Heating Effect of Electric Current and Electric Power Questions in English

(N/A) The resistivity of an alloy is generally much higher than that of its constituent pure metals.
Due to this high resistivity,alloys produce more heat when an electric current passes through them.
Furthermore,alloys do not oxidize (burn) or melt readily even at high temperatures.
Therefore,alloys are preferred for heating elements in appliances like electric toasters and electric irons.

(B) The heating element of an electric heater is a resistor.
According to Joule's law of heating,the heat produced $(H)$ is given by $H = I^2Rt$,where $I$ is the current,$R$ is the resistance,and $t$ is the time.
The resistance of the heating element is very high,which causes it to produce a large amount of heat,making it glow red.
In contrast,the cord is made of good conductors like copper,which have very low resistance.
Therefore,the heat produced in the cord is negligible,and it does not glow.

(C) The amount of heat $(H)$ produced is given by the formula $H = V \times Q$,where $V$ is the potential difference and $Q$ is the total charge transferred.
Given:
Potential difference,$V = 50\, V$
Charge,$Q = 96000\, C$
Using the formula:
$H = V \times Q$
$H = 50\, V \times 96000\, C$
$H = 4800000\, J$
$H = 4.8 \times 10^{6}\, J$
Therefore,the heat generated is $4.8 \times 10^{6}\, J$.

(D) The amount of heat $(H)$ produced is given by Joule's law of heating as $H = I^2Rt$.
Given:
Resistance,$R = 20\, \Omega$
Current,$I = 5\, A$
Time,$t = 30\, s$
Using the formula:
$H = (5)^2 \times 20 \times 30$
$H = 25 \times 20 \times 30$
$H = 500 \times 30$
$H = 15000\, J$
$H = 1.5 \times 10^4\, J$
Therefore,the amount of heat developed in the electric iron is $1.5 \times 10^4\, J$.

(B) The rate of consumption of electric energy in an electric appliance is defined as electric power.
Mathematically,$P = \frac{W}{t} = VI$.
Therefore,the rate at which energy is delivered by a current is known as electric power.

(B) Power $(P)$ is given by the expression,
$P = V \times I$
Where,
Voltage,$V = 220\, V$
Current,$I = 5\, A$
$P = 220 \times 5 = 1100\, W$
Energy consumed $(E)$ by the motor is given by,
$E = P \times t$
Where,
Time,$t = 2\, h = 2 \times 60 \times 60 = 7200\, s$
$E = 1100 \times 7200 = 7,920,000\, J = 7.92 \times 10^6\, J$
Therefore,the power of the motor is $1100\, W$ and the energy consumed is $7.92 \times 10^6\, J$.

(D) Electrical power is given by the expression,$P = VI$ $(i)$.
According to Ohm's law,$V = IR$ $(ii)$.
Where,
$V =$ Potential difference
$I =$ Current
$R =$ Resistance
From equation $(i)$,substituting $V = IR$,we get:
$P = (IR) \times I = I^2R$.
From equation $(ii)$,substituting $I = \frac{V}{R}$,we get:
$P = V \times \frac{V}{R} = \frac{V^2}{R}$.
Therefore,the expressions for power are $P = VI = I^2R = \frac{V^2}{R}$.
The term $IR^2$ does not represent electrical power.

(A) The power consumed by an electrical appliance is given by the formula $P = \frac{V^2}{R}$.
First,we calculate the resistance $R$ of the bulb using its rated values:
$R = \frac{V^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484\, \Omega$.
The resistance of the bulb remains constant regardless of the operating voltage.
Now,when the bulb is operated at a new voltage $V' = 110\, V$,the new power consumed $P'$ is:
$P' = \frac{(V')^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25\, W$.
Therefore,the power consumed will be $25\, W$.

(B) The power rating of each bulb is $P_1 = 10\, W$ at a voltage of $V = 220\, V$.
The current drawn by each bulb is $I_1 = \frac{P_1}{V} = \frac{10}{220} = \frac{1}{22}\, A$.
The maximum allowable current in the circuit is $I_{total} = 5\, A$.
Since the bulbs are connected in parallel,the total current is the sum of the currents through each bulb.
If $n$ is the number of bulbs,then $I_{total} = n \times I_1$.
$5 = n \times \frac{1}{22}$.
$n = 5 \times 22 = 110$.
Therefore,$110$ electric bulbs can be connected in parallel.

(A) Both the bulbs are connected in parallel. Therefore,the potential difference across each of them will be $220\, V$,because the voltage remains the same in a parallel circuit.
For the first bulb $(100\, W, 220\, V)$:
$I_1 = \frac{P_1}{V} = \frac{100}{220}\, A$
For the second bulb $(60\, W, 220\, V)$:
$I_2 = \frac{P_2}{V} = \frac{60}{220}\, A$
Total current drawn from the line $(I)$ is the sum of currents through both bulbs:
$I = I_1 + I_2 = \frac{100}{220} + \frac{60}{220} = \frac{160}{220}\, A$
$I \approx 0.727\, A$
Rounding to two decimal places,we get $0.73\, A$.

(D) The power dissipated in a resistor is given by the formula $P = I^2 R$, where $I$ is the current and $R$ is the resistance.
Initially, let the current be $I_1 = I$ and the power be $P_1 = I^2 R$.
When the current is increased by $100\%$, the new current $I_2$ becomes $I + 100\% \text{ of } I = I + I = 2I$.
The new power dissipated $P_2$ is $P_2 = (I_2)^2 R = (2I)^2 R = 4I^2 R = 4P_1$.
The increase in power is $P_2 - P_1 = 4P_1 - P_1 = 3P_1$.
The percentage increase in power is given by $\frac{P_2 - P_1}{P_1} \times 100\% = \frac{3P_1}{P_1} \times 100\% = 300\%$.

(B) When bulbs are connected in parallel,the potential difference $(V)$ across each bulb remains the same.
The power consumed by a bulb is given by the formula $P = V^2 / R$.
Since $V$ is constant,the power consumed is directly proportional to the brightness of the bulb.
Bulb $C$ has the highest power rating $(100 \, W)$,followed by bulb $B$ $(60 \, W)$,and bulb $A$ $(40 \, W)$.
Therefore,the brightness of bulb $C$ will be the maximum,and the brightness of bulb $A$ will be the minimum.
Comparing the options,the brightness of bulb $B$ $(60 \, W)$ is indeed more than that of bulb $A$ $(40 \, W)$.

(C) Given: Resistors $R_1 = 2\,\Omega$ and $R_2 = 4\,\Omega$ are connected in series.
Voltage $V = 6\,V$.
Time $t = 5\,s$.
Total resistance $R_{eq} = R_1 + R_2 = 2\,\Omega + 4\,\Omega = 6\,\Omega$.
The current $I$ flowing through the circuit is $I = V / R_{eq} = 6\,V / 6\,\Omega = 1\,A$.
Since the resistors are in series,the same current $I = 1\,A$ flows through the $4\,\Omega$ resistor.
The heat dissipated $H$ is given by the formula $H = I^2 R t$.
Substituting the values: $H = (1)^2 \times 4 \times 5 = 1 \times 4 \times 5 = 20\,J$.

(D) The power consumed by the electric kettle is $P = 1\, kW = 1000\, W$.
The voltage at which it operates is $V = 220\, V$.
Using the formula for electric power,$P = V \times I$,we can find the current $I$ as:
$I = P / V$
$I = 1000 / 220 \approx 4.54\, A$.
Since the current drawn is approximately $4.54\, A$,a fuse wire with a rating slightly higher than this value is required to prevent the circuit from breaking during normal operation.
Therefore,a fuse of $5\, A$ rating is the most appropriate choice among the given options.

(B) Electric power $(P)$ is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit.
Mathematically,$P = V \times I$,where $V$ is the potential difference measured in volts $(V)$ and $I$ is the electric current measured in amperes $(A)$.
Therefore,the $SI$ unit of power,the watt $(W)$,can also be expressed as the product of volts and amperes $(V \cdot A)$.
- Kilowatt hour $(kWh)$ is a unit of electrical energy.
- Watt second $(Ws)$ is also a unit of energy (since $Power \times Time = Energy$).
- Joule second $(Js)$ is a unit of action.
Thus,the correct expression for electric power is volt ampere.

(A) The maximum power $P_{max}$ that each resistor can withstand is $18\,W$,and the resistance $R$ of each is $2\,\Omega$.
Using the formula $P = I^2R$,the maximum current $I_{max}$ that any individual resistor can withstand is:
$I_{max} = \sqrt{\frac{P_{max}}{R}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3\,A$.
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 $I = 3\,A$,the current through $B$ and $C$ (which are equal) will be $I_B = I_C = \frac{I}{2} = 1.5\,A$.
Since $1.5\,A < 3\,A$,resistors $B$ and $C$ will not melt.
Thus,the maximum current that can flow through the circuit is $3\,A$.

(N/A) The commercial unit of electrical energy is the kilowatt-hour $(kWh)$.
To convert $1 \; kWh$ into joules $(J)$:
$1 \; kWh = 1 \; kW \times 1 \; h$
$1 \; kW = 1000 \; W = 1000 \; J/s$
$1 \; h = 60 \times 60 \; s = 3600 \; s$
Therefore,$1 \; kWh = 1000 \; J/s \times 3600 \; s = 3.6 \times 10^6 \; J$.

(N/A) Joule's heating effect states that when an electric current $I$ flows through a conductor of resistance $R$ for time $t$,the heat $H$ produced is given by the formula $H = I^2Rt$.
Experimental demonstration: Connect a resistor of known resistance $R$ in series with an ammeter,a battery,and a plug key. Connect a voltmeter in parallel across the resistor. Measure the current $I$ and potential difference $V$ across the resistor. The heat produced can be calculated by measuring the temperature rise in a known mass of water or by using the formula $H = VIt$.
Applications in daily life:
$1$. Electric iron
$2$. Electric toaster
$3$. Electric heater
$4$. Electric kettle

(N/A) Tungsten is used for making the filament of an electric lamp because it has a very high melting point (approximately $3380 ^\circ C$), which allows it to glow at high temperatures without melting. Additionally, it has a high resistivity, which helps in generating sufficient heat when an electric current passes through it.

(N/A) The wire used in the element of an electric heater is made of an alloy (like Nichrome) which has a very high resistance and a high melting point,allowing it to produce heat without melting. In contrast,a fuse wire is made of a material with a low melting point and appropriate resistance,designed to melt and break the circuit when the current exceeds a safe limit.

(A) The formula for energy consumed is given by $E = P \times t$,where $P$ is power and $t$ is time.
Given: Power $(P)$ = $60\, W$,Time $(t)$ = $1\, s$.
Substituting the values: $E = 60\, W \times 1\, s = 60\, J$.
Therefore,the energy consumed is $60\, J$.

(A) Energy consumed by an electrical appliance is given by the formula, $H = P \times t$.
For the $TV$ set:
Power $(P)$ = $250 \,W$
Time $(t)$ = $1 \,hr = 3600 \,s$
Energy = $250 \,W \times 3600 \,s = 900,000 \,J = 9 \times 10^5 \,J$.
For the toaster:
Power $(P)$ = $1200 \,W$
Time $(t)$ = $10 \,minutes = 10 \times 60 \,s = 600 \,s$
Energy = $1200 \,W \times 600 \,s = 720,000 \,J = 7.2 \times 10^5 \,J$.
Comparing the two values, $9 \times 10^5 \,J > 7.2 \times 10^5 \,J$.
Therefore, the $250 \,W$ $TV$ set consumes more energy.

(C) The rate at which heat is developed in an electric heater is defined as its electric power $(P)$.
The formula for electric power is given by:
$P = I^2 R$
Given:
Resistance $(R)$ = $8 \, \Omega$
Current $(I)$ = $15 \, A$
Substituting the values into the formula:
$P = (15)^2 \times 8$
$P = 225 \times 8$
$P = 1800 \, W$
Since $1 \, W = 1 \, J/s$,the rate at which heat is developed is $1800 \, J/s$ or $1800 \, W$.

(N/A) The melting point and resistivity of tungsten are very high. It does not oxidize or burn readily at high temperatures. Electric lamps glow at very high temperatures,and tungsten can withstand these temperatures without melting. Hence,tungsten is used as the filament of electric bulbs.
$(b)$ The conductors of electric heating devices,such as bread-toasters and electric irons,are made of an alloy because the resistivity of an alloy is significantly higher than that of pure metals. Furthermore,alloys do not oxidize (burn) readily at high temperatures even when they are red hot. This property allows them to produce a large amount of heat efficiently.

(B) One kilowatt-hour $(kWh)$ is the amount of energy consumed by an electrical device of power $1 \text{ kW}$ in $1 \text{ hour}$.
$1 \text{ kW} = 1000 \text{ W} = 1000 \text{ J/s}$.
$1 \text{ hour} = 3600 \text{ seconds}$.
Therefore, $1 \text{ kWh} = 1000 \text{ J/s} \times 3600 \text{ s} = 3,600,000 \text{ J}$.
This can be expressed in scientific notation as $3.6 \times 10^{6} \text{ J}$.

(N/A) Electric power is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit. Mathematically,it is the rate of doing work,given by $P = \frac{W}{t} = VI = I^2R = \frac{V^2}{R}$. The $SI$ unit of electric power is the watt $(W)$,which is equivalent to $1 \text{ joule per second} (J/s).$

(C) The commercial unit of electric energy is the kilowatt-hour $(kWh)$.
$1 \ kWh$ is equivalent to $3.6 \times 10^6 \ J$ of energy.
It is commonly known as a 'unit' of electricity used in billing domestic and industrial power consumption.

(A) The commercial unit of electrical energy is the kilowatt-hour $(kWh)$.
It is also commonly known as the Board of Trade Unit $(BOTU)$ or simply a 'unit' in electricity billing.

(B) The power rating of a bulb is given by the formula $P = V^2 / R$,where $P$ is power,$V$ is voltage,and $R$ is resistance.
Assuming both bulbs are designed for the same voltage $V$,the resistance is inversely proportional to the power $(R = V^2 / P)$.
For a $100$ $W$ bulb,$R_1 = V^2 / 100$.
For a $60$ $W$ bulb,$R_2 = V^2 / 60$.
Since $60 < 100$,it follows that $V^2 / 60 > V^2 / 100$,which means $R_2 > R_1$.
Therefore,a $60$ $W$ bulb has a greater resistance.

(A) According to Joule's law of heating,the heat $(H)$ dissipated by a conductor is given by the formula $H = I^2Rt$.
Therefore,the factors on which heat dissipated depends are:
$(i)$ The square of the current $(I^2)$ flowing through the conductor.
$(ii)$ The resistance $(R)$ of the conductor.
$(iii)$ The time $(t)$ for which the current is passed through the conductor.

(N/A) When an electric current flows through a conductor,the free electrons move in a specific direction. During their motion,these electrons frequently collide with the atoms and ions of the conductor's lattice. Due to these collisions,the electrons lose some of their kinetic energy. This lost kinetic energy is converted into thermal energy,which is dissipated as heat across the conductor.

(D) The heating effect of current is utilized in various electrical appliances where electrical energy is converted into heat energy.
Two common practical applications are:
$1$. Electric Iron: It uses a heating element to heat the base plate for ironing clothes.
$2$. Electric Fuse: It uses a wire with a low melting point that melts and breaks the circuit when excessive current flows,protecting the appliance.

(N/A) The power expended by a source is said to be $1$ watt,if $1$ ampere of current flows through it under a potential difference of $1$ volt. Mathematically,$P = V \times I$,where $P$ is power in watts,$V$ is potential difference in volts,and $I$ is current in amperes. Thus,$1 \text{ watt} = 1 \text{ volt} \times 1 \text{ ampere}$.

(N/A) fuse wire is designed to protect electrical appliances from damage caused by excessive current. When the current flowing through the circuit exceeds the safe limit,the heat generated $(H = I^2Rt)$ increases significantly. Because the fuse wire has a low melting point,it melts quickly when this excess heat is produced,thereby breaking the circuit and preventing further flow of current. This action protects the connected devices from potential damage or fire hazards.

(N/A) Joule's law of heating states that the heat produced $(H)$ in a conductor is directly proportional to the square of the current $(I)$ flowing through it, the resistance $(R)$ of the conductor, and the time $(t)$ for which the current flows.
Mathematically, it is expressed as: $H = I^{2}Rt$
Where:
$H$ = Heat produced in Joules $(J)$
$I$ = Current in Amperes $(A)$
$R$ = Resistance in Ohms $(\Omega)$
$t$ = Time in seconds $(s)$

(N/A) The filament of an electric bulb is made of tungsten because it has a very high melting point (approximately $3380 ^\circ C$).
This high melting point allows the filament to become white-hot and emit light without melting when a large amount of heat is produced due to the flow of electric current.
Additionally, tungsten does not easily oxidize or react with air at high temperatures, which ensures the longevity of the bulb.

One watt hour is defined as the amount of electrical energy consumed by an electrical appliance of power $1$ watt when it operates for a duration of $1$ hour.
Mathematically:
$1 \text{ watt hour} = 1 \text{ watt} \times 1 \text{ hour}$
Since $1 \text{ watt} = 1 \text{ J s}^{-1}$ and $1 \text{ hour} = 3600 \text{ s}$,
$1 \text{ watt hour} = (1 \text{ J s}^{-1}) \times (3600 \text{ s}) = 3600 \text{ J}$.

(N/A) Tungsten is selected for making the filaments of incandescent lamps because it has a very high melting point (approximately $3380 \, ^\circ C$).
This high melting point allows the filament to become white-hot and emit light without melting away when a large amount of electric current passes through it.

(C) According to Joule's law of heating,the heat generated $(H)$ in a conductor is given by $H = I^2Rt$,where $I$ is the current,$R$ is the resistance,and $t$ is the time.
Electric cables are typically made of thick copper wires,which have a very low resistance $(R)$.
In contrast,the filaments of electric bulbs are made of thin tungsten wire,which has a very high resistance $(R)$.
Since the heat generated is directly proportional to the resistance $(H \propto R)$,the low resistance of the cables results in significantly less heat generation compared to the high resistance of the bulb filaments.

(B) The resistance $R$ of an electrical appliance is given by the formula $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power rating.
Assuming the voltage $V$ remains constant for both bulbs,the relationship between resistance and power is inversely proportional,i.e.,$R \propto \frac{1}{P}$.
For a $50 \, W$ bulb: $R_1 = \frac{V^2}{50}$.
For a $25 \, W$ bulb: $R_2 = \frac{V^2}{25}$.
Comparing the two: $\frac{R_2}{R_1} = \frac{V^2/25}{V^2/50} = \frac{50}{25} = 2$.
Therefore,the $25 \, W$ lamp has double ($2$ times) the resistance of the $50 \, W$ lamp.

(A) Given:
Voltage $(V)$ = $2.5 \, V$
Current $(I)$ = $500 \, mA = 500 \times 10^{-3} \, A = 0.5 \, A$
Formula for power $(P)$:
$P = V \times I$
Calculation:
$P = 2.5 \, V \times 0.5 \, A = 1.25 \, W$
Therefore,the power of the bulb is $1.25 \, W$.

(A) We know that the formula for resistance is $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power.
Assuming the voltage $V$ is constant for both appliances,the resistance $R$ is inversely proportional to the power $P$,i.e.,$R \propto \frac{1}{P}$.
For the toaster,$P_1 = 1\, kW$.
For the electric heater,$P_2 = 2\, kW$.
Since $P_1 < P_2$,it follows that $R_1 > R_2$.
Therefore,the $1\, kW$ toaster has a greater resistance than the $2\, kW$ electric heater.

(N/A) Alloys are commonly used in electrical heating devices because of the following reasons:
$1$. They have high resistivity,which allows them to produce more heat when an electric current passes through them.
$2$. They do not oxidize (burn) readily even at high temperatures,which increases the durability and lifespan of the heating element.

(B) When a high-power appliance like a geyser is switched on,it draws a large amount of current from the source.
Due to the internal resistance of the supply lines and the source,this high current causes a significant potential drop (voltage drop) across the supply lines.
As a result,the voltage available to the bulb decreases momentarily.
Since the power consumed by the bulb is given by $P = V^2 / R$,a decrease in voltage $(V)$ leads to a decrease in the power output of the bulb,making it appear dimmer.

(N/A) The heat generated in a conductor is given by Joule's law of heating,$H = I^2Rt$,where $I$ is the current,$R$ is the resistance,and $t$ is the time.
Since the electric cables are made of good conductors like copper or aluminum,they have very low electrical resistance $(R)$.
In contrast,the filament of an electric bulb is made of tungsten,which has a very high melting point and high electrical resistance $(R)$ to produce light through incandescence.
Because the heat generated $(H)$ is directly proportional to the resistance $(R)$,the high resistance of the filament results in significant heat production,whereas the low resistance of the cables results in minimal heat generation.

(N/A) Nichrome is an alloy that possesses high electrical resistivity,which allows it to generate a significant amount of heat when an electric current passes through it.
Additionally,it has a very high melting point,which prevents it from melting or burning out even when it becomes extremely hot during operation.

(N/A) Alloys do not oxidise readily at high temperatures,which makes them more resistant to corrosion.
$(b)$ Alloys have higher electrical resistivity (lower electrical conductivity) than pure metals,which allows them to generate more heat when current passes through them.

(N/A) The material used for the heating element of an electric iron must possess the following two special features:
$(a)$ High resistivity: This ensures that the material produces a large amount of heat when an electric current passes through it,according to Joule's law of heating $(H = I^2Rt)$.
$(b)$ High melting point: This prevents the element from melting or breaking down even when it becomes extremely hot during continuous operation.

(N/A) $(i)$ Tungsten is used in making the filament of an electric bulb because:
$(a)$ Tungsten has a very high melting point $(3380 \ ^\circ C)$,which allows it to glow at high temperatures without melting.
$(b)$ It has high resistivity,which allows it to produce a large amount of heat when current passes through it.
$(ii)$ Copper and aluminium are used for the transmission of electric current because:
$(a)$ They have very low electrical resistivity,which minimizes energy loss during transmission.
$(b)$ They are excellent conductors of electricity,allowing current to flow efficiently over long distances.

(A) The power $P$ and voltage $V$ are given as $P = 100 \, W$ and $V = 250 \, V$.
Using the formula for electrical power,$P = \frac{V^2}{R}$,we can rearrange it to solve for resistance $R$:
$R = \frac{V^2}{P}$.
Substituting the given values:
$R = \frac{250 \times 250}{100} = \frac{62500}{100} = 625 \, \Omega$.
Therefore,the resistance of the bulb is $625 \, \Omega$.