Electrical Energy and Power Questions in English

(A) The rate of heat generation is given by the electrical power $P = i^2 R$.
We know that the resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$, where $A$ is the cross-sectional area.
The area of the cross-section of the wire is $A = \pi r^2$.
Substituting the value of $A$ into the resistance formula, we get $R = \frac{\rho l}{\pi r^2}$.
Now, substituting $R$ into the power formula, we get $P = i^2 \left( \frac{\rho l}{\pi r^2} \right) = \frac{i^2 l \rho}{\pi r^2}$.

(B) The unit of power is $Watt$ $(W)$.
$1 \ W = 1 \ J/s$ (Joule per second).
From the formula $P = VI$,$1 \ W = 1 \ V \times 1 \ A$ (Volt-Ampere).
From the formula $P = I^2R$,$1 \ W = (1 \ A)^2 \times 1 \ \Omega$ (Ampere$^2$-Ohm).
Comparing these with the options,$Amp / Volt$ is not equal to $Watt$ because $Amp / Volt = 1 / Ohm$ (Siemens),which is the unit of conductance.
Therefore,the correct option is $B$.

(B) The rate of heat generated is defined as electrical power $(P)$.
By definition,power is the product of voltage $(V)$ and current $(I)$: $P = V \times I$.
According to Ohm's Law,the voltage across a resistor is given by $V = I \times R$.
Substituting this into the power formula,we get: $P = (I \times R) \times I = I^2R$.
Therefore,the rate of heat generated is $I^2R$.

(A) The power dissipated in a resistor is given by the formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Given: $V = 220\,V$ and $P = 40\,W$.
Rearranging the formula for $R$,we get $R = \frac{V^2}{P}$.
Substituting the values: $R = \frac{(220)^2}{40} = \frac{48400}{40} = 1210\,\Omega$.
Therefore,the resistance is $1210\,\Omega$.

(B) The power $P$ consumed by an electrical appliance is given by the formula $P = V \times i$,where $V$ is the voltage and $i$ is the current.
Given: Power $P = 60\,W$ and Voltage $V = 220\,V$.
Rearranging the formula to solve for current $i$: $i = \frac{P}{V}$.
Substituting the given values: $i = \frac{60}{220} = \frac{6}{22} = \frac{3}{11}\,A$.
Therefore,the current flowing through the bulb is $3/11\,A$.

(C) The rate of heat dissipation (power) is given by $P = \frac{V^2}{R}$. Since the voltage $V$ is the same, $P \propto \frac{1}{R}$.
Resistance $R$ is given by $R = \rho \frac{l}{A}$. Since the mass $m = \text{density} \times \text{volume} = \rho_d \times A \times l$, we have $A = \frac{m}{\rho_d \times l}$.
Substituting $A$ in the resistance formula: $R = \rho \frac{l^2 \rho_d}{m}$.
For wires of the same material and mass, $R \propto l^2$.
Therefore, $\frac{P_A}{P_B} = \frac{R_B}{R_A} = \frac{l_B^2}{l_A^2}$.
Given $\frac{l_A}{l_B} = \frac{1}{2}$, we have $\frac{P_A}{P_B} = (\frac{2}{1})^2 = 4$.
Given $P_B = 5\,W$, then $P_A = 4 \times 5\,W = 20\,W$.

(A) The power $P$ of an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Since the voltage $V$ is the same for both bulbs,we have $R = \frac{V^2}{P}$.
Therefore,the resistance is inversely proportional to the power,i.e.,$R \propto \frac{1}{P}$.
For the two bulbs with powers $P_1 = 40\,W$ and $P_2 = 60\,W$,the ratio of their resistances is $\frac{R_1}{R_2} = \frac{P_2}{P_1}$.
Substituting the values,we get $\frac{R_1}{R_2} = \frac{60}{40} = \frac{3}{2}$.
Thus,the ratio of their resistances is $3:2$.

(B) The power consumed by an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $R$ is the resistance of the bulb.
Since the resistance $R$ of the bulb is constant,we have $P \propto V^2$.
Therefore,the ratio of power at voltage $V$ to the power at voltage $V_0$ is given by $\frac{P}{P_0} = \left( \frac{V}{V_0} \right)^2$.
Rearranging this,we get $P = \left( \frac{V}{V_0} \right)^2 P_0$.

(D) The heat produced in a resistor of resistance $R$ when a current $I$ flows through it for a time $t$ is given by Joule's Law of heating.
According to this law,the heat energy $H$ is expressed as $H = I^2Rt$.
Therefore,the correct option is $D$.

(C) The heat dissipated $H$ in a resistor is given by the formula $H = \frac{V^2}{R} t$,where $V$ is the voltage,$R$ is the resistance,and $t$ is the time.
Since the voltage $V$ and time $t$ are the same for both wires,the heat dissipated is inversely proportional to the resistance: $H \propto \frac{1}{R}$.
Therefore,the ratio of heat dissipated is $\frac{H_1}{H_2} = \frac{R_2}{R_1}$.
Given $R_1 = 2 \ \Omega$ and $R_2 = 4 \ \Omega$,we have $\frac{H_1}{H_2} = \frac{4}{2} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.
For the two bulbs,the resistances are $R_1 = \frac{V^2}{P_1}$ and $R_2 = \frac{V^2}{P_2}$.
When connected in parallel to a voltage source $V$,the total power $P_p$ consumed is the sum of the power consumed by each bulb.
Since the voltage across each bulb remains $V$,the power consumed by each bulb is its rated power.
Therefore,the total power $P_p = P_1 + P_2$.

(B) Let the resistance of each bulb be $R$ and the voltage of the source be $V$.
The power rating of each bulb is given by $P = V^2 / R$,which implies the resistance of each bulb is $R = V^2 / P$.
When $n$ such bulbs are connected in series,the total resistance of the circuit becomes $R_t = nR$.
The current flowing through the series circuit is $I = V / R_t = V / (nR)$.
The total power drawn by the series combination is $P_t = I^2 R_t$.
Substituting the values of $I$ and $R_t$,we get $P_t = (V / nR)^2 \times (nR) = (V^2 / n^2 R^2) \times (nR) = V^2 / (nR)$.
Since $V^2 / R = P$,we can write $P_t = (V^2 / R) / n = P / n$.

(D) The power rating of the bulb is $P_1 = 40\, W$ at a voltage $V_1 = 200\, V$.
Since the resistance $R$ of the bulb remains constant,we use the formula $P = \frac{V^2}{R}$.
Therefore,the ratio of the new power $P_2$ to the rated power $P_1$ is given by $\frac{P_2}{P_1} = \frac{V_2^2}{V_1^2}$.
Given $V_2 = 100\, V$,we have $\frac{P_2}{40} = \left( \frac{100}{200} \right)^2$.
$\frac{P_2}{40} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
$P_2 = \frac{40}{4} = 10\, W$.

(B) The power $P$ of the bulb is $1\, kW = 1000\, W$.
The voltage $V$ is $250\, V$.
The formula for resistance $R$ in terms of power and voltage is given by $R = \frac{V^2}{P}$.
Substituting the given values: $R = \frac{(250)^2}{1000} = \frac{62500}{1000} = 62.5\,\Omega$.
Therefore,the resistance of the bulb is $62.5\,\Omega$.

(C) The power $P$ and voltage $V$ of an electric bulb are related to its resistance $R$ by the formula: $P = \frac{V^2}{R}$.
Rearranging for $R$,we get $R = \frac{V^2}{P}$.
Given $V = 220\,V$ and $P = 60\,W$.
Substituting the values: $R = \frac{220 \times 220}{60} = \frac{48400}{60} \approx 806.67\,\Omega$.
Rounding to the nearest whole number,we get $R = 807\,\Omega$.

(D) The resistance $R$ of the bulb is constant and is given by $R = \frac{V^2}{P}$.
For the given bulb,$R = \frac{220^2}{1000} = \frac{48400}{1000} = 48.4\, \Omega$.
When connected to a $110\, V$ supply,the power consumed $P'$ is given by $P' = \frac{V'^2}{R}$.
$P' = \frac{110^2}{48.4} = \frac{12100}{48.4} = 250\, W$.
Alternatively,using the ratio formula: $\frac{P_1}{P_2} = \left( \frac{V_1}{V_2} \right)^2$.
$\frac{1000}{P_2} = \left( \frac{220}{110} \right)^2 = 2^2 = 4$.
$P_2 = \frac{1000}{4} = 250\, W$.

(D) The heat produced $H$ in an electric circuit is given by the formula $H = \frac{P \times t}{J}$,where $P$ is the power in watts,$t$ is the time in seconds,and $J$ is the mechanical equivalent of heat $(J \approx 4.2 \, J/cal)$.
Given: Power $P = 210 \, W$,Time $t = 5 \, \text{minutes} = 5 \times 60 = 300 \, s$.
Substituting the values: $H = \frac{210 \times 300}{4.2} = \frac{63000}{4.2} = 15000 \, cal$.

(C) Using the principle of conservation of energy,the electrical energy supplied to the conductor is converted into heat energy.
$H = I^2Rt = mS\Delta T$
Here,$I$ is the current,$R$ is the resistance,$t$ is time,$m$ is mass,$S$ is specific heat,and $\Delta T$ is the rise in temperature.
Since $R$,$t$,$m$,and $S$ are constant,we have $\Delta T \propto I^2$.
Given that the initial rise in temperature $\Delta T_1 = 5\, ^oC$ for current $I_1 = I$.
When the current is doubled,$I_2 = 2I$.
Then,$\frac{\Delta T_2}{\Delta T_1} = \left(\frac{I_2}{I_1}\right)^2 = \left(\frac{2I}{I}\right)^2 = 4$.
Therefore,$\Delta T_2 = 4 \times \Delta T_1 = 4 \times 5\, ^oC = 20\, ^oC$.

(B) The electrical energy consumed is calculated by the formula: $E = P \times t$.
Given power $P = 2\, kW$.
Time used per day $t_{day} = 1\, hour$.
Total time for $30$ days $t = 1 \times 30 = 30\, hours$.
Therefore, total energy consumed $E = 2\, kW \times 30\, hours = 60\, kWh$.
Since $1\, kWh = 1\, unit$, the total energy consumed is $60\, units$.

(A) The resistors $R$ and $2R$ are connected in parallel to the voltage source $E$.
In a parallel connection,the potential difference $V$ across each resistor is the same.
The heat generated $H$ in a resistor is given by the formula $H = \frac{V^2}{R} t$.
Since $V$ and $t$ are constant for both resistors,the heat generated is inversely proportional to the resistance: $H \propto \frac{1}{R}$.
Therefore,the ratio of heat generated in $R$ and $2R$ is $\frac{H_R}{H_{2R}} = \frac{2R}{R} = \frac{2}{1}$ or $2:1$.

$(A)$ Given: Current $I = 4\, A$, Potential difference $V = 250\, V$, Time $t = 1\, \text{minute} = 60\, s$.
Power $P$ is given by the formula $P = V \times I$.
$P = 250\, V \times 4\, A = 1000\, W = 1\, kW$.
Energy $E$ consumed is given by $E = P \times t$.
$E = 1\, kW \times 60\, s = 60\, kJ$ (or $1000\, W \times 60\, s = 60,000\, J = 60\, kJ$).
Thus, the power is $1\, kW$ and the energy consumed is $60\, kJ$.

(C) The power dissipated in a resistor is given by the formula $P = \frac{V^2}{R}$.
Given that the voltage $V = 25\, V$ and the heat produced per second (power) $P = 25\, J/s = 25\, W$.
Substituting these values into the formula:
$25 = \frac{25^2}{R}$
$R = \frac{625}{25} = 25\,\Omega$.
Therefore,the value of resistance $R$ is $25\,\Omega$.

(D) The power rating of an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Since both bulbs are rated at the same voltage $V = 220\, V$,the resistance $R$ is inversely proportional to the power $P$,i.e.,$R \propto \frac{1}{P}$.
Therefore,the ratio of their resistances is given by $\frac{R_1}{R_2} = \frac{P_2}{P_1}$.
Given $P_1 = 40\, W$ and $P_2 = 60\, W$,we have $\frac{R_1}{R_2} = \frac{60}{40} = \frac{3}{2}$.
Thus,the ratio of their resistances is $3:2$.

(A) The heat energy produced in a resistor is given by the formula $H = \frac{V^2}{R} \times t$.
Given:
Voltage $V = 10\, V$
Resistance $R = 50\,\Omega$
Time $t = 1\, \text{hour} = 3600\, \text{seconds}$.
Substituting the values into the formula:
$H = \frac{10^2}{50} \times 3600$
$H = \frac{100}{50} \times 3600$
$H = 2 \times 3600 = 7200\, J$.
Therefore, the heat energy produced is $7200\, J$.

(C) The formula for electrical energy $E$ consumed is given by $E = P \times t$,where $P$ is power and $t$ is time.
Since power $P = \frac{V^2}{R}$,the energy formula becomes $E = \frac{V^2}{R} \times t$.
Given values are $V = 200\,V$,$R = 80\,\Omega$,and $t = 2\,h$.
Substituting these values into the formula:
$E = \frac{200 \times 200}{80} \times 2$
$E = \frac{40000}{80} \times 2$
$E = 500 \times 2 = 1000\,Wh$.

(A) The heat produced $(H)$ by an electrical appliance is given by the formula $H = P \times t$, where $P$ is the power and $t$ is the time in seconds.
Given: Power $P = 100\, W$, Time $t = 2\, minutes = 2 \times 60\, s = 120\, s$.
Substituting the values: $H = 100\, W \times 120\, s = 12000\, J$.
This can be expressed in scientific notation as $12 \times 10^3\, J$.

(C) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage.
Since both bulbs are connected in parallel to the same source,the voltage $V$ across each bulb is the same.
From the relation $P = \frac{V^2}{R}$,we have $R = \frac{V^2}{P}$.
For bulb $A$,$P_A = 60\, W$,so $R_A = \frac{V^2}{60}$.
For bulb $B$,$P_B = 100\, W$,so $R_B = \frac{V^2}{100}$.
The current drawn by a bulb is given by $I = \frac{V}{R}$.
Substituting $R$,we get $I = \frac{V}{V^2/P} = \frac{P}{V}$.
Since $V$ is constant,$I \propto P$.
Because $P_B > P_A$ $(100\, W > 60\, W)$,the current drawn by bulb $B$ is greater than the current drawn by bulb $A$ $(I_B > I_A)$.

(B) Given power $P = 2.2\,kW = 2.2 \times 10^3\,W$.
Voltage $V = 22,000\,V$.
Resistance $R = 100\,\Omega$.
The current $i$ flowing through the line is given by $P = Vi$,so $i = \frac{P}{V}$.
$i = \frac{2.2 \times 10^3}{22,000} = \frac{2200}{22000} = 0.1\,A$.
The power loss in the form of heat is given by $P_{loss} = i^2R$.
$P_{loss} = (0.1)^2 \times 100 = 0.01 \times 100 = 1\,W$.

(C) The power of the lamp is $P = 60\, W = 0.06\, kW$.
The total time of usage is $T = 8\, \text{hours/day} \times 30\, \text{days} = 240\, \text{hours}$.
The total energy consumed in $kWh$ is $E = P \times T = 0.06\, kW \times 240\, h = 14.4\, kWh$.
The cost of $1\, kWh$ is Rs. $1.25$.
Therefore, the total cost is $14.4 \times 1.25 = 18\, \text{Rs}$.

(A) The heat generated in a fuse wire of length $l$,radius $r$,and resistivity $\rho$ due to current $I$ is given by $H = I^2 R t = I^2 (\rho l / \pi r^2) t$.
This heat is dissipated through the surface area of the wire via radiation,which is proportional to the surface area $A = 2\pi rl$ and the temperature difference $\Delta T$,following Stefan's Law: $H_{dissipated} \propto (2\pi rl) \Delta T$.
At the point of blowing out,the heat generated equals the heat dissipated: $I^2 (\rho l / \pi r^2) \propto 2\pi rl$.
Rearranging for $I^2$,we get $I^2 \propto r^3$,which implies $I \propto r^{3/2}$.
Therefore,the maximum current is proportional to $r^{3/2}$.

(D) The power $P$ and voltage $V$ of the lamp are given as $P = 50\, W$ and $V = 250\, V$.
Using the formula for electrical power,$P = V \times I$,where $I$ is the current.
Rearranging the formula to solve for current: $I = \frac{P}{V}$.
Substituting the given values: $I = \frac{50}{250} = \frac{1}{5} = 0.2\, A$.
Therefore,the current flowing through the lamp is $0.2\, A$.

(C) The heat generated by a heater is given by the formula $H = \frac{V^2 t}{R}$,where $V$ is the voltage,$t$ is the time,and $R$ is the resistance of the coil.
When the coil is cut into two equal parts,the resistance of each part becomes $R' = \frac{R}{2}$.
Since the voltage $V$ remains constant,the new heat generated $H'$ is given by $H' = \frac{V^2 t}{R'} = \frac{V^2 t}{R/2} = 2 \times \frac{V^2 t}{R}$.
Therefore,$H' = 2H$.
Thus,the heat generated is doubled.

(A) The chemical energy reduced in the battery is equal to the electrical energy supplied by the battery to the external circuit.
Electrical energy $E = V \times I \times t$
Given:
Voltage $V = 6.0 \, V$
Current $I = 5.0 \, A$
Time $t = 6.0 \, \text{minutes} = 6.0 \times 60 \, \text{seconds} = 360 \, \text{s}$
Substituting the values:
$E = 6.0 \, V \times 5.0 \, A \times 360 \, \text{s}$
$E = 30 \times 360 \, J$
$E = 10800 \, J$
$E = 1.08 \times 10^{4} \, J$
Therefore, the chemical energy of the battery is reduced by $1.08 \times 10^{4} \, J$.

(C) The heat generated per second is equivalent to the power consumed by the lamp.
The power $P$ is given by the formula $P = V \times I$.
Given: Voltage $V = 15\,V$ and Current $I = 2.1\,A$.
$P = 15\,V \times 2.1\,A = 31.5\,W$ (or $31.5\,J/s$).
To convert the heat generated in Joules to calories,we use the conversion factor $1\,cal = 4.2\,J$.
Heat in calories per second $= \frac{31.5\,J}{4.2\,J/cal} = 7.5\,cal/s$.
Therefore,the heat generated per second in each lamp is $7.5\,cal$.

(D) The Joule effect refers to the heating of a conductor when an electric current passes through it,given by $H = I^2Rt$. This process is irreversible because the heat generated is dissipated into the surroundings and cannot be converted back into electrical energy simply by reversing the current.
The Peltier effect refers to the production or absorption of heat at the junction of two dissimilar metals when an electric current flows through them. This effect is reversible; if the direction of the current is reversed,the heat that was previously absorbed will be released,and vice versa.
Therefore,the Joule effect is irreversible,while the Peltier effect is reversible.

(D) The power supplied by the circuit is given by $P = V \times I$.
Given,$V = 220 \, V$ and $I = 9 \, A$.
Total power $P = 220 \times 9 = 1980 \, W$.
Let $n$ be the number of $60 \, W$ lamps connected in parallel.
The total power consumed by $n$ lamps is $n \times 60 \, W$.
For the fuse not to blow,the total power consumed must be less than or equal to the supply power:
$n \times 60 \leq 1980$.
$n \leq \frac{1980}{60} = 33$.
Therefore,the maximum number of lamps that can be turned on is $33$.

(C) The power consumed by a heating element is given by $P = \frac{V^2}{R}$.
Since the resistance $R$ of the heating unit remains constant,the power is proportional to the square of the voltage: $P \propto V^2$.
Therefore,the ratio of the new power $P_{new}$ to the rated power $P_{rated}$ is given by $\frac{P_{new}}{P_{rated}} = \left( \frac{V_{new}}{V_{rated}} \right)^2$.
Given $P_{rated} = 500\, W$,$V_{rated} = 115\, V$,and $V_{new} = 110\, V$,we have:
$P_{new} = \left( \frac{110}{115} \right)^2 \times 500 = (0.9565)^2 \times 500 \approx 0.9149 \times 500 = 457.46\, W$.
The drop in power is $\Delta P = P_{rated} - P_{new} = 500 - 457.46 = 42.54\, W$.
The percentage drop in heat output is $\frac{\Delta P}{P_{rated}} \times 100 = \frac{42.54}{500} \times 100 = 8.508\% \approx 8.6\% $.

(D) The heat energy $H$ required to heat a volume of water is constant.
The heat produced by a heater is given by $H = \frac{V^2}{R} t$, where $V$ is the voltage, $R$ is the resistance of the heater, and $t$ is the time.
Since the same volume of water is heated, the heat required $H$ is the same in both cases.
Assuming the resistance $R$ of the heater remains constant, we have $H = \frac{V_1^2}{R} t_1 = \frac{V_2^2}{R} t_2$.
Given $V_1 = 220\, V$, $t_1 = 5\, \text{min}$, and $V_2 = 110\, V$.
Substituting the values: $\frac{220^2}{R} \times 5 = \frac{110^2}{R} \times t_2$.
$t_2 = 5 \times \left( \frac{220}{110} \right)^2 = 5 \times (2)^2 = 5 \times 4 = 20\, \text{min}$.

(B) The power consumed by the electric kettle is $P = V \times I = 220\, V \times 4\, A = 880\, W$.
To boil $1\, kg$ of water,the heat required is $H = m \times c \times \Delta T$.
Given $m = 1\, kg$,$c = 4200\, J/kg\, ^\circ C$,and $\Delta T = (100 - 20)\, ^\circ C = 80\, ^\circ C$.
$H = 1 \times 4200 \times 80 = 336000\, J$.
Since $H = P \times t$,the time $t$ in seconds is $t = H / P = 336000 / 880 \approx 381.82\, s$.
Converting to minutes: $t = 381.82 / 60 \approx 6.36\, minutes$.
Rounding to the nearest provided option,the correct answer is $6.3\, minutes$.

(C) The electrical power dissipated in the wire is given by $P = \frac{V^2}{R}$.
Substituting the given values: $P = \frac{(210)^2}{20} = \frac{44100}{20} = 2205\,W$ (or $J/s$).
The rate of melting of ice $m$ is given by the formula $P = m \times L$,where $L$ is the latent heat of fusion of ice.
Given $L = 80\,cal/g$ and $1\,cal = 4.2\,J$,we have $L = 80 \times 4.2 = 336\,J/g$.
Therefore,$m = \frac{P}{L} = \frac{2205}{336} = 6.56\,g/s$.

(D) The heat produced in a wire is given by $H = I^2 R t$.
Since the wires are connected in series, the current $I$ and time $t$ are the same for both.
Resistance $R = \rho \frac{l}{A}$.
Since the mass $m = \text{density} \times \text{volume} = \rho_d \times A \times l$ is constant, $l = \frac{m}{\rho_d A}$.
Substituting this into the resistance formula: $R = \rho \frac{m}{\rho_d A^2} \propto \frac{1}{A^2}$.
Since the area $A = \pi r^2$, we have $R \propto \frac{1}{(r^2)^2} = \frac{1}{r^4}$.
Therefore, $H \propto \frac{1}{r^4}$.
For the two wires with radii $r_1 = 1 \, mm$ and $r_2 = 2 \, mm$, the ratio of heat produced is:
$\frac{H_1}{H_2} = \left( \frac{r_2}{r_1} \right)^4 = \left( \frac{2}{1} \right)^4 = \frac{16}{1}$.

(A) According to Faraday's first law of electrolysis,the mass $m$ deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalent of silver $(Ag)$.
For silver,the atomic mass is $108 \, g/mol$ and its valency is $1$. Thus,$Z = \frac{108}{96500} \, g/C$.
Given: $m = 2.68 \, g$,$t = 10 \, \min = 600 \, s$.
Calculating the current $I$:
$I = \frac{m}{Zt} = \frac{2.68}{\left(\frac{108}{96500}\right) \times 600} = \frac{2.68 \times 96500}{108 \times 600} \approx 4 \, A$.
Now,the heat $H$ developed in the resistor $R = 20 \, \Omega$ is given by Joule's law of heating:
$H = I^2 Rt = (4)^2 \times 20 \times 600 = 16 \times 20 \times 600 = 192000 \, J = 192 \, kJ$.

(D) The thermal energy $U$ produced in a resistor of resistance $R$ in a given time $t$ is given by Joule's law of heating: $U = i^2 Rt$.
Since $R$ and $t$ are constants,we have $U \propto i^2$.
This relationship represents a parabola opening upwards along the $U$-axis.
Among the given plots,the curve $d$ represents a parabolic variation where $U$ increases more rapidly as $i$ increases,which is consistent with the $U \propto i^2$ relationship.
Therefore,the correct plot is $d$.

(D) The rate of production of thermal energy in a resistor is given by the power $P = \frac{dU}{dt} = i^2 R$.
The resistance $R$ of a conductor depends on temperature $T$ as $R = R_0(1 + \alpha \Delta T)$,where $R_0$ is the initial resistance,$\alpha$ is the temperature coefficient of resistance,and $\Delta T$ is the change in temperature.
As current $i$ flows through the resistor,it heats up,so the temperature $T$ increases with time $t$. Assuming the temperature rise is proportional to the time elapsed (i.e.,$\Delta T \propto t$),we can write $R(t) = R_0(1 + \alpha' t)$,where $\alpha'$ is a constant.
Substituting this into the power equation:
$P(t) = i^2 R_0(1 + \alpha' t) = i^2 R_0 + (i^2 R_0 \alpha') t$.
This equation is of the form $y = mx + c$,which represents a straight line with a positive slope $(i^2 R_0 \alpha')$ and a positive y-intercept $(i^2 R_0)$.
Among the given plots,curve $d$ shows a linear increase starting from a non-zero value,which correctly represents the relationship.

(C) When $n$ identical electrical appliances,each of power $P$,are connected in parallel,the total power consumed is given by the formula $P_{total} = n \times P$.
Given that $n = 2$ and $P = 40\,W$.
Therefore,$P_{total} = 2 \times 40\,W = 80\,W$.
Thus,the total power consumed by the combination is $80\,W$.

(D) The power consumed by a resistor (bulb) is given by the formula $P = \frac{V^2}{R}$,where $R$ is the resistance of the bulb.
Initially,the bulb operates at voltage $V_0$ and power $P_0$. Therefore,the resistance $R$ is given by:
$R = \frac{V_0^2}{P_0} \quad \dots(1)$
When the bulb is connected to a new voltage $V$,it consumes a new power $P$. Since the resistance $R$ of the bulb remains constant,we have:
$P = \frac{V^2}{R}$
Substituting the value of $R$ from equation $(1)$ into this expression:
$P = \frac{V^2}{(V_0^2 / P_0)}$
$P = \left( \frac{V^2}{V_0^2} \right) P_0$
$P = \left( \frac{V}{V_0} \right)^2 P_0$

(D) Let the voltage of the source be $V$ and the resistance of each bulb be $R$.
Since each bulb is designed to consume power $P$ at voltage $V$,we have $P = V^2/R$,which implies $R = V^2/P$.
When $n$ such bulbs are connected in series,the total resistance of the circuit becomes $R_{eq} = nR$.
The total power consumed by the series combination is $P_{total} = V^2 / R_{eq}$.
Substituting $R_{eq} = nR$,we get $P_{total} = V^2 / (nR) = (1/n) * (V^2/R)$.
Since $P = V^2/R$,the total power is $P_{total} = P/n$.

(D) In a parallel connection,the voltage $(V)$ across each bulb is the same.
The power consumed by a bulb is given by the formula $P = \frac{V^2}{R}$.
Since $P \propto \frac{1}{R}$,the bulb with lower resistance will consume more power.
For a given rated voltage,$R = \frac{V^2}{P}$,which means $R \propto \frac{1}{P}$.
Therefore,the $40\, W$ bulb has higher resistance than the $60\, W$ bulb.
However,in parallel,the power dissipated is $P_{dissipated} = \frac{V^2}{R}$.
Since the $60\, W$ bulb has lower resistance,it will dissipate more power $(P = \frac{V^2}{R})$ and thus glow brighter.

(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.
For the two bulbs,their resistances are $R_1 = \frac{V^2}{P_1}$ and $R_2 = \frac{V^2}{P_2}$.
When connected in parallel to the same voltage $V$,the total power consumed $P_{total}$ is the sum of the individual powers consumed by each bulb.
Since each bulb is connected to its rated voltage $V$,each bulb consumes its rated power.
Therefore,$P_{total} = P_1 + P_2$.

(B) Power $(P)$ is defined as the rate of doing work or the rate of energy consumption. The $SI$ unit of power is Watt $(W)$,which is equivalent to $Joule/sec$ $(J/s)$.
From Ohm's law,$V = IR$. The formula for power is $P = VI = I^2R = V^2/R$.
$1$. $P = I^2R$: The unit is $(Amp)^2 \times \Omega$.
$2$. $P = VI$: The unit is $Amp \times Volt$.
$3$. $P = W/t$: The unit is $Joule/sec$.
Comparing these with the options,$Amp / Volt$ is not a unit of power. In fact,$Amp / Volt$ is the unit of conductance $(1/Resistance)$,which is Siemens $(S)$.