Heating Effect of Electric Current and Electric Power Questions in English

(B) Step $1$: Calculate the total power consumption.
Total power = (Number of tubes) $\times$ (Power per tube) = $5 \times 40 \, W = 200 \, W$.
Step $2$: Convert power to kilowatts $(kW)$.
$200 \, W = 200 / 1000 \, kW = 0.2 \, kW$.
Step $3$: Calculate the total energy consumed in kilowatt-hours ($kWh$ or units).
Energy = Power $(kW)$ $\times$ Time $(h)$ = $0.2 \, kW \times 20 \, h = 4 \, kWh$ (or $4$ units).
Step $4$: Calculate the total cost.
Total cost = (Total units) $\times$ (Cost per unit) = $4 \times ₹ 0.50 = ₹ 2.00$.

(C) In a household circuit, bulbs are connected in parallel.
In a parallel connection, the voltage $(V)$ across each bulb is the same.
The power consumed by a bulb is given by the formula $P = V^2 / R$.
Since $V$ is constant, the power $P$ is inversely proportional to the resistance $R$ $(P \propto 1/R)$.
Therefore, the bulb that glows brighter consumes more power, which means it must have a lower resistance.
Conversely, the bulb that glows dimmer consumes less power, which means it must have a higher resistance.
Thus, the dim bulb has a larger resistance.

(D) The heat generated in a circuit is given by the formula $H = \frac{V^2}{R} \times t$. Since the voltage $V$ and time $t$ are constant,the heat generated is inversely proportional to the resistance $(H \propto \frac{1}{R})$. To maximize heat,we must minimize the total resistance $R$.
Let the resistance of the full wire be $R$.
$A$: Resistance = $R/2$.
$B$: Two parts of $R/2$ in parallel give $R_{eq} = \frac{(R/2) \times (R/2)}{(R/2) + (R/2)} = R/4$.
$C$: Resistance = $R$.
$D$: Four parts of $R/4$ in parallel give $R_{eq} = \frac{R/4}{4} = R/16$.
Comparing the resistances: $R/16 < R/4 < R/2 < R$. Since $R/16$ is the smallest resistance,the heat generated is the largest in case $D$.

(B) The rate of generation of heat $(H)$ in a conductor is given by the formula $H = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
If the potential difference is doubled,the new potential difference becomes $V' = 2V$.
The new rate of heat generation $(H')$ will be $H' = \frac{(V')^2}{R} = \frac{(2V)^2}{R} = \frac{4V^2}{R}$.
Since $H = \frac{V^2}{R}$,we have $H' = 4H$.
Therefore,the rate of generation of heat becomes four times the original value.

(C) heater filament works on the principle of the heating effect of electric current $(H = I^2Rt)$.
To produce a significant amount of heat, the material must have high electrical resistivity $(R = \rho L/A)$ so that it generates heat efficiently when current passes through it.
Additionally, the filament must have a high melting point so that it does not melt or break when it becomes extremely hot during operation.
Therefore, the essential requirements are high resistivity and a high melting point.

(A) Electric power is defined as the rate at which electrical energy is dissipated or consumed in an electric circuit.
Mathematically,$P = \frac{W}{t}$,where $P$ is power,$W$ is work done (or energy transferred),and $t$ is time.
Since the work done in maintaining an electric current is equivalent to the electrical energy consumed,the statement is correct.

(A) The statement is True.
$1 \text{ kWh}$ (kilowatt-hour) is a unit of energy.
$1 \text{ kW} = 1000 \text{ W} = 1000 \text{ J/s}$.
$1 \text{ hour} = 3600 \text{ seconds}$.
Therefore,$1 \text{ kWh} = 1000 \text{ J/s} \times 3600 \text{ s} = 3,600,000 \text{ J} = 3.6 \times 10^6 \text{ J}$.

(B) The statement is False.
$BOTU$ stands for $Board$ $of$ $Trade$ $Unit$, which is another name for $1$ $kWh$ (kilowatt-hour), the unit of electrical energy, not electric power.
Electric power is measured in $Watts$ $(W)$ or $Kilowatts$ $(kW)$.

(A) According to Joule's law of heating,the heat $(H)$ produced 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 for which the current flows.
From this relation,it is clear that $H \propto R$ when $I$ and $t$ are constant.
Therefore,the heat dissipated is directly proportional to the resistance of the conductor.

(A) The electric bulb contains a filament made of a high-resistance material,such as tungsten. When an electric current passes through this filament,the resistance causes the filament to heat up to a very high temperature,resulting in the emission of light. This phenomenon is known as the heating effect of electric current.

(A) The relationship between power $(P)$,voltage $(V)$,and current $(I)$ is given by the formula $P = V \times I$.
Assuming the voltage $(V)$ remains constant (as in a household supply),the power $(P)$ is directly proportional to the current $(I)$: $P \propto I$.
Therefore,an electrical appliance with higher power will draw a larger current from the source.

(B) The unit $kWh$ stands for kilowatt-hour.
Power is defined as the rate of doing work or energy consumption,measured in kilowatts $(kW)$.
Energy is defined as the product of power and time,expressed as $Energy = Power \times Time$.
Therefore,the unit of energy is $kW \times h = kWh$.
Since work is a form of energy,$kWh$ is a unit of work or energy.

(A) $1 \; kWh$ is the unit of electrical energy.
$1 \; kWh = 1 \; \text{kilowatt} \times 1 \; \text{hour}$.
Since $1 \; \text{kilowatt} = 1000 \; \text{watts}$ and $1 \; \text{hour} = 3600 \; \text{seconds}$.
Therefore,$1 \; kWh = 1000 \; \text{W} \times 3600 \; \text{s}$.
$1 \; kWh = 3,600,000 \; \text{joules}$.
In scientific notation,this is $3.6 \times 10^{6} \; \text{joules}$.

(D) Given:
Power $(P)$ = $1.1 \,kW = 1100 \,W$
Voltage $(V)$ = $220 \,V$
We know the formula for electric power is $P = V \times I$,where $I$ is the current.
Rearranging the formula to solve for current: $I = P / V$.
Substituting the values: $I = 1100 \,W / 220 \,V = 5 \,A$.
Therefore,the current flowing through the heater is $5 \,A$.

(D) The formula for electric power $(P)$ is given by the product of electric potential $(V)$ and electric current $(I)$.
$P = V \times I$
Since the unit of electric potential $(V)$ is Volt and the unit of electric current $(I)$ is Ampere,the product $Volt \times Ampere$ represents the unit of Electric Power,which is Watt.

(C) The heating elements of electrical heating appliances,such as electric irons,toasters,and heaters,are made of alloys because alloys have a much higher resistivity than their constituent metals.
Nichrome,an alloy of nickel and chromium,is commonly used for this purpose because it does not oxidize (burn) easily even at high temperatures when it is red hot.

(B) According to Joule's law of heating,the heat energy $H$ (or electrical energy $W$) produced in a resistor is given by the formula $W = I^{2} R t$.
Here,$I$ is the current in amperes,$R$ is the resistance in ohms,and $t$ is the time in seconds.
Therefore,the correct expression for electrical energy consumed is $I^{2} R t$.

(C) Joule's law of heating states that the heat produced $(H)$ in a conductor is directly proportional to the square of the current $(I^2)$,the resistance of the conductor $(R)$,and the time $(t)$ for which the current flows.
Mathematically,this is expressed as $H = I^2Rt$.
Here,$H$ is the heat energy,$I$ is the electric current,$R$ is the resistance,and $t$ is the time.

(A) The heating effect of electric current is described by Joule's Law of Heating,which states that the heat produced $(H)$ in a conductor is given by $H = I^2Rt$.
Devices like water heaters,electric irons,and toasters work on this principle,where electrical energy is converted into heat energy due to the resistance of the heating element.
Electroplating and voltameters work on the chemical effect of electric current,while electric fans work on the magnetic effect of electric current.

(D) The formula for electric energy is $Energy = Power \times Time$.
Since the unit of power is $Watt$ $(W)$ and the unit of time is $second$ $(s)$,the unit of energy is $Watt \times second$ $(W \cdot s)$.
$1 \ W \cdot s = 1 \ Joule$ $(J)$.
Therefore,$Watt-second$ is a unit of electric energy.

(B) The unit of electric power is defined as the rate at which electric energy is consumed or dissipated.
Power $(P)$ is given by the formula $P = W / t$,where $W$ is the work done (or energy consumed) in Joules $(J)$ and $t$ is the time in seconds $(s)$.
Therefore,the unit of power is $Joule/second$ $(J/s)$,which is also known as $Watt$ $(W)$.
Thus,$Joule/second$ is the unit of electric power.

(D) The total power consumed by $10$ bulbs is calculated as: $P = 10 \times 100 \ W = 1000 \ W = 1 \ kW$.
Energy consumed is given by the formula: $E = P \times t$.
Here, $P = 1 \ kW$ and $t = 1 \ \text{hour}$.
Therefore, $E = 1 \ kW \times 1 \ h = 1 \ kWh$.
Since $1 \ kWh$ is equal to $1 \ \text{unit}$ of electrical energy, the total energy consumed is $1 \ \text{unit}$.

(D) The commercial unit of electrical energy is the kilowatt-hour $(kWh)$,which is commonly referred to as $1$ unit.
$1 \text{ unit} = 1 \text{ kWh}$.
Since $1 \text{ kW} = 1000 \text{ W}$ and $1 \text{ hour} = 3600 \text{ seconds}$,we can calculate the value in watt-seconds as follows:
$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^{6} \text{ watt-seconds}$.
Therefore,$1$ unit is equal to $3.6 \times 10^{6} \text{ watt-seconds}$.

(B) To convert $1 \; kWh$ into $\text{watt-seconds}$,we follow these steps:
$1 \; kWh = 1 \; kW \times 1 \; h$
Since $1 \; kW = 1000 \; W$ and $1 \; h = 3600 \; s$ (seconds),
$1 \; kWh = 1000 \; W \times 3600 \; s$
$1 \; kWh = 3,600,000 \; \text{watt-seconds}$
$1 \; kWh = 3.6 \times 10^{6} \; \text{watt-seconds}$.

(A) The $SI$ unit of energy is the Joule $(J)$. However,for commercial and practical purposes,the electrical energy consumed is measured in a larger unit called the Kilowatt-hour $(kWh)$.
$1 \ kWh$ is defined as the amount of electrical energy consumed by an appliance of $1 \ kW$ power rating when used for $1 \ hour$.
Therefore,the practical unit for electrical energy is the Kilowatt-hour $(kWh)$.

(A) Electric power $P$ is defined as the rate at which electrical energy is dissipated or consumed in an electric circuit.
Mathematically,power is the product of potential difference $V$ and electric current $I$.
Therefore,the formula is $P = V \times I$.
Option $A$ is the correct formula for electric power.

(C) Electrical energy is defined as the product of power and time $(E = P \times t)$.
$1$. Watt-second $(W \cdot s)$ is equivalent to Joule $(J)$,which is a unit of energy.
$2$. 'Unit' in common parlance refers to $1 \text{ kWh}$,which is a unit of electrical energy.
$3$. Kilowatt-hour $(kWh)$ is the commercial unit of electrical energy.
$4$. Watt $(W)$ is the $SI$ unit of power,not energy.
Therefore,Watt is not a unit of electrical energy.

(B) Electroplating is a process in which a thin layer of a desired metal is deposited onto another material using an electric current.
This process relies on the chemical effect of electric current,where the passage of electricity through an electrolyte causes a chemical reaction (electrolysis),leading to the deposition of metal ions at the cathode.
Therefore,the correct option is $B$.

(D) In electrical devices,the heating effect of electric current is often a desired outcome,such as in a water heater,electric heater,or oven,where the primary function is to produce heat. However,in an electric motor,the primary function is to convert electrical energy into mechanical energy. The heat produced in an electric motor is a byproduct of energy loss (due to resistance in the windings,friction,etc.) and is considered undesirable or 'not appreciated' because it reduces the efficiency of the device.

(D) $1$. Calculate the total power consumption per hour: $P_{\text{total}} = 100 \; W + 60 \; W + 40 \; W = 200 \; W$.
$2$. Calculate the total energy consumed in $30 \; \text{days}$ in Watt-hours: $E = P_{\text{total}} \times \text{time} \times \text{days} = 200 \; W \times 2 \; \text{hours/day} \times 30 \; \text{days} = 12,000 \; Wh$.
$3$. Convert Watt-hours to units (kilowatt-hours): $1 \; \text{unit} = 1 \; kWh = 1,000 \; Wh$.
$4$. Therefore, $E = 12,000 / 1,000 = 12 \; \text{units}$.

(C) The power $(P)$ of an electrical appliance is given by the formula: $P = V \times I$.
Given: Voltage $(V)$ = $220\;V$,Current $(I)$ = $5\;A$.
Substituting the values: $P = 220\;V \times 5\;A = 1100\;W$.
To convert the power into kilowatts $(kW)$,divide by $1000$: $1100\;W / 1000 = 1.1\;kW$.
Therefore,the power of the iron is $1.1\;kW$.

(C) Given: Voltage $(V)$ = $220 \; V$, Energy $(E)$ = $4.4 \; kWh = 4400 \; Wh$, Time $(t)$ = $1 \; hour$.
First, calculate the power $(P)$ of the geyser using the formula $P = E / t$.
$P = 4.4 \; kWh / 1 \; h = 4.4 \; kW = 4400 \; W$.
Next, use the power formula $P = V^2 / R$ to find the resistance $(R)$.
$R = V^2 / P$.
$R = (220 \; V)^2 / 4400 \; W$.
$R = (220 \times 220) / 4400$.
$R = 48400 / 4400 = 11 \; \Omega$.
Therefore, the resistance of the geyser is $11 \; \Omega$.

(D) The power rating of the bulb is given by $P = 100\;W$ and the voltage rating is $V = 220\;V$.
We know the relationship between power,voltage,and resistance is given by the formula $P = \frac{V^2}{R}$.
Rearranging the formula to solve for resistance $R$,we get $R = \frac{V^2}{P}$.
Substituting the given values: $R = \frac{(220)^2}{100}$.
$R = \frac{48400}{100}$.
$R = 484\;\Omega$.
Therefore,the resistance of the bulb is $484\;\Omega$.

(B) The power $P$ of 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 $V = 240 \ V$.
Rearranging the formula to solve for current: $I = P / V$.
Substituting the values: $I = 60 \ W / 240 \ V = 1 / 4 \ A = 0.25 \ A$.
Converting the current to milliamperes: $0.25 \ A = 0.25 \times 1000 \ mA = 250 \ mA$.
Therefore,the current flowing through the bulb is $250 \ mA$.

(A) An electric fuse is a safety device used in electrical circuits to prevent damage from excessive current.
It consists of a wire made of a material with a low melting point (such as an alloy of lead and tin).
When the current in the circuit exceeds a safe limit,the heat generated causes the fuse wire to melt and break the circuit,thereby protecting the appliances.
Therefore,a fuse is a conductor with a low melting point.

(A, C, D) Power rating refers to the amount of electrical energy an appliance consumes per unit of time $(P = V \times I)$.
Appliances like $TV$s generally have a lower power rating (typically $50 \ W$ to $200 \ W$).
Appliances like Geysers,Heaters,and Air conditioners are high-power appliances that require significant electrical energy to generate heat or cooling,often ranging from $1000 \ W$ to $3000 \ W$ or more.
Among the given options,$TV$ is the only appliance with a low power rating,while Geysers,Heaters,and Air conditioners are high-power appliances. However,the question asks which one is $NOT$ of lower power rating. Since Geysers,Heaters,and Air conditioners are all high-power,this question implies identifying the high-power category. Given the standard classification,Geysers,Heaters,and Air conditioners are all high-power. If we must choose one that is typically the highest or most representative of high-power,Air conditioners often have the highest rating,but all three ($A$,$C$,$D$) are high-power compared to a $TV$.

(C) fuse wire is a safety device used in electrical circuits to protect appliances from high current.
It must have a low melting point so that it melts and breaks the circuit when the current exceeds a safe limit.
Nichrome is an alloy with a very high melting point and high resistance,which is why it is used as a heating element in appliances like heaters and toasters,not as a fuse wire.
Pure tin,lead,and alloys of tin and lead have low melting points,making them suitable for fuse wires.