Electrical Energy and Power Questions in English

(D) The unit of power is the watt $(W)$.
Power is defined as the rate of doing work,so $P = \frac{W}{t}$,which gives the unit $\text{Joule/second}$.
From the electrical power formula $P = VI$,the unit is $\text{Ampere} \times \text{volt}$.
Using Ohm's law $V = IR$,we can substitute $V$ to get $P = I(IR) = I^{2}R$,which gives the unit $\text{Ampere}^{2} \times \text{ohm}$.
However,$\text{Ampere/volt}$ is the unit of conductance (Siemens),not power.
Therefore,the correct option is $D$.

(B) The $Kilowatt-hour$ $(kWh)$ is a unit of energy.
By definition,$1 \ kWh$ is the amount of energy consumed by a device with a power rating of $1 \ kilowatt$ $(1000 \ W)$ operating for $1 \ hour$ $(3600 \ s)$.
Since $Power = \frac{Energy}{Time}$,it follows that $Energy = Power \times Time$.
Therefore,$1 \ kWh = 1000 \ W \times 3600 \ s = 3.6 \times 10^6 \ Joules$.
Thus,the correct option is $B$.

(B) The unit $kWh$ (kilowatt-hour) is a unit of energy.
$1 \, kWh = 1 \, kW \times 1 \, h$
Since $1 \, kW = 1000 \, W = 1000 \, J/s$ and $1 \, h = 3600 \, s$,
$1 \, kWh = 1000 \, J/s \times 3600 \, s = 3,600,000 \, J$
$1 \, kWh = 36 \times 10^5 \, J$.

(B) Power of the motor $P = 12 \text{ HP} = 12 \times 746 \text{ W} = 8952 \text{ W} = 8.952 \text{ kW}$.
Total time of operation $t = 10 \text{ days} \times 8 \text{ hours/day} = 80 \text{ hours}$.
Total energy consumed $E = P \times t = 8.952 \text{ kW} \times 80 \text{ h} = 716.16 \text{ kWh}$.
Given rate is $50 \text{ paisa/kWh} = 0.5 \text{ Rs/kWh}$.
Total cost = $\text{Energy} \times \text{Rate} = 716.16 \text{ kWh} \times 0.5 \text{ Rs/kWh} = 358.08 \text{ Rs}$.
Rounding to the nearest integer,the cost is $358 \text{ Rs}$.

(D) When an electric current $I$ flows through a conductor of resistance $R$ for a time $t$,heat is produced in the conductor according to Joule's law of heating.
The amount of heat produced is given by the formula $H = I^2Rt$.
This heat energy causes the internal energy of the conductor to rise,which results in an increase in its temperature.
Therefore,the correct option is $D$.

(A) Given:
$I = 2\, A$
$t = 6\, \text{minutes} = 6 \times 60 = 360\, s$
$W = 1000\, J$
We know that the work done $W$ in a circuit is given by $W = V \times I \times t$, where $V$ is the $e.m.f.$ of the source.
Rearranging the formula to solve for $V$:
$V = \frac{W}{I \times t}$
Substituting the given values:
$V = \frac{1000}{2 \times 360}$
$V = \frac{1000}{720}$
$V \approx 1.38\, V$
Therefore, the $e.m.f.$ of the source is $1.38\, V$.

(B) The work done $W$ in moving a charge $q$ through a potential difference $V$ is given by the formula $W = qV$.
Given:
Charge $q = 6 \mu C = 6 \times 10^{-6} C$
Potential difference $V = 9 V$
Substituting these values into the formula:
$W = (6 \times 10^{-6} C) \times (9 V)$
$W = 54 \times 10^{-6} J$
Therefore,the work required is $54 \times 10^{-6} J$.

(C) The power delivered to an external circuit by a battery with emf $E$ and internal resistance $r$ is given by $P = I^2 R$,where $I = \frac{E}{R+r}$.
Thus,$P = \left( \frac{E}{R+r} \right)^2 R$.
For maximum power transfer,the external resistance $R$ must be equal to the internal resistance $r$ $(R = r)$.
The formula for maximum power is $P_{\text{max}} = \frac{E^2}{4r}$.
Given $E = 2\, V$ and $r = 0.5\,\Omega$,we substitute these values:
$P_{\text{max}} = \frac{(2)^2}{4 \times 0.5} = \frac{4}{2} = 2\, W$.

(B) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$,where $R$ is the external resistance.
Power $P$ delivered to the external resistance is $P = i^2 R = \frac{E^2 R}{(r + R)^2}$.
To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = E^2 \left[ \frac{(r + R)^2 - R(2)(r + R)}{(r + R)^4} \right] = 0$.
This simplifies to $(r + R) - 2R = 0$,which gives $R = r$.
Substituting $R = r$ into the power equation: $P_{\max} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$.

(A) The unit of energy is kilowatt hour $(kWh)$.
$1 \; kWh = 1 \; kW \times 1 \; h$
Since $1 \; kW = 1000 \; W$ and $1 \; h = 3600 \; s$,
$1 \; kWh = 1000 \; W \times 3600 \; s = 3,600,000 \; W \cdot s$.
Since $1 \; W \cdot s = 1 \; J$,we have $1 \; kWh = 3.6 \times 10^6 \; J$.
This can also be written as $36 \times 10^5 \; J$.

(B) The power $P$ of a bulb operating at a constant voltage $V$ is given by the formula $P = \frac{V^2}{R}$.
Since the voltage $V$ is the same for both bulbs,we have $P \propto \frac{1}{R}$.
This implies that $P_1 R_1 = P_2 R_2$,or $\frac{P_1}{P_2} = \frac{R_2}{R_1}$.
Given $P_1 = 200 \ W$ and $P_2 = 100 \ W$,we substitute these values into the ratio:
$\frac{200}{100} = \frac{R_2}{R_1}$
$2 = \frac{R_2}{R_1}$
Therefore,$R_2 = 2R_1$.

(C) The power rating of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
For the first bulb $(40 \; W, 200 \; V)$:
$R_{40} = \frac{V^2}{P_1} = \frac{(200)^2}{40} = \frac{40000}{40} = 1000 \; \Omega$.
For the second bulb $(100 \; W, 200 \; V)$:
$R_{100} = \frac{V^2}{P_2} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \; \Omega$.
Comparing the two values,we find that $1000 \; \Omega > 400 \; \Omega$,which implies $R_{40} > R_{100}$.

(C) fuse wire is a safety device used to protect electrical circuits from excessive current.
According to Joule's law of heating,the heat produced is given by $H = I^2Rt$,where $I$ is the current,$R$ is the resistance,and $t$ is the time.
To ensure the fuse melts quickly when the current exceeds a safe limit,it must have a high resistance (so that more heat is generated) and a low melting point (so that it melts easily).
Therefore,the material of a fuse wire should have a high specific resistance and a low melting point.

(A) The power $P$ dissipated by a heater coil connected to a constant voltage $V$ is given by $P = \frac{V^2}{R}$,where $R$ is the resistance of the coil.
Since $P \propto \frac{1}{R}$ and the resistance $R$ is directly proportional to the length $l$ of the wire $(R = \rho \frac{l}{A})$,we have $R \propto l$.
Therefore,$P \propto \frac{1}{l}$.
Let the original length be $l$ and the original resistance be $R$. The new length is $l' = \frac{l}{2}$,so the new resistance is $R' = \frac{R}{2}$.
The power produced by the original coil is $P = \frac{V^2}{R}$.
The power produced by the half coil is $P' = \frac{V^2}{R'} = \frac{V^2}{R/2} = 2 \frac{V^2}{R} = 2P$.
Thus,the ratio of the heat produced by the half coil to that by the original coil is $P' : P = 2:1$.

(A) In a series circuit,the current $I$ flowing through both bulbs is the same.
The power dissipated in a resistor is given by the formula $P = I^2 R$.
Since $I$ is constant for both bulbs,the power dissipated is directly proportional to the resistance,i.e.,$P \propto R$.
Therefore,the ratio of the powers dissipated is equal to the ratio of their resistances:
$\frac{P_1}{P_2} = \frac{R_1}{R_2} = \frac{1}{2}$.

(D) The formula for energy consumed in kilowatt-hour $(kWh)$ is given by: $E = \frac{P \times t}{1000}$,where $P$ is power in watts and $t$ is time in hours.
Given:
Number of bulbs = $10$
Power of each bulb = $50\,W$
Total power $(P)$ = $10 \times 50\,W = 500\,W$
Time per day = $10\,h$
Number of days = $30$
Total time $(t)$ = $10 \times 30 = 300\,h$
Energy consumed $(E)$ = $\frac{500\,W \times 300\,h}{1000} = \frac{150000}{1000} = 150\,kWh$.
Therefore,the total energy consumed is $150\,kWh$.

(C) When electric appliances are connected in parallel,the total power consumed by the combination is the sum of the power consumed by each individual appliance.
Given that there are $n = 2$ lamps,each with a power rating of $P = 40\, W$.
The total power $P_p$ in a parallel combination is given by $P_p = P_1 + P_2 + ... + P_n$.
Since all lamps have the same power rating,$P_p = n \times P$.
Substituting the values: $P_p = 2 \times 40\, W = 80\, W$.
Therefore,the correct option is $C$.

(D) The resistance $R$ of the bulb is constant and is calculated using the rated values: $R = \frac{V_{rated}^2}{P_{rated}} = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega$.
When the bulb is operated at a new voltage $V' = 110\, V$,the power consumed $P'$ is given by: $P' = \frac{(V')^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\, W$.

(C) The power $P$ of an electrical appliance is related to its resistance $R$ and the operating voltage $V$ by the formula $P = \frac{V^2}{R}$.
Since both appliances operate at the same voltage $V = 220 \ V$,the resistance is given by $R = \frac{V^2}{P}$.
This implies that $R \propto \frac{1}{P}$,meaning resistance is inversely proportional to power.
For the fan,$P_f = 100 \ W$,so $R_f = \frac{V^2}{100}$.
For the heater,$P_h = 1000 \ W$,so $R_h = \frac{V^2}{1000}$.
Comparing the two,since $1000 > 100$,it follows that $R_h < R_f$.
Therefore,the resistance of the heater is less than that of the fan.

(D) The power dissipated as heat in a conductor is given by $P = \frac{V^2}{R}$.
We know that the resistance $R$ of a conductor is given by $R = \frac{\rho l}{A}$,where $\rho$ is the specific resistance (resistivity),$l$ is the length,and $A$ is the cross-sectional area.
Substituting the expression for $R$ into the power formula,we get $P = \frac{V^2}{(\rho l / A)} = \frac{A V^2}{\rho l}$.
Given that the potential difference $V$,length $l$,and cross-sectional area $A$ are constant,the term $\frac{A V^2}{l}$ is a constant.
Therefore,$P \propto \frac{1}{\rho}$.
Thus,the heat produced is directly proportional to $\frac{1}{\rho}$.

(B) When bulbs are connected in parallel,the potential difference $(V)$ across each bulb is the same.
The power consumed by a bulb is given by the formula $P = \frac{V^2}{R}$.
Since $V$ is constant,the power consumed is directly proportional to the rated power of the bulb $(P \propto P_{rated})$.
Therefore,the bulb with the higher rated power will consume more electrical energy per unit time and glow more brightly.
Thus,the $100\, W$ bulb will glow more brightly than the $25\, W$ bulb.

(B) The power $P$ consumed by an electric bulb is given by the formula $P = i^2R$,where $i$ is the current and $R$ is the resistance of the bulb.
Assuming the resistance $R$ remains constant,we can use the method of relative errors:
$\frac{\Delta P}{P} = 2 \frac{\Delta i}{i}$.
To find the percentage change,we multiply both sides by $100$:
$\left( \frac{\Delta P}{P} \times 100 \right) = 2 \times \left( \frac{\Delta i}{i} \times 100 \right)$.
Given that the percentage change in current is $1\%$,we have:
$\% \text{ change in power} = 2 \times 1\% = 2\%$.
Therefore,the power will change by $2\%$.

(A) The rate of heat developed (power) in a wire is given by $P = \frac{V^2}{R}$, where $V$ is the constant voltage and $R$ is the resistance.
Since $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$, we have $P \propto \frac{1}{R} \propto \frac{r^2}{l}$.
Let the initial length be $l_1 = l$ and radius be $r_1 = r$. Then $P_1 \propto \frac{r^2}{l}$.
When both length and radius are doubled, $l_2 = 2l$ and $r_2 = 2r$.
The new power is $P_2 \propto \frac{r_2^2}{l_2} = \frac{(2r)^2}{2l} = \frac{4r^2}{2l} = 2 \left( \frac{r^2}{l} \right)$.
Therefore, $P_2 = 2P_1$. The rate of heat developed will be doubled.

(D) The power $P$ dissipated by a heating coil is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Given that the voltage $V = 220\, V$ is constant for all coils,the power $P$ is inversely proportional to the resistance $R$ $(P \propto \frac{1}{R})$.
To obtain the maximum power,the resistance $R$ must be the minimum value among the given options.
Comparing the resistances $55\,\Omega$,$110\,\Omega$,$220\,\Omega$,and $440\,\Omega$,the minimum resistance is $55\,\Omega$.
Therefore,the heater with $55\,\Omega$ resistance will produce the maximum power.

(C) According to Joule's law of heating,the heat produced $(H)$ in a conductor is given by the formula $H = I^2Rt$,where $I$ is the current,$R$ is the resistance,and $t$ is the time.
From this formula,it is clear that $H \propto R$,$H \propto I^2$,and $H \propto t$.
Since $I = \frac{q}{t}$,we can substitute this into the equation: $H = (\frac{q}{t})^2 Rt = \frac{q^2 R}{t}$.
Therefore,$H$ is proportional to the square of the charge $(q^2)$ and inversely proportional to time $(t)$,not directly proportional to the charge $(q)$.
Thus,the statement that heat produced is proportional to charge is false.

(B) The power rating of the heater is $P = 1100\, W$.
The time for which it is used is $t = 4\, hours$.
The energy consumed $E$ is given by the formula $E = P \times t$.
Since we need the energy in $kWh$, we convert the power from $W$ to $kW$ by dividing by $1000$.
$P = \frac{1100}{1000} = 1.1\, kW$.
Now, $E = 1.1\, kW \times 4\, h = 4.4\, kWh$.
Therefore, the correct option is $B$.

(B) The heat produced $H$ in a conductor is given by the formula $H = \frac{V^2}{R} t$,where $V$ is the potential difference,$R$ is the resistance,and $t$ is the time.
Since the potential difference $V$ is constant,the heat produced is inversely proportional to the resistance $R$.
Therefore,$H \propto \frac{1}{R}$.

(A) The power $P$ consumed by an electric motor is given by the formula $P = V \times I$,where $V$ is the voltage and $I$ is the current.
Given: $V = 200\, V$ and $I = 3.75\, A$.
Substituting the values: $P = 200\, V \times 3.75\, A = 750\, W$.
Since $1\, H.P. \approx 746\, W$,the power rating is approximately $1\, H.P.$

(A) The power $P$ consumed by an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance of the filament.
Rearranging the formula to solve for resistance $R$,we get $R = \frac{V^2}{P}$.
Given values are $P = 100\, W$ and $V = 220\, V$.
Substituting these values into the formula: $R = \frac{220 \times 220}{100}$.
$R = \frac{48400}{100} = 484\,\Omega$.
Therefore,the resistance of the filament is $484\,\Omega$.

(B) The power $P$ delivered by the generator is $250\,kW = 250 \times 10^3\,W$.
The voltage $V$ is $10000\,V$.
The current $I$ flowing through the cable is given by $I = P/V = (250 \times 10^3) / 10000 = 25\,A$.
The power lost in the cable due to resistance $R = 10\,\Omega$ is given by $P_{loss} = I^2 R$.
Substituting the values: $P_{loss} = (25)^2 \times 10 = 625 \times 10 = 6250\,W$.
Converting to $kW$: $6250\,W = 6.25\,kW$.

(A) The electrical energy $E$ dissipated as heat in a resistor $R$ carrying current $I$ for time $t$ is given by Joule's law of heating: $H = I^2 Rt \text{ Joules}$.
To convert this energy from Joules to calories,we use the mechanical equivalent of heat,where $1 \text{ calorie} \approx 4.2 \text{ Joules}$.
Therefore,the heat $H$ in calories is given by $H = \frac{I^2 Rt}{4.2} \text{ cal}$.

(B) The resistance of a bulb is given by the formula $R = \frac{V^2}{P}$.
For the first bulb: $R_1 = \frac{200^2}{100} = \frac{40000}{100} = 400\,\Omega$.
For the second bulb: $R_2 = \frac{100^2}{200} = \frac{10000}{200} = 50\,\Omega$.
The maximum current rating is given by $i = \frac{P}{V}$.
For the first bulb: $i_1 = \frac{100}{200} = 0.5\, A$.
For the second bulb: $i_2 = \frac{200}{100} = 2.0\, A$.
The ratio of maximum current ratings is $\frac{i_1}{i_2} = \frac{0.5}{2.0} = \frac{1}{4}$.

(B) The devices are connected in parallel to a $220\, V$ supply,which is the rated voltage for all devices.
Total power consumed by the circuit is the sum of the power of the kettle and the three bulbs.
$P_{\text{total}} = P_{\text{kettle}} + 3 \times P_{\text{bulb}}$
$P_{\text{total}} = 800\, W + 3 \times 100\, W = 800\, W + 300\, W = 1100\, W$.
Using the formula for power $P = V \times I$,where $V = 220\, V$:
$I = \frac{P_{\text{total}}}{V} = \frac{1100\, W}{220\, V} = 5.0\, A$.
Therefore,the total current is $5.0\, A$.

(C) The power $P$ consumed by a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the bulb and $R$ is its resistance.
Assuming the voltage $V$ is constant for all bulbs,we have $R = \frac{V^2}{P}$.
This shows that resistance $R$ is inversely proportional to power $P$ $(R \propto \frac{1}{P})$.
Therefore,the bulb with the highest power rating will have the lowest resistance.
Comparing the given values ($25\, W$,$40\, W$,and $60\, W$),the $60\, W$ bulb has the highest power.
Thus,the $60\, W$ bulb has the lowest resistance.

(C) The power loss $(P_L)$ in transmission lines is given by the formula $P_L = I^2 R$,where $I$ is the current and $R$ is the resistance of the wire.
Since the power transmitted is $P = VI$,we can write $I = P/V$.
Substituting this into the power loss formula,we get $P_L = (P/V)^2 R = \frac{P^2 R}{V^2}$.
From this expression,it is clear that $P_L \propto \frac{1}{V^2}$.
Therefore,by increasing the transmission voltage $(V)$,the power loss $(P_L)$ is significantly reduced.

(C) The rate of heat production is given by $P = \frac{H}{t} = 800 \, cal/sec$.
Converting this to $SI$ units (Joules per second or Watts): $P = 800 \times 4.2 \, J/sec = 3360 \, W$.
The formula for electrical power in terms of voltage $V$ and resistance $R$ is $P = \frac{V^2}{R}$.
Substituting the given values: $3360 = \frac{20^2}{R}$.
$3360 = \frac{400}{R}$.
$R = \frac{400}{3360} = \frac{40}{336} \approx 0.119 \, \Omega$.
Rounding to two decimal places,we get $R \approx 0.12 \, \Omega$.

(D) The energy consumed $(E)$ is given by the product of power $(P)$ and time $(t)$:
$E = P \times t$
Given:
Power $(P)$ = $1\, kW = 1000\, W$
Time $(t)$ = $30\, s$
Substituting the values:
$E = 1000\, W \times 30\, s = 30,000\, J$
$E = 3 \times 10^4\, J$

(B) The energy liberated (heat) $H$ is given by the formula $H = \frac{V^2 t}{R}$.
Given:
Resistance $R = 6 \, \Omega$
Voltage $V = 120 \, \text{V}$
Time $t = 10 \, \text{minutes} = 10 \times 60 \, \text{s} = 600 \, \text{s}$.
Substituting the values:
$H = \frac{(120)^2 \times 600}{6}$
$H = \frac{14400 \times 600}{6}$
$H = 14400 \times 100 = 1,440,000 \, \text{J}$.
$H = 14.4 \times 10^5 \, \text{J}$.
Therefore, the correct option is $B$.

(D) The resistance of a conductor is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Since the conductors are made of the same material,$\rho$ is constant.
The cross-sectional area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,where $d$ is the diameter.
Thus,$R \propto \frac{l}{d^2}$.
The power delivered to a conductor connected across a constant potential difference $V$ is $P = \frac{V^2}{R}$.
Therefore,$P \propto \frac{1}{R} \propto \frac{d^2}{l}$.
Given for conductor $A$: $d_A = 2d_B$ and $l_A = 2l_B$.
Calculating the ratio: $\frac{P_A}{P_B} = \frac{d_A^2 / l_A}{d_B^2 / l_B} = \left( \frac{d_A}{d_B} \right)^2 \times \left( \frac{l_B}{l_A} \right) = (2)^2 \times \left( \frac{1}{2} \right) = 4 \times 0.5 = 2$.
Hence,$P_A/P_B = 2$.

(C) The heat produced in a conductor due to the flow of current is given by Joule's law of heating: $H = i^2Rt$.
Since the rise in temperature $\Delta T$ is directly proportional to the heat produced $H$,we have $\Delta T \propto i^2$.
Given that the initial rise in temperature $\Delta T_1 = 5\,^{\circ}\text{C}$ for current $i_1 = i$.
When the current is doubled,$i_2 = 2i$.
The new rise in temperature $\Delta T_2$ is given by the ratio:
$\frac{\Delta T_2}{\Delta T_1} = \left(\frac{i_2}{i_1}\right)^2 = \left(\frac{2i}{i}\right)^2 = 4$.
Therefore,$\Delta T_2 = 4 \times \Delta T_1 = 4 \times 5\,^{\circ}\text{C} = 20\,^{\circ}\text{C}$.

(A) The Watt-hour meter is an electrical instrument used to measure the total amount of electric energy consumed by an electrical circuit over a period of time.
Since $Energy = Power \times Time$, the unit $Watt-hour$ $(Wh)$ represents the energy consumed by a device of $1 \ W$ power in $1 \ hour$.
Therefore, the correct option is $A$.

(D) The power rating of the lamp is $P = 60\, W = 0.06\, kW$.
The time for which the lamp is used is $t = 8\, hours$.
The total energy consumed $E$ is given by $E = P \times t = 0.06\, kW \times 8\, h = 0.48\, kWh$.
The cost per unit is Rs. $1.25$ per $kWh$.
Therefore, the total cost $= 0.48\, kWh \times 1.25\, Rs./kWh = 0.60\, Rs.$

(A) The rated power of each bulb is $P = 40\, W$ at a rated voltage of $V = 250\, V$.
The resistance of each bulb is given by $R = \frac{V^2}{P} = \frac{250^2}{40} = \frac{62500}{40} = 1562.5\, \Omega$.
When $n = 4$ identical bulbs are connected in series,the total resistance is $R_{eq} = nR = 4 \times 1562.5 = 6250\, \Omega$.
The total power consumed by the series combination connected to the $250\, V$ mains is $P_{total} = \frac{V^2}{R_{eq}} = \frac{250^2}{6250} = \frac{62500}{6250} = 10\, W$.
Alternatively,for $n$ identical bulbs in series,the total power is $P_{total} = \frac{P}{n} = \frac{40}{4} = 10\, W$.

(D) The power $P$ consumed by an electrical appliance is given by the formula $P = \frac{V^2}{R}$, where $V$ is the voltage and $R$ is the resistance.
Given: $V = 225\, V$ and $R = 50\, \Omega$.
Substituting the values into the formula:
$P = \frac{(225)^2}{50} = \frac{50625}{50} = 1012.5\, W$.
Rounding to the nearest given option, the power is approximately $1000\, W$.

(B) The relationship between power $(P)$,potential difference $(V)$,and current $(i)$ is given by the formula: $P = V \times i$.
To find the current $(i)$,we rearrange the formula as: $i = \frac{P}{V}$.
Given values are $P = 100\, W$ and $V = 200\, V$.
Substituting these values into the formula: $i = \frac{100}{200} = 0.5\, A$.
Therefore,the current flowing is $0.5\, A$.

(B) The heat produced in a conductor is given by Joule's law of heating: $H = I^2Rt$.
Given:
$I = 2\, A$
$H = 80\, J$
$t = 10\, s$
Rearranging the formula to solve for resistance $R$:
$R = \frac{H}{I^2t}$
Substituting the values:
$R = \frac{80}{(2)^2 \times 10} = \frac{80}{4 \times 10} = \frac{80}{40} = 2\,\,\Omega$.
Therefore,the resistance of the conductor is $2\,\,\Omega$.

(D) The rate at which heat is developed in an electric heater is given by the electric power $P$.
The formula for electric power in terms of voltage $V$ and resistance $R$ is $P = \frac{V^2}{R}$.
Given: Voltage $V = 110\, V$ and Resistance $R = 10\, \Omega$.
Substituting the values into the formula:
$P = \frac{(110)^2}{10} = \frac{12100}{10} = 1210\, W$.
Therefore, the rate at which it develops heat is $1210\, watts$.

(A) The resistance $R$ of the bulb is calculated using its rated power $P_R$ and rated voltage $V_R$:
$R = \frac{V_R^2}{P_R} = \frac{200^2}{100} = \frac{40000}{100} = 400\,\Omega$.
When connected to a supply voltage $V = 160\, V$,the power consumed $P$ is given by:
$P = \frac{V^2}{R} = \frac{160^2}{400} = \frac{25600}{400} = 64\, W$.
Alternatively,using the ratio formula:
$P_{consumed} = \left( \frac{V}{V_R} \right)^2 \times P_R = \left( \frac{160}{200} \right)^2 \times 100 = (0.8)^2 \times 100 = 0.64 \times 100 = 64\, W$.

(D) The power dissipated in a resistor is given by the formula $P = I^2 R$,where $P$ is the power in watts,$I$ is the current in amperes,and $R$ is the resistance in ohms.
Given:
$I = 15\, A$
$P = 22.5\, W$
Substituting these values into the formula:
$22.5 = (15)^2 \times R$
$22.5 = 225 \times R$
$R = \frac{22.5}{225}$
$R = 0.10\,\,\Omega$
Therefore,the resistance of the fuse wire is $0.10\,\,\Omega$.

(A) The rate of energy dissipation is defined as the electrical power $P$ consumed by the device.
Given:
Current $I = 2\,A$
Voltage $V = 250\,V$
The formula for electrical power is $P = V \times I$.
Substituting the given values:
$P = 250\,V \times 2\,A = 500\,W$.
Therefore,the rate of energy dissipation is $500\,W$.