Mix Examples - Electricity Questions in English

(A) Electrostatic potential is defined as the amount of work done in moving a unit positive charge from infinity to a specific point in an electric field.
Since work is a scalar quantity and charge is a scalar quantity,their ratio,which defines the potential,is also a scalar quantity.
Therefore,electrostatic potential is a scalar quantity.

(N/A) Potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit positive charge from one point to the other.
It is a scalar quantity because it only has magnitude and no specific direction.
The $SI$ unit of potential difference is the volt $(V)$,which is equivalent to joules per coulomb $(J/C)$.

(A) The potential difference between two points is defined as the work done per unit charge in moving a charge from one point to another.
Mathematically,$V = W / Q$.
If $W = 1 \text{ joule}$ and $Q = 1 \text{ coulomb}$,then $V = 1 \text{ joule} / 1 \text{ coulomb} = 1 \text{ volt}$.
Therefore,the potential difference between two points is said to be $1$ volt if $1$ joule of work is done to move a charge of $1$ coulomb between those two points.

(N/A) To maintain a constant potential difference across the ends of a conductor,we need to connect it to a source of electrical energy,such as a cell or a battery.
This source of electromotive force $(emf)$ performs work to move charges,thereby sustaining the flow of current through the conductor.

(A) The potential difference between two points is defined as the work done in moving a unit positive charge from one point to another. The $SI$ unit of potential difference is the volt $(V)$,where $1 \text{ volt} = \frac{1 \text{ joule}}{1 \text{ coulomb}}$. Therefore,the single word for $\frac{1 \text{ joule}}{1 \text{ coulomb}}$ is $1 \text{ volt}$.

(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) The potential difference across the ends of a conductor is measured using a device called a voltmeter.
$A$ voltmeter is always connected in parallel across the points between which the potential difference is to be measured.
It has a high resistance to ensure that it draws a negligible amount of current from the circuit.

(B) The electric current in a circuit is measured using a device called an $Ammeter$.
An $Ammeter$ is always connected in series with the component through which the current is to be measured.

(B) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area of the wire.
From this relation,it is clear that the resistance $R$ is inversely proportional to the cross-sectional area $A$ $(R \propto \frac{1}{A})$.
Therefore,as the cross-sectional area increases,the resistance of the wire decreases.

(B) voltmeter is a device used to measure the potential difference between two points in an electrical circuit.
To measure this potential difference accurately without significantly altering the current flowing through the circuit,it must have a very high resistance.
Therefore,it is always connected in parallel across the component or the points where the potential difference needs to be measured.

(A) An ammeter is a device used to measure the electric current flowing through a circuit. To measure the total current passing through the components,it must be connected in series with the circuit elements so that the same current flows through it.

(N/A) Electric current is defined as the rate of flow of electric charge through a conductor. Mathematically,it is the amount of charge $Q$ flowing through a cross-section of a conductor in time $t$,expressed as $I = Q/t$.
The $SI$ unit of electric current is the ampere $(A)$,which is equivalent to one coulomb per second $(1 \ A = 1 \ C/s).$

(N/A) Ohm's law states that the electric current flowing through a metallic conductor is directly proportional to the potential difference $(V)$ applied across its ends,provided that its physical conditions,such as temperature,remain constant.
Mathematically,this is expressed as $V \propto I$ or $V = IR$,where $R$ is the constant of proportionality known as resistance.

(N/A) Resistance is the property of a conductor to oppose the flow of current (electrons) through it. It is defined as the ratio of the potential difference $(V)$ applied across the conductor to the current $(I)$ flowing through it, expressed by the formula $R = V/I$. The $SI$ unit of resistance is the ohm $(\Omega)$.

(C) The $SI$ unit of electrical resistance is the ohm,denoted by the Greek letter omega $(\Omega)$.
Resistance is defined as the opposition to the flow of electric current in a conductor.
According to Ohm's Law,$R = V / I$,where $R$ is resistance in ohms,$V$ is potential difference in volts,and $I$ is current in amperes.

(N/A) The resistance of a conductor is said to be $1$ ohm if a current of $1$ ampere flows through it when a potential difference of $1$ volt is applied across its ends. This is based on Ohm's Law, which states $R = V / I$. Therefore, $1 \Omega = 1 \ V / 1 \ A$.

(B) According to Ohm's Law,$V = IR$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance of the conductor.
If the resistance $R$ remains constant,then $I = V/R$.
If the potential difference $V$ is doubled $(V' = 2V)$,the new current $I'$ becomes $I' = V'/R = (2V)/R = 2(V/R) = 2I$.
Therefore,the current flowing through the conductor is also doubled.

(D) The resistance $(R)$ of a conductor depends on the following factors:
$(i)$ Length of the conductor $(l)$: Resistance is directly proportional to the length of the conductor $(R \propto l)$.
$(ii)$ Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section $(R \propto 1/A)$.
$(iii)$ Nature of the material: Resistance depends on the resistivity $(\rho)$ of the material.
$(iv)$ Temperature: Resistance of a conductor generally increases with an increase in temperature.

(N/A) Equivalent resistance is defined as the single resistance value that,when connected in a circuit,would draw the same amount of current from the source as the original combination of multiple resistors connected in series,parallel,or a combination of both. It effectively replaces the entire network of resistors with one equivalent component while maintaining the same electrical characteristics for the rest of the circuit.

(B) Resistors are said to be connected in series when they are joined end-to-end such that the same amount of electric current flows through each resistor.
In a series circuit,there is only one path for the current to flow,which ensures that the current remains constant throughout the circuit.

(B) Resistors are said to be connected in parallel when the potential difference $(V)$ across each resistor is the same.
In a parallel circuit,the current $(I)$ divides among the branches,but the voltage remains constant across all components connected between the same two points.

(C) The resistivity $(\rho)$ of a material is an intrinsic property that depends on the following factors:
$1$. Nature of the material: Different materials have different atomic structures, which affect the mobility of charge carriers.
$2$. Temperature: For most conductors, resistivity increases with an increase in temperature due to increased lattice vibrations, which hinder the flow of electrons.

(B) According to Ohm's Law,$V = IR$,which can be rearranged as $I = V/R$.
If the resistance $R$ is doubled $(R' = 2R)$ while the voltage $V$ remains constant,the new current $I'$ will be $I' = V / (2R) = 1/2 * (V/R) = I/2$.
Therefore,the current becomes half of its original value.

(C) According to Ohm's Law,$V = IR$,which implies $R = V/I$.
Resistance $(R)$ is a property of the conductor that depends on its material,length,cross-sectional area,and temperature.
It does not depend on the current $(I)$ flowing through the circuit or the potential difference $(V)$ applied across it.
Therefore,if the current through the circuit is doubled,the resistance will remain unchanged.

(A) During their motion through a conductor,electrons collide with other electrons and the positive ions of the conductor lattice.
These collisions hinder the flow of electrons,creating an opposition to the electric current.
This opposition to the flow of charge is defined as electrical resistance.

(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) The resistivity of a material is defined as the resistance offered by a conductor of that material having a unit length $(1 \ m)$ and a unit cross-sectional area $(1 \ m^2)$.
Mathematically,it is expressed as $\rho = \frac{RA}{l}$,where $R$ is the resistance,$A$ is the cross-sectional area,and $l$ is the length of the conductor.
Its $SI$ unit is $\Omega \cdot m$ (ohm-meter).

(A) The formula for resistivity is $\rho = \frac{R \cdot A}{L}$.
Given:
Resistance $(R)$ = $20\, \Omega$
Area of cross-section $(A)$ = $2\, cm^{2}$
Length $(L)$ = $10\, cm$
Substituting the values into the formula:
$\rho = \frac{20 \times 2}{10} = \frac{40}{10} = 4\, \Omega \cdot cm$.
Thus,the resistivity is $4\, \Omega \cdot cm$.

(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.

(A) When a wire of resistance $10 \, \Omega$ is bent into a circle,the total resistance is distributed uniformly along the circumference.
Any diameter divides the circle into two equal semicircular parts.
Each semicircle will have a resistance of $R_1 = R_2 = 10 / 2 = 5 \, \Omega$.
These two semicircular parts are connected in parallel between the two ends of the diameter.
The effective resistance $R_{eq}$ is given by the formula for parallel resistors: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.
Substituting the values: $\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$.
Therefore,$R_{eq} = \frac{5}{2} = 2.5 \, \Omega$.

(A) The total resistance of the wire is $R = 5 \, \Omega$.
When the wire is bent into a circle,the resistance is distributed uniformly along the circumference.
When we consider two points at the ends of a diameter,the circle is divided into two equal semicircular parts.
Each semicircular part has a resistance of $R' = R / 2 = 5 / 2 = 2.5 \, \Omega$.
These two semicircular parts are connected in parallel between the two points.
The equivalent resistance $R_{eq}$ is given by the formula for parallel resistors: $1 / R_{eq} = 1 / R' + 1 / R'$.
$1 / R_{eq} = 1 / 2.5 + 1 / 2.5 = 2 / 2.5$.
$R_{eq} = 2.5 / 2 = 1.25 \, \Omega$.

(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.
Since both bulbs are rated for the same voltage $(V = 220\, V)$,the resistance $R$ is inversely proportional to the power $P$ $(R \propto \frac{1}{P})$.
Therefore,the bulb with the lower power rating will have a higher resistance.
Comparing the two bulbs,the $60\, W$ bulb has a lower power rating than the $100\, W$ bulb.
Thus,the $60\, W$ bulb has higher resistance.

(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.

(A) According to Ohm's Law,the relationship between voltage $(V)$,current $(I)$,and resistance $(R)$ is given by the formula: $V = I \times R$.
To find the resistance,we rearrange the formula as: $R = \frac{V}{I}$.
Given values are: Voltage $(V)$ = $220\, V$ and Current $(I)$ = $20\, A$.
Substituting these values into the formula: $R = \frac{220}{20} = 11\, \Omega$.
Therefore,the resistance of the electric arc lamp is $11\, \Omega$.

(A) $(i)$ False: Current is defined as the rate of flow of charge,i.e.,$I = Q/t$. The statement incorrectly suggests $I = Q \times t$.
$(ii)$ True: $A$ cell converts chemical energy into electrical energy through internal chemical reactions,which drives the charge flow.
$(iii)$ True: For pure metals,the resistivity increases as temperature rises due to increased collisions between electrons and vibrating ions.
$(iv)$ False: Ohm's law states that the potential difference across a conductor is directly proportional to the current flowing through it $(V \propto I)$,provided temperature remains constant. It does not relate to power.
$(v)$ True: In a series circuit,there is only one path for current,whereas a parallel circuit provides multiple branches for current flow.
$(vi)$ False: Resistance in a wire is primarily due to the collisions of electrons with the atoms/ions of the conductor's lattice,not due to electron-electron repulsion.