Heating Effect of Electric Current and Electric Power Questions in English

(N/A) The heating effect of electric current is utilized in various electrical appliances to produce heat for specific purposes.
Some common practical applications include:
$1$. Electric Iron: Used for pressing clothes.
$2$. Electric Toaster: Used for browning bread.
$3$. Electric Oven: Used for baking and cooking.
$4$. Electric Kettle: Used for boiling water.
$5$. Electric Geyser: Used for heating water.
$6$. Electric Fuse: $A$ safety device that melts and breaks the circuit when excessive current flows,protecting appliances from damage.
$7$. Electric Bulb: Uses the heating effect to produce light through a tungsten filament.

(N/A) Electric power is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit.
Mathematically,it is expressed as:
$P = \frac{W}{t} = VI = I^2R = \frac{V^2}{R}$
Where:
$P$ = Electric power
$W$ = Work done or energy consumed
$t$ = Time taken
$V$ = Potential difference
$I$ = Current
$R$ = Resistance
The $SI$ unit of electric power is the watt $(W)$,which is equivalent to $1 \text{ joule/second}$ $(1 \text{ J/s})$.

(N/A) The commercial unit of electrical energy is kilowatt-hour $(kWh)$.
$1 kWh = 1 kW \times 1 \text{ hour}$
$= 1000 W \times 3600 s$
$= 1000 J/s \times 3600 s$
$= 3.6 \times 10^6 J$
It is also known as the Board of Trade Unit $(BOTU)$.

(N/A) Joule's law of heating states that the heat produced $(H)$ in a conductor is:
$(i)$ Directly proportional to the square of the current $(I)$ flowing through it, i.e., $H \propto I^2$.
$(ii)$ Directly proportional to the resistance $(R)$ of the conductor, i.e., $H \propto R$.
$(iii)$ Directly proportional to the time $(t)$ for which the current flows through the conductor, i.e., $H \propto t$.
Combining these three factors, we get the mathematical expression:
$H = I^2Rt$
Where $H$ is the heat energy in Joules, $I$ is the current in Amperes, $R$ is the resistance in Ohms, and $t$ is the time in seconds.

(A) Given:
Voltage $(V) = 5.0 \text{ V}$
Current $(I) = 100 \text{ mA} = 100 \times 10^{-3} \text{ A} = 0.1 \text{ A}$
$(i)$ Power $(P)$ is calculated as:
$P = V \times I = 5.0 \text{ V} \times 0.1 \text{ A} = 0.5 \text{ W}$
$(ii)$ Resistance $(R)$ is calculated using Ohm's Law:
$R = \frac{V}{I} = \frac{5.0 \text{ V}}{0.1 \text{ A}} = 50 \, \Omega$

(C) The total power $P$ consumed by two geysers is $P = 2 \times 1.1 \, kW = 2.2 \, kW = 2200 \, W$.
Given supply voltage $V = 220 \, V$.
The current $I$ flowing through the circuit is calculated using the formula $I = P / V$.
$I = 2200 \, W / 220 \, V = 10 \, A$.
To safely operate the appliances,the fuse rating must be slightly higher than the current drawn by the circuit to prevent nuisance tripping while protecting against overcurrent.
Therefore,the minimum rating of the fuse should be greater than $10 \, A$.

(N/A) $(i)$ The family consumed electrical energy.
$(ii)$ The energy consumed was measured in kilowatt-hour $(kWh)$.

(N/A)

Name of appliance	Power rating, Voltage rating, and Daily use
Fluorescent tube	$40 \, W, 220 \, V, 5 \, \text{hours}$
Fan	$60 \, W, 220 \, V, 15 \, \text{hours}$
Geyser	$2000 \, W, 220 \, V, 1.5 \, \text{hours}$

(FALSE) The statement is False.
Black surfaces are actually excellent absorbers of heat because they absorb most of the radiation that falls on them,rather than reflecting it. This is why dark-colored objects heat up faster than light-colored ones when exposed to sunlight.

(N/A) The heat produced $(H)$ in a current-carrying conductor is given by Joule's Law of Heating: $H = I^2Rt$.
$(i)$ The factors on which the heat produced depends are:
$(a)$ The square of the current flowing through the conductor $(I^2)$.
$(b)$ The resistance of the conductor $(R)$.
$(c)$ The time for which the current flows through the conductor $(t)$.
$(ii)$ $A$ practical application of this heating effect is the electric iron,which converts electrical energy into heat energy to press clothes.

(N/A) The filament of an electric bulb is made of $Tungsten$.
$(b)$ The characteristics that make $Tungsten$ suitable for this purpose are:
$1$. It has a very high melting point (approximately $3380^{\circ}C$),which prevents it from melting even when it becomes white-hot during the emission of light.
$2$. It has high resistivity,which allows it to produce a large amount of heat when an electric current passes through it,enabling it to glow brightly.

(C) Sample $C$ is suitable for heating elements of electrical appliances.
Reason: Heating elements require materials with high resistivity so that they can produce a significant amount of heat when an electric current passes through them, according to Joule's Law of heating $(H = I^2Rt)$. Among the given samples, Sample $C$ has the highest resistivity $(100 \times 10^{-6} \Omega \cdot m)$, making it the most efficient choice for generating heat.

(N/A) Tungsten is used for making the filament of electric lamps because:
$(i)$ It has a very high melting point (approximately $3380^{\circ}C$),which allows it to glow at high temperatures without melting.
$(ii)$ It has high resistivity,which allows it to produce a large amount of heat when current passes through it.
$(b)$ Heating elements are made of alloys because:
$(i)$ Alloys have higher resistivity than pure metals,which generates more heat.
$(ii)$ Alloys do not oxidize (burn) easily even at high temperatures,which increases the longevity of the appliance.

(C) The wires of the connecting cord of an electric heater are typically made of copper $(Cu)$,which has extremely low electrical resistance. According to Joule's law of heating,$H = I^2Rt$,the heat produced is directly proportional to the resistance $(R)$. Since the resistance of copper wires is negligible,very little heat is produced,and they do not glow.
Conversely,the heating element is made of an alloy like nichrome,which has a very high electrical resistance. Due to this high resistance,a large amount of heat is generated when current flows through it,causing it to glow red-hot.

(N/A) $1$. Tungsten has a very high melting point (approximately $3380 ^\circ C$), which allows it to glow at high temperatures without melting.
$2$. It has a high resistance, which allows it to produce a large amount of heat and light when an electric current passes through it.

(N/A) The electrical fuse works on the heating effect of electric current. When the current exceeds the safe limit,the heat generated melts the fuse wire,breaking the circuit.
$(b)$ An electrical fuse is always connected in series with the live wire of the domestic circuit to protect appliances from high current surges.

(N/A) Electrical resistance is the property of a conductor to oppose the flow of electric current through it.
$(i)$ When a low current passes through a conductor for a short duration, the heat generated $(H = I^2Rt)$ is negligible. Therefore, there is no significant rise in temperature, and the resistance remains effectively unchanged.
$(ii)$ When a heavy current passes through a conductor for about $30$ seconds, a significant amount of heat is produced. This leads to a considerable rise in the temperature of the conductor. Since the resistance of most metallic conductors increases with an increase in temperature, the resistance of the conductor will increase.

(D) The power $P$ and voltage $V$ are given for both lamps. The current $I$ drawn by each lamp can be calculated using the formula $P = VI$,or $I = P/V$.
For the first lamp,$I_1 = P_1 / V = 100 \, W / 220 \, V = 10/22 \, A = 5/11 \, A$.
For the second lamp,$I_2 = P_2 / V = 60 \, W / 220 \, V = 6/22 \, A = 3/11 \, A$.
Since the lamps are connected in parallel,the total current $I$ drawn from the supply is the sum of the individual currents:
$I = I_1 + I_2 = 5/11 \, A + 3/11 \, A = 8/11 \, A$.
$I \approx 0.727 \, A$.

(N/A) Given: Power, $P = 2 kW = 2000 W$, Time, $t = 2 h$ daily.
Energy consumed per day, $E = P \times t = 2 kW \times 2 h = 4 kWh$.
In $SI$ units (Joules): $4 kWh = 4 \times 3.6 \times 10^6 J = 14.4 \times 10^6 J$.
In commercial units: $4 kWh$ (or $4$ units).
$(b)$ Total energy consumed in June ($30$ days):
Total Energy $= 4 kWh/day \times 30 days = 120 kWh$.
Cost of electricity $= 120 \text{ units} \times ₹ 3.00/unit = ₹ 360$.

(A) $(i)$ For bulb $A$:
Resistance,$R_{A} = \frac{V^{2}}{P_{A}} = \frac{(120)^{2}}{90} = 160 \, \Omega$.
Therefore,current $I_{A} = \frac{V}{R_{A}} = \frac{120}{160} = 0.75 \, A$.
For bulb $B$:
Resistance,$R_{B} = \frac{V^{2}}{P_{B}} = \frac{(120)^{2}}{60} = 240 \, \Omega$.
Therefore,current $I_{B} = \frac{V}{R_{B}} = \frac{120}{240} = 0.5 \, A$.
$(ii)$ Since the bulbs are connected in parallel to a $120 \, V$ source,each bulb operates at its rated voltage. Bulb $A$ has a higher power rating $(90 \, W)$ compared to bulb $B$ $(60 \, W)$,therefore bulb $A$ will consume more energy.

(B) The filament of the bulb and the connecting wires are connected in series,so the same current flows through both.
According to Joule's law of heating,the heat produced is given by $H = I^2Rt$.
Since the resistance $(R)$ of the connecting wires is negligibly small compared to the resistance of the filament,the heat produced in the wires is negligible.
Conversely,the filament has a high resistance,which leads to a significant amount of heat production,causing it to glow.
Therefore,the bulb glows while the connecting wires remain cool.

(B) Let the resistance of each wire be $R$.
When connected in series,the total resistance is $R_{S} = R + R = 2R$.
The heat produced by the series combination is given by $H = \frac{V^{2}}{R_{S}} t = \frac{V^{2}}{2R} t$.
The heat produced by an individual resistor (if connected alone to the same voltage source) is $H' = \frac{V^{2}}{R} t$.
Comparing the two,we find $H = \frac{1}{2} \left( \frac{V^{2}}{R} t \right) = \frac{H'}{2}$.
Thus,the heat produced by the series combination is half the heat produced by a single wire.

(N/A) Tungsten is a metal that has a very high melting point and high resistivity,which allows it to glow at high temperatures without melting,making it ideal for bulb filaments.
$(b)$ Copper and aluminium are excellent conductors of electricity because they possess very low resistivity,which minimizes energy loss during the transmission of electric current.
$(c)$ Alloys are used in electrical heating devices because their resistivity is significantly higher than that of their constituent pure metals,and they do not oxidize (burn) readily at high temperatures.

(N/A) Suppose a current $I$ flows through a conductor of resistance $R$ for time $t$ under a potential difference of $V$ volts.
The charge flowing through the conductor is $Q = I \times t$.
The work done $(W)$ in moving a charge $Q$ across a potential difference $V$ is given by:
$W = V \times Q$
Substituting $Q = I \times t$ into the equation:
$W = V \times I \times t$
$(b)$ Given:
For bulb $1$: $P_1 = 40 \ W$,$V_1 = 240 \ V$
For bulb $2$: $P_2 = 25 \ W$,$V_2 = 240 \ V$
We know that $P = \frac{V^2}{R}$,so $R = \frac{V^2}{P}$.
$R_1 = \frac{240^2}{40} = \frac{57600}{40} = 1440 \ \Omega$
$R_2 = \frac{240^2}{25} = \frac{57600}{25} = 2304 \ \Omega$
Comparing $R_1$ and $R_2$,$R_2 > R_1$.
Ratio: $\frac{R_2}{R_1} = \frac{2304}{1440} = 1.6$
Thus,the $25 \ W$ bulb has higher resistance,and it is $1.6$ times the resistance of the $40 \ W$ bulb.

(N/A) The heating effect of electric current is the phenomenon where heat is generated in a conductor when an electric current passes through it.
When an electric current flows through a resistor,the electrons move through the conductor and collide with the atoms or ions of the resistor. During these collisions,the kinetic energy of the electrons is transferred to the atoms of the resistor,causing them to vibrate more vigorously. This increase in the vibrational energy of the atoms manifests as an increase in the temperature of the resistor,resulting in the production of heat.
Two devices based on the heating effect of electric current are:
$(i)$ Electric iron
$(ii)$ Electric geyser

(N/A) The commercial unit of electric energy is kilowatt-hour $(kWh)$.
The $SI$ unit of electrical energy is Joule $(J)$.
$1 \, kWh = 1000 \, W \times 3600 \, s = 3.6 \times 10^6 \, J$.
$(b)$ According to Joule's law of heating,$H = I^2Rt$.
If the current $I$ becomes $3I$,the new heat $H' = (3I)^2Rt = 9I^2Rt = 9H$.
Thus,the heat produced will become nine times the initial value.
$(c)$ Initial heat $H = I^2Rt$.
If the resistance $R$ is replaced by a conductor of double resistance $(2R)$,the new heat $H' = I^2(2R)t = 2I^2Rt = 2H$.
Thus,the heat produced will become twice the initial value.

(N/A) In a series arrangement,if any one component stops working,the circuit breaks,and none of the other devices will function.
$(b)$ Tungsten is used because it has a very high melting point $(3380 ^\circ C)$ and high resistivity,allowing it to emit light without melting.
$(c)$ Alloys have higher resistivity than pure metals and do not oxidize (burn) easily at high temperatures,making them ideal for heating elements.
$(d)$ Copper and aluminium have very low resistivity,which minimizes energy loss during the transmission of electricity.
$(e)$ The cord has very low resistance,so it produces negligible heat,whereas the heating element has high resistance,causing it to glow due to the heating effect of current.

(N/A) Power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.
$1$ watt is the power consumed by a device that carries a current of $1 \ A$ when operated at a potential difference of $1 \ V$, whereas $1$ watt hour is the amount of energy consumed when $1$ watt of power is used for $1$ hour.
The commercial unit of electric energy is $kWh$.
$1 \ kWh = 1000 \ W \times 3600 \ s = 3.6 \times 10^6 \ J$.
Total energy consumed by the heater in $30$ days:
$E = \text{Power} \times \text{Time} \times \text{Days} = 1000 \ W \times 2 \ h \times 30 = 60000 \ Wh = 60 \ kWh$.
Cost of energy = $60 \ kWh \times ₹ 5.00/kWh = ₹ 300$.

(N/A) $(i)$ Current $(I)$ is calculated using the formula $P = V \times I$,so $I = P / V = 750\, W / 200\, V = 3.75\, A$.
$(ii)$ Resistance $(R)$ is calculated using Ohm's Law $V = I \times R$,so $R = V / I = 200\, V / 3.75\, A \approx 53.33\, \Omega$.
$(iii)$ Energy consumed $(E)$ is calculated as $E = P \times t = 750\, W \times 2\, h = 1500\, Wh = 1.5\, kWh$.

(N/A) We know that the power input is $P = V \times I$.
Thus,the current is $I = P / V$.
$(a)$ When heating is at the maximum rate $(P = 840\, W)$:
$I = 840 / 220 = 3.82\, A$.
The resistance is $R = V / I = 220 / 3.82 = 57.60\, \Omega$.
$(b)$ When heating is at the minimum rate $(P = 360\, W)$:
$I = 360 / 220 = 1.64\, A$.
The resistance is $R = V / I = 220 / 1.64 = 134.15\, \Omega$.

(B) Given: Heat produced $H = 100 \, J$,Resistance $R = 4 \, \Omega$,Time $t = 1 \, s$.
We know that the formula for heat produced is $H = I^2 R t$.
Rearranging to find current $I$:
$I = \sqrt{\frac{H}{R t}} = \sqrt{\frac{100}{4 \times 1}} = \sqrt{25} = 5 \, A$.
Now,using Ohm's Law to find the potential difference $V$:
$V = I \times R$
$V = 5 \, A \times 4 \, \Omega = 20 \, V$.
Therefore,the potential difference across the resistor is $20 \, V$.

(A) Given: Potential difference $V = 220 \ V$ and current $I = 0.50 \ A$.
The formula for electric power is $P = V \times I$.
Substituting the given values: $P = 220 \ V \times 0.50 \ A$.
Therefore,the power of the bulb is $P = 110 \ W$.

(A) Given: Power $P = 400 \ W$,Time $t = 8 \ h$ per day for $30$ days,Rate $= ₹ 3.00$ per $kWh$.
Total time in hours $= 8 \times 30 = 240 \ h$.
Total energy consumed in $kWh = \frac{P \times t}{1000} = \frac{400 \times 240}{1000} = 96 \ kWh$.
Cost of energy $= \text{Total energy} \times \text{Rate} = 96 \times 3 = ₹ 288.00$.
Therefore,the cost to operate the refrigerator for $30$ days is ₹ $288.00$.

(A) Given:
Resistance $R = 480 \Omega$
Voltage $V = 220 \text{ V}$
Time $t = 10 \text{ s}$
The formula for electrical energy consumed is $E = \frac{V^2 t}{R}$.
Substituting the given values:
$E = \frac{(220)^2 \times 10}{480}$
$E = \frac{48400 \times 10}{480}$
$E = \frac{484000}{480}$
$E \approx 1008.33 \text{ J}$.

(25.2 KWH) Energy consumed by one $100 \text{ W}$ bulb working for $2 \text{ hours}$ daily:
$= \frac{100 \text{ W}}{1000} \times 2 \text{ h} = 0.2 \text{ kWh}$
Energy consumed by four $40 \text{ W}$ bulbs working for $4 \text{ hours}$ daily:
$= \frac{4 \times 40 \text{ W}}{1000} \times 4 \text{ h} = 0.64 \text{ kWh}$
Total energy consumed everyday:
$= 0.2 \text{ kWh} + 0.64 \text{ kWh} = 0.84 \text{ kWh}$
Total energy consumed in $30 \text{ days}$:
$= 0.84 \text{ kWh} \times 30 = 25.2 \text{ kWh}$

(N/A) Given: Voltage $V = 200 \ V$, Power $P = 100 \ W$, Number of bulbs $n = 5$, Time $t = 4 \ hours$.
$(a)$ We know that $P = \frac{V^2}{R}$, therefore, the resistance of each bulb is:
$R = \frac{V^2}{P} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \ \Omega$.
$(b)$ Electrical energy consumed by one bulb in $4 \ hours$ is:
$E_1 = P \times t = 100 \ W \times 4 \ h = 400 \ Wh$.
Total energy consumed by $5$ bulbs in $4 \ hours$ is:
$E_{total} = 5 \times 400 \ Wh = 2000 \ Wh = 2 \ kWh$.
$(c)$ Cost of consumed electricity:
Since $1 \ unit = 1 \ kWh$ and the rate is $50 \ paise = ₹ 0.50$ per unit:
$Cost = 2 \ units \times ₹ 0.50/unit = ₹ 1.00$.

$1$. Energy consumed by $5$ tube lights of $40 \text{ W}$ each for $5 \text{ hours}$ per day:
$E_1 = 5 \times 40 \text{ W} \times 5 \text{ h} = 1000 \text{ Wh} = 1 \text{ kWh}$.
$2$. Energy consumed by an electric press of $500 \text{ W}$ for $4 \text{ hours}$ per day:
$E_2 = 500 \text{ W} \times 4 \text{ h} = 2000 \text{ Wh} = 2 \text{ kWh}$.
$3$. Total energy consumed per day:
$E_{\text{total/day}} = 1 \text{ kWh} + 2 \text{ kWh} = 3 \text{ kWh}$.
$4$. Total energy consumed in a month of $30 \text{ days}$:
$E_{\text{month}} = 3 \text{ kWh/day} \times 30 \text{ days} = 90 \text{ kWh}$.

(N/A) $(i)$ The resistance $R$ of the lamp is calculated using the formula $R = \frac{V^2}{P}$. Substituting the given values: $R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega$.
$(ii)$ The energy consumed $E$ is calculated using the formula $E = P \times t$. Given $P = 100 \, W$ and $t = 5 \, h$,we get $E = 100 \, W \times 5 \, h = 500 \, Wh$. Converting to $kWh$,$E = \frac{500}{1000} \, kWh = 0.5 \, kWh$.

(N/A) Given: Power $P = 1100 \ W$,Voltage $V = 220 \ V$.
$1$. Calculation of Resistance:
Using the formula $R = \frac{V^2}{P}$,we have:
$R = \frac{(220)^2}{1100} = \frac{48400}{1100} = 44 \ \Omega$.
$2$. Calculation of Energy Consumed:
The month of November has $30$ days.
Total time $t = 4 \ \text{hours/day} \times 30 \ \text{days} = 120 \ \text{hours}$.
Energy $E = P \times t = 1100 \ \text{W} \times 120 \ \text{h} = 132000 \ \text{Wh}$.
To convert into $kWh$,divide by $1000$:
$E = \frac{132000}{1000} = 132 \ kWh$.

(3 A) The power dissipated by a resistor is given by the formula $P = I^2 R$.
Given that the maximum power each resistor can withstand is $P_{max} = 18 \,W$ and the resistance is $R = 2 \,\Omega$.
First,calculate the maximum current $I_{max}$ that any single resistor can handle:
$I_{max}^2 = P_{max} / R = 18 / 2 = 9$
$I_{max} = \sqrt{9} = 3 \,A$.
In the given circuit,resistor $A$ is in series with the parallel combination of resistors $B$ and $C$.
The total current $I$ flows through resistor $A$. Therefore,the maximum current $I$ that can flow through the circuit is limited by the maximum current capacity of resistor $A$,which is $3 \,A$.
If $I = 3 \,A$ flows through the circuit,then at the parallel junction,the current splits equally between $B$ and $C$ (since $R_B = R_C = 2 \,\Omega$).
Thus,current through $B$ is $I_B = 1.5 \,A$ and current through $C$ is $I_C = 1.5 \,A$.
Since $1.5 \,A < 3 \,A$,both $B$ and $C$ are safe.
Therefore,the maximum current that can flow through the circuit is $3 \,A$.

(B) Given:
Power $(P)$ = $600 \ W = 0.6 \ kW$
Time $(t)$ = $6 \ hours/day \times 30 \ days = 180 \ hours$
Energy consumed $(E)$ = $P \times t = 0.6 \ kW \times 180 \ h = 108 \ kWh$
Since $1 \ unit = 1 \ kWh$, the total energy consumed is $108 \ units$.
Cost = Energy consumed $\times$ Rate per unit
Cost = $108 \times 3 = ₹ \ 324$.

(B) Power of the electric heater $= 1200 \ W = 1.2 \ kW$.
Time used per day $= 2 \ h$.
Energy consumed per day $= \text{Power} \times \text{Time} = 1.2 \ kW \times 2 \ h = 2.4 \ kWh$.
Total energy consumed in $30$ days $= 2.4 \ kWh/day \times 30 \ days = 72 \ kWh$.
Since $1 \ kWh = 1 \ \text{unit}$, total energy $= 72 \ \text{units}$.
Cost of energy $= \text{Total units} \times \text{Rate per unit} = 72 \times 5 = ₹ \ 360$.

(N/A) Electrical resistance is the property of a conductor to oppose the flow of electric current through it.
$(i)$ When a low current passes through a conductor for a short duration, the temperature of the conductor does not change significantly. Therefore, the resistance remains constant.
$(ii)$ When a heavy current passes through a conductor for about $30$ seconds, the conductor heats up due to the heating effect of electric current $(H = I^2Rt)$. Since the resistance of a metallic conductor increases with an increase in temperature, the resistance of the conductor will increase.

(B) The heat produced per unit time is given by the power formula $P = \frac{V^2}{R}$, where $V$ is the potential difference and $R$ is the resistance.
Since both wires are connected to the same battery, $V$ is constant for both.
Therefore, $P \propto \frac{1}{R}$, meaning the wire with lower resistance will produce more heat.
In a $V-I$ graph, the slope represents the resistance $(R = \frac{V}{I})$. The slope of wire $A$ is greater than the slope of wire $B$, which means $R_A > R_B$.
Since $R_B < R_A$, wire $B$ will have a lower resistance and thus will produce more heat per unit time.

(C) Heating elements in devices like toasters, heaters, and irons require materials with high electrical resistivity so that they can generate significant heat when current passes through them (Joule's heating effect, $H = I^2Rt$).
Comparing the given values:
$A$: $6.84 \times 10^{-8} \, \Omega m$ (Conductor)
$B$: $1.60 \times 10^{-8} \, \Omega m$ (Conductor)
$C$: $1.00 \times 10^{-4} \, \Omega m$ (Alloy/Resistive material)
$D$: $2.50 \times 10^{12} \, \Omega m$ (Insulator)
$E$: $4.40 \times 10^{-5} \, \Omega m$ (Alloy/Resistive material)
$F$: $2.30 \times 10^{17} \, \Omega m$ (Insulator)
Materials $C$ and $E$ have significantly higher resistivity compared to conductors like $A$ and $B$, making them suitable for heating elements. Among the options provided, $C$ is a standard choice for such applications.

(A) The resistance $R$ of an electric bulb is calculated using the formula $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power.
For bulb $A$: $R_A = \frac{220^2}{60} \approx 806.67\, \Omega$.
For bulb $B$: $R_B = \frac{220^2}{100} = 484\, \Omega$.
Since the voltage $V$ is constant for both bulbs,the resistance is inversely proportional to the power $(R \propto \frac{1}{P})$.
Therefore,the bulb with the lower power rating $(60\, W)$ will have a higher resistance.
Thus,bulb $A$ has greater resistance.

(B) The heat produced per second in a resistor is given by the power formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Since the resistance $R$ remains constant,the heat produced per second is directly proportional to the square of the potential difference,i.e.,$P \propto V^2$.
If the heat produced per second increases by a factor of $16$,then $P' = 16P$.
Substituting this into the proportionality,we get $16P \propto (V')^2$.
Since $P \propto V^2$,we have $16V^2 = (V')^2$.
Taking the square root of both sides,we get $V' = 4V$.
Therefore,the applied potential difference changes by a factor of $4$.

(B) When bulbs are connected in parallel,the potential difference across each bulb remains the same,which is equal to the supply voltage $(220\, V)$.
The brightness of a bulb depends on the power consumed by it,which is given by the formula $P = V^2 / R$.
Since the voltage $V$ is constant for both bulbs,the power consumed is directly proportional to the rated power of the bulb.
The bulb with the higher wattage rating $(100\, W)$ will consume more power and therefore glow brighter than the $60\, W$ bulb.

(B) The rate of heat produced in a conductor is given by the formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Since both conductors are connected across the same $dc$ source,the potential difference $V$ remains constant for both.
Therefore,the rate of heat produced is inversely proportional to the resistance: $Q \propto \frac{1}{R}$.
For the first conductor with resistance $R_1 = R$,the rate of heat is $Q_1 = \frac{V^2}{R}$.
For the second conductor with resistance $R_2 = 2R$,the rate of heat is $Q_2 = \frac{V^2}{2R}$.
Calculating the ratio $Q_1 / Q_2$:
$\frac{Q_1}{Q_2} = \frac{V^2 / R}{V^2 / 2R} = \frac{V^2}{R} \times \frac{2R}{V^2} = 2$.
Thus,the value of $Q_1 / Q_2$ is $2$.

(A) The power $P$ of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Since both lamps are designed to operate on the same voltage $V$,we have $R = \frac{V^2}{P}$.
For the first lamp,$P_{1} = 400 \ W$,so $R_{1} = \frac{V^2}{400}$.
For the second lamp,$P_{2} = 200 \ W$,so $R_{2} = \frac{V^2}{200}$.
Dividing $R_{2}$ by $R_{1}$,we get $\frac{R_{2}}{R_{1}} = \frac{V^2 / 200}{V^2 / 400} = \frac{400}{200} = 2$.
Therefore,$R_{2} = 2R_{1}$.