/\text{Q}$.
We are given $\text{V} = 6 \text{ V}$ and $\text{Q} = 1 \text{ C}$.
Energy (Work done, $\text{W}$) = $\text{V} \times \text{Q} = 6 \text{ V} \times 1 \text{ C} = \underline{6 \text{ J}}$.
$\underline{6 \text{ J}}$ of energy is given to each coulomb of charge passing through a 6 V battery.

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Intext Questions (Page 181)

Question 1: On what factors does the resistance of a conductor depend?

Answer-
Precise measurements based on Ohm’s law show that the resistance ($\text{R}$) of a uniform metallic conductor depends on three factors:
(i) Directly on its length ($\text{l}$).
(ii) Inversely on its area of cross-section ($\text{A}$).
(iii) On the nature of its material (via its electrical resistivity, $\rho$).

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?

Answer-
Current will flow more easily through a thick wire.
Reason: The resistance ($\text{R}$) of a conductor is inversely proportional to its area of cross-section ($\text{A}$). A thick wire has a larger area of cross-section ($\text{A}$), meaning it will offer a lower resistance ($\text{R}$). Since current ($\text{I}$) is inversely proportional to resistance ($\text{I} = \text{V}/\text{R}$), a lower resistance allows the current to flow more easily.

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?

Answer-
According to Ohm’s law, $I = V/R$. If the resistance ($\text{R}$) remains constant, the current ($\text{I}$) is directly proportional to the potential difference ($\text{V}$).
If the potential difference ($\text{V}$) is decreased to half its former value, the current ($\text{I}$) will also decrease to half its former value.

Question 4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer-
Coils of electric toasters and electric irons are made of alloys (like Nichrome) rather than pure metals for two main reasons:

1. High Resistivity: The resistivity of an alloy is generally higher than that of its constituent metals. This higher resistance means more heat is generated for the same current ($H = I^2Rt$).
2. High Temperature Tolerance: Alloys, such as Nichrome, do not oxidise (burn) readily at high temperatures, allowing them to operate safely and effectively in heating devices without melting or deteriorating.

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?

Answer-
(a) A better conductor has a lower electrical resistivity.

• Iron resistivity: $10.0 \times 10^{-8} \text{ }\Omega \text{ m}$.
• Mercury resistivity: $94.0 \times 10^{-8} \text{ }\Omega \text{ m}$.
  Since iron has a lower resistivity than mercury, Iron is a better conductor.

(b) The best conductor is the material with the lowest resistivity.
From the list of conductors in Table 11.2, Silver has the lowest resistivity ($1.60 \times 10^{-8} \text{ }\Omega \text{ m}$).

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Intext Questions (Page 188)

Question 1: Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and $10^6$ Ω, (b) 1 Ω and $10^3$ Ω, and $10^6$ Ω.

Answer-
When resistors are connected in parallel, the reciprocal of the equivalent resistance ($\text{R}_p$) is the sum of the reciprocals of the individual resistances. The equivalent resistance ($\text{R}_p$) is less than the value of the smallest individual resistance in the combination.

(a) For 1 $\Omega$ and $10^6 \Omega$:
$\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^6} \approx 1$
$R_p$ will be slightly less than 1 $\Omega$.

(b) For 1 $\Omega$, $10^3 \Omega$, and $10^6 \Omega$:
$\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} \approx 1$
$R_p$ will also be slightly less than 1 $\Omega$, which is the smallest resistance.

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?

Answer-

1. Total Resistance ($\text{R}_p$) of the three appliances in parallel:
   $\frac{1}{R_p} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} = \frac{5 + 10 + 1}{500} = \frac{16}{500} \Omega^{-1}$
   $R_p = \frac{500}{16} = \underline{31.25 \Omega}$

2. Total Current ($\text{I}$) drawn by the three appliances:
   The total current is drawn from a 220 V source ($\text{V} = 220 \text{ V}$).
   $I = \frac{V}{R_p} = \frac{220 \text{ V}}{31.25 \Omega} = \underline{7.04 \text{ A}}$

3. Resistance of the Electric Iron ($\text{R}_{iron}$):
   The iron takes the same current ($7.04 \text{ A}$) from the same source ($220 \text{ V}$).
   $R_{iron} = \frac{V}{I} = \frac{220 \text{ V}}{7.04 \text{ A}} = \underline{31.25 \Omega}$
   The resistance of the electric iron is $\underline{31.25 \Omega}$ and the current through it is $\underline{7.04 \text{ A}}$.

Question 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer-
Connecting electrical devices in parallel offers several advantages over connecting them in series:

1. Independent Operation: In a parallel circuit, if one component fails, the circuit is not broken, and the other components continue to work. In a series circuit, if one component fails, none of the components works.
2. Current Division: A parallel circuit divides the current through the electrical gadgets. This is crucial because different gadgets (like a bulb and a heater) require currents of widely different values to operate properly, which is impracticable in a series circuit where the current is constant throughout.
3. Voltage Supply: Each component in a parallel circuit receives the full potential difference of the battery/source ($\text{V}$).
4. Overall Resistance: The total resistance in a parallel circuit is decreased, which is helpful when driving multiple devices simultaneously.

Question 4: How can three resistors of resistances 2 $\Omega$, 3 $\Omega$, and 6 $\Omega$ be connected to give a total resistance of (a) 4 $\Omega$, (b) 1 $\Omega$?

Answer-
(a) To obtain a total resistance of $\underline{4 \Omega}$:
We need a combination where $\text{R} > 3 \Omega$ but $\text{R} < (2+3+6) = 11 \Omega$.
Connect the $2 \Omega$ resistor in series with the parallel combination of the $3 \Omega$ and $6 \Omega$ resistors:
$R_p (3 \text{ and } 6) = \frac{1}{\frac{1}{3} + \frac{1}{6}} = \frac{1}{\frac{2+1}{6}} = \frac{6}{3} = 2 \Omega$
Total resistance $R = 2 \Omega + R_p = 2 \Omega + 2 \Omega = \underline{4 \Omega}$.

(b) To obtain a total resistance of $\underline{1 \Omega}$:
Since $1 \Omega$ is smaller than the smallest individual resistor ($2 \Omega$), all three must be connected in parallel.
$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 \Omega^{-1}$
Total resistance $R = \underline{1 \Omega}$.

Question 5: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 $\Omega$, 8 $\Omega$, 12 $\Omega$, 24 $\Omega$?

Answer-
(a) Highest Total Resistance: The highest resistance is secured when all resistors are connected in series.
$R_s = R_1 + R_2 + R_3 + R_4 = 4 \Omega + 8 \Omega + 12 \Omega + 24 \Omega = \underline{48 \Omega}$

(b) Lowest Total Resistance: The lowest resistance is secured when all resistors are connected in parallel.
$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2} \Omega^{-1}$
$R_p = \underline{2 \Omega}$

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Intext Questions (Page 190)

Question 1: Why does the cord of an electric heater not glow while the heating element does?

Answer-
The cord and the heating element are both part of the circuit and carry the same current. The difference lies in their resistance and material.

1. Heating Element: The heating element is made of an alloy (like Nichrome), which has a very high resistance and a high melting point. Due to its high resistance, it generates a large amount of heat ($H = I^2Rt$), causing it to become incandescent and glow.
2. Connecting Cord: The cord is made of materials like copper or aluminium. These metals have very low resistivity (they are good conductors). Due to their negligible resistance, they generate very little heat and therefore do not glow.

Question 2: Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer-
Given: Charge $Q = 96000 \text{ C}$; Time $t = 1 \text{ hour} = 3600 \text{ s}$; Potential difference $V = 50 \text{ V}$.
The heat generated ($\text{H}$) is equal to the energy supplied by the source, $H = VIt$.
First, calculate the current $I = Q/t$:
$I = \frac{96000 \text{ C}}{3600 \text{ s}} \approx 26.67 \text{ A}$
Now calculate the heat generated $H = VIt$:
$H = 50 \text{ V} \times 26.67 \text{ A} \times 3600 \text{ s} = \underline{4,800,000 \text{ J}}$
(Alternatively, $H = VQ$: $H = 50 \text{ V} \times 96000 \text{ C} = \underline{4,800,000 \text{ J}}$).

Question 3: An electric iron of resistance 20 $\Omega$ 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 $t = 30 \text{ s}$.
The heat developed ($\text{H}$) is given by Joule’s law of heating, $H = I^2Rt$.
$H = (5 \text{ A})^2 \times (20 \Omega) \times (30 \text{ s})$
$H = 25 \times 20 \times 30 = \underline{15,000 \text{ J}}$

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Intext Questions (Page 192)

Question 1: What determines the rate at which energy is delivered by a current?

Answer-
The rate at which electric energy is delivered (dissipated or consumed) by a current is called electric power ($\text{P}$). Power is determined by the expressions:
$P = V I$
$P = I^2R$
$P = V^2/R$

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.

Answer-
Given: Voltage $V = 220 \text{ V}$; Current $I = 5 \text{ A}$; Time $t = 2 \text{ h}$.

1. Power of the motor ($\text{P}$):
   $P = V I = 220 \text{ V} \times 5 \text{ A} = \underline{1100 \text{ W}}$

2. Energy Consumed ($\text{E}$): Energy is the product of power and time. Since the time is in hours, we calculate the energy in watt hour ($\text{W h}$) and kilowatt hour ($\text{kW h}$).
   $\text{E} = \text{P} \times \text{t} = 1100 \text{ W} \times 2 \text{ h} = 2200 \text{ W h}$
   $\text{E} = \underline{2.2 \text{ kW h}}$
   (If the energy is requested in Joules: $E = 2.2 \times 3.6 \times 10^6 \text{ J} = 7.92 \times 10^6 \text{ J}$).

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Exercise Questions (Page 193-194)

Question 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) 1/25 (b) 1/5 (c) 5 (d) 25

Answer-

1. Resistance of each part: $\text{r} = \text{R}/5$.
2. Parallel combination ($\text{R}'$): $\frac{1}{R'} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}$.
   $\text{R}' = r/5 = (\text{R}/5)/5 = \text{R}/25$
3. Ratio $\text{R}/\text{R}'$:
   $\frac{R}{R'} = \frac{R}{R/25} = \underline{25}$
   The correct option is (d) 25.

Question 2: Which of the following terms does not represent electrical power in a circuit?
(a) $I^2R$ (b) $IR^2$ (c) $VI$ (d) $V^2/R$

Answer-
Electrical power ($\text{P}$) is represented by the equations $P = VI$, $P = I^2R$, and $P = V^2/R$. The term $IR^2$ is not a standard representation of electrical power.
The correct option is (b) $<u>IR^2</u>$.

Question 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-

1. Resistance of the bulb ($\text{R}$): Resistance is constant. Use $P = V^2/R$:
   $R = \frac{V_{rated}^2}{P_{rated}} = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} = 484 \Omega$
2. Power consumed at 110 V ($\text{P}'$):
   $P' = \frac{V_{new}^2}{R} = \frac{(110 \text{ V})^2}{484 \Omega} = \frac{12100}{484} = \underline{25 \text{ W}}$
   The power consumed will be (d) 25 W.

Question 4: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then 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-
Let the resistance of each wire be $R$.

1. Series Resistance ($\text{R}_s$): $R_s = R + R = 2R$.
2. Parallel Resistance ($\text{R}_p$): $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = R/2$.
3. Heat Production Ratio ($\text{H} = V^2t/R$): Since $\text{V}$ and $\text{t}$ are the same, $H \propto 1/R$.
   $\frac{H_s}{H_p} = \frac{V^2t/R_s}{V^2t/R_p} = \frac{R_p}{R_s}$
   $\frac{H_s}{H_p} = \frac{R/2}{2R} = \frac{1}{4}$
   The ratio of heat produced in series and parallel is (c) 1:4.

Question 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 points between which the potential difference is to be measured.

Question 6: A copper wire has diameter 0.5 mm and resistivity of $1.6 \times 10^{-8} \Omega$ m. What will be the length of this wire to make its resistance 10 $\Omega$? How much does the resistance change if the diameter is doubled?

Answer-
Given: $d = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$; $\rho = 1.6 \times 10^{-8} \Omega \text{ m}$; $R = 10 \Omega$.

1. Calculate the length ($\text{l}$): We use the resistance formula $R = \rho l/A$.
   Area of cross-section $A = \pi (d/2)^2 = \pi (0.25 \times 10^{-3} \text{ m})^2 \approx 1.964 \times 10^{-7} \text{ m}^2$.
   $l = \frac{R A}{\rho} = \frac{(10 \Omega) \times (1.964 \times 10^{-7} \text{ m}^2)}{1.6 \times 10^{-8} \Omega \text{ m}} \approx \underline{122.7 \text{ m}}$

2. Change in resistance if diameter is doubled: Resistance is inversely proportional to the area of cross-section ($\text{R} \propto 1/\text{A}$). Since Area $A \propto d^2$, resistance $R \propto 1/d^2$.
   If the diameter ($\text{d}$) is doubled ($2\text{d}$), the area increases by a factor of $2^2 = 4$.
   The new resistance ($\text{R}'$) will be one-fourth (1/4) of the original resistance.
   $R' = R/4 = 10 \Omega / 4 = \underline{2.5 \Omega}$

Question 7: The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –
I (amperes) 0.5 1.0 2.0 3.0 4.0
V (volts) 1.6 3.4 6.7 10.2 13.2
Plot a graph between V and I and calculate the resistance of that resistor.

Answer-
A graph plotted between $\text{V}$ (y-axis) and $\text{I}$ (x-axis) will show that the plot is approximately a straight line passing through the origin, which verifies Ohm’s law.
The resistance ($\text{R}$) is calculated from the ratio $\text{R} = \text{V}/\text{I}$.
Calculate $\text{V}/\text{I}$ for each point: 3.2, 3.4, 3.35, 3.4, 3.3.
To calculate the resistance, we take the average value of $\text{V}/\text{I}$:
$R_{avg} = \frac{3.2 + 3.4 + 3.35 + 3.4 + 3.3}{5} = \frac{16.65}{5} = \underline{3.33 \Omega}$
The resistance of the resistor is approximately $\underline{3.33 \Omega}$.

Question 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 $I = 2.5 \text{ mA} = 2.5 \times 10^{-3} \text{ A}$.
Using Ohm's law, $R = V/I$:
$R = \frac{12 \text{ V}}{2.5 \times 10^{-3} \text{ A}} = \underline{4800 \Omega} \text{ or } \underline{4.8 \text{ k}\Omega}$

Question 9: A battery of 9 V is connected in series with resistors of 0.2 $\Omega$, 0.3 $\Omega$, 0.4 $\Omega$, 0.5 $\Omega$ and 12 $\Omega$, respectively. How much current would flow through the 12 $\Omega$ resistor?

Answer-
In a series combination of resistors, the current is the same in every part of the circuit and the same current flows through each resistor.

1. Total Resistance ($\text{R}_s$): $R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = \underline{13.4 \Omega}$.
2. Total Current ($\text{I}$): Using Ohm’s law, $I = V/R_s$:
   $I = \frac{9 \text{ V}}{13.4 \Omega} \approx \underline{0.67 \text{ A}}$
   The current flowing through the $12 \Omega$ resistor is the total circuit current, which is approximately $\underline{0.67 \text{ A}}$.

Question 10: How many 176 $\Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer-
Given: $R_{individual} = 176 \Omega$; $I_{total} = 5 \text{ A}$; $V = 220 \text{ V}$.
Let $n$ be the number of resistors connected in parallel.

1. Total required resistance ($\text{R}_p$): Using Ohm’s law for the whole circuit:
   $R_p = \frac{V}{I_{total}} = \frac{220 \text{ V}}{5 \text{ A}} = 44 \Omega$
2. Number of resistors ($\text{n}$): In parallel, $\frac{1}{R_p} = \frac{n}{R_{individual}}$.
   $n = R_{individual} \times \frac{1}{R_p} = 176 \Omega \times \frac{1}{44 \Omega} = \underline{4}$
   $\underline{4}$ resistors are required.

Question 11: Show how you would connect three resistors, each of resistance 6 $\Omega$, so that the combination has a resistance of (i) 9 $\Omega$, (ii) 4 $\Omega$.

Answer-
Let $R_1 = R_2 = R_3 = 6 \Omega$.

(i) To get $9 \Omega$: This requires a value larger than $6 \Omega$ but smaller than $18 \Omega$.
Connect two $6 \Omega$ resistors in parallel, and connect this combination in series with the third $6 \Omega$ resistor.
$R_p (6 \text{ and } 6) = \frac{1}{1/6 + 1/6} = \frac{1}{2/6} = 3 \Omega$
$R_{total} = R_p + 6 \Omega = 3 \Omega + 6 \Omega = \underline{9 \Omega}$

(ii) To get $4 \Omega$: This requires a value smaller than $6 \Omega$.
Connect three $6 \Omega$ resistors in parallel. (Note: $4 \Omega$ is not possible by combining all three. We must use a combination of series and parallel, or verify if the source means using only some resistors, but since the question specifies "three resistors," a combination must be used that yields $4 \Omega$).
Let's try a different combination for $4 \Omega$: Connect two $6 \Omega$ resistors in series ($12 \Omega$), and connect this combination in parallel with the third $6 \Omega$ resistor.
$\frac{1}{R_p} = \frac{1}{R_s} + \frac{1}{R_3} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} = \frac{1}{4} \Omega^{-1}$
$R_{total} = \underline{4 \Omega}$

Question 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 220 V line if the maximum allowable current is 5 A?

Answer-
Given: $V = 220 \text{ V}$; $P_{lamp} = 10 \text{ W}$; $I_{max} = 5 \text{ A}$.

1. Maximum Total Power ($\text{P}_{max}$):
   $P_{max} = V \times I_{max} = 220 \text{ V} \times 5 \text{ A} = 1100 \text{ W}$
2. Number of lamps ($\text{n}$): Since the lamps are connected in parallel, the total power is the sum of individual powers.
   $n = \frac{P_{max}}{P_{lamp}} = \frac{1100 \text{ W}}{10 \text{ W}} = \underline{110}$
   $\underline{110}$ lamps can be connected in parallel.

Question 13: A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 $\Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer-
Given: $V = 220 \text{ V}$; $R_A = R_B = 24 \Omega$.

1. Used Separately ($\text{R} = 24 \Omega$):
   $I_{sep} = \frac{V}{R} = \frac{220 \text{ V}}{24 \Omega} \approx \underline{9.17 \text{ A}}$

2. Used in Series ($\text{R}_s = R_A + R_B = 24 + 24 = 48 \Omega$):
   $I_{series} = \frac{V}{R_s} = \frac{220 \text{ V}}{48 \Omega} \approx \underline{4.58 \text{ A}}$

3. Used in Parallel ($\frac{1}{R_p} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} \implies R_p = 12 \Omega$):
   $I_{parallel} = \frac{V}{R_p} = \frac{220 \text{ V}}{12 \Omega} \approx \underline{18.33 \text{ A}}$

Question 14: Compare the power used in the 2 $\Omega$ resistor in each of the following circuits: (i) a 6 V battery in series with 1 $\Omega$ and 2 $\Omega$ resistors, and (ii) a 4 V battery in parallel with 12 $\Omega$ and 2 $\Omega$ resistors.

Answer-
(i) Series Circuit (6 V, $1 \Omega, 2 \Omega$):
Total resistance $R_s = 1 \Omega + 2 \Omega = 3 \Omega$.
Current through the $2 \Omega$ resistor $I = V/R_s = 6 \text{ V} / 3 \Omega = 2 \text{ A}$.
Power used in $2 \Omega$ resistor $P_i = I^2R = (2 \text{ A})^2 \times 2 \Omega = \underline{8 \text{ W}}$.

(ii) Parallel Circuit (4 V, $12 \Omega, 2 \Omega$):
In parallel, the potential difference across the $2 \Omega$ resistor is $V = 4 \text{ V}$.
Power used in $2 \Omega$ resistor $P_{ii} = V^2/R = (4 \text{ V})^2 / 2 \Omega = 16 \text{ V}^2 / 2 \Omega = \underline{8 \text{ W}}$.

Comparison: The power used in the $2 \Omega$ resistor is 8 W in both cases ($P_i : P_{ii}$ is $8 \text{ W} : 8 \text{ W}$, or $1:1$).

Question 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-
Since the lamps are connected in parallel across the $220 \text{ V}$ supply, the total power consumed ($\text{P}_{total}$) is the sum of their individual power ratings.
$P_{total} = 100 \text{ W} + 60 \text{ W} = 160 \text{ W}$
The total current ($\text{I}$) drawn is calculated using $P_{total} = V I$:
$I = \frac{P_{total}}{V} = \frac{160 \text{ W}}{220 \text{ V}} \approx \underline{0.73 \text{ A}}$

Question 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer-
Energy ($\text{E}$) is the product of Power ($\text{P}$) and Time ($\text{t}$).

1. TV Set: $P = 250 \text{ W}$; $t = 1 \text{ hr}$.
   $E_{TV} = 250 \text{ W} \times 1 \text{ h} = \underline{250 \text{ W h}}$

2. Toaster: $P = 1200 \text{ W}$; $t = 10 \text{ minutes} = 10/60 = 1/6 \text{ hr}$.
   $E_{Toaster} = 1200 \text{ W} \times (1/6) \text{ h} = \underline{200 \text{ W h}}$

Comparison: The 250 W TV set used for 1 hour uses more energy ($250 \text{ W h}$) than the 1200 W toaster used for 10 minutes ($200 \text{ W h}$).

Question 17: An electric heater of resistance 44 $\Omega$ draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer-
The rate at which heat is developed is equivalent to the power ($\text{P}$).
Given: Resistance $R = 44 \Omega$; Current $I = 5 \text{ A}$.
$P = I^2R = (5 \text{ A})^2 \times 44 \Omega = 25 \times 44 = \underline{1100 \text{ W}}$
The rate at which heat is developed is $\underline{1100 \text{ W}}$ (or $1100 \text{ J/s}$). (The time information, 2 hours, is extra here, as the rate is requested, not the total energy).

Question 18: Explain the following.
(a) Why is the tungsten used almost exclusively for 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?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer-
(a) Tungsten is used for electric lamp filaments because it is a strong metal with a very high melting point (3380°C). This high melting point allows the filament to retain as much heat as possible, enabling it to get very hot and emit light without melting.

(b) Coils of electric heating devices are made of an alloy (like Nichrome) because alloys have a generally higher electrical resistivity than their constituent metals. Furthermore, alloys do not oxidise (burn) readily at high temperatures, making them suitable for constant heat generation.

(c) The series arrangement is not used for domestic circuits for two major reasons:

1. Failure Dependency: If one component fails, the circuit is broken and none of the other appliances works.
2. Voltage and Current Issues: Appliances need to be operated independently at the full supply voltage, which is not possible in a series circuit where voltage is divided. Also, household appliances require currents of widely different values to operate properly, which is impossible in a series circuit where current is constant.

(d) The resistance ($\text{R}$) of a conductor varies inversely proportional to its area of cross-section ($\text{A}$). That is, $R \propto 1/A$. This means a thicker wire (larger A) has a lower resistance, and a thinner wire has a higher resistance.

(e) Copper and aluminium wires are usually employed for electricity transmission because they are good conductors of electricity. They have a very low electrical resistivity (in the range of $10^{-8} \Omega \text{ m}$), minimizing energy loss as heat during transmission.