Mix Examples - Electricity Questions in English

(A) The power $P$ of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Since both lamps are designed to operate on the same voltage $V$,we have $R = \frac{V^2}{P}$.
For the first lamp,$P_{1} = 400 \ W$,so $R_{1} = \frac{V^2}{400}$.
For the second lamp,$P_{2} = 200 \ W$,so $R_{2} = \frac{V^2}{200}$.
Dividing $R_{2}$ by $R_{1}$,we get $\frac{R_{2}}{R_{1}} = \frac{V^2 / 200}{V^2 / 400} = \frac{400}{200} = 2$.
Therefore,$R_{2} = 2R_{1}$.

(D) When resistors are connected in parallel, the potential difference (voltage) across each resistor remains the same.
According to Ohm's Law, $V = IR$, which implies $I = V/R$.
Since $V$ is constant for both resistors, the current $I$ is inversely proportional to the resistance $R$ $(I \propto 1/R)$.
Therefore, the resistor with the smaller resistance will have a larger current, and the resistor with the larger resistance will have a smaller current.
Thus, the voltage is the same across both resistors.

(A) In a series combination of resistors,the equivalent resistance $(R_{eq})$ is the sum of individual resistances: $R_{eq} = R_1 + R_2 + R_3 + ... + R_n$.
Since each individual resistance is positive,the sum $R_{eq}$ will always be greater than any single individual resistance present in the circuit.
Therefore,the equivalent resistance is larger than the largest individual resistance.

(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 $D$.

(A) The power rating of a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
From this formula,we can express resistance as $R = \frac{V^2}{P}$.
Since the voltage $V$ is constant $(220 \, V)$ for all three bulbs,the resistance $R$ is inversely proportional to the power $P$ $(R \propto \frac{1}{P})$.
Therefore,the bulb with the highest power rating will have the lowest resistance.
Comparing the given power ratings: $60 \, W > 40 \, W > 25 \, W$.
Thus,the $60 \, W$ bulb has the lowest resistance.

(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.

(D) The resistance of a wire 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.
Since both wires are of the same metal, $\rho$ is constant. Given that $L$ is also the same, we have $R \propto \frac{1}{A}$.
Let $A_1$ be the area of the thicker wire and $A_2$ be the area of the thinner wire. Given $A_1 : A_2 = 3 : 1$, so $A_1 = 3A$ and $A_2 = A$.
The resistance of the thicker wire $(R_1)$ is $10\, \Omega$. Since $R_1 = \frac{\rho L}{3A} = 10\, \Omega$, then $\frac{\rho L}{A} = 30\, \Omega$.
The resistance of the thinner wire $(R_2)$ is $R_2 = \frac{\rho L}{A} = 30\, \Omega$.
When joined in series, the total resistance $R_{total} = R_1 + R_2 = 10\, \Omega + 30\, \Omega = 40\, \Omega$.

(A) Let the resistance of each resistor be $R$. We can combine these three resistors in the following ways:
$1$. All three in series: $R_{eq} = R + R + R = 3R$.
$2$. All three in parallel: $1/R_{eq} = 1/R + 1/R + 1/R = 3/R$,so $R_{eq} = R/3$.
$3$. Two in series and one in parallel with them: $R_{eq} = (2R \times R) / (2R + R) = 2R^2 / 3R = (2/3)R$.
$4$. Two in parallel and one in series with them: $R_{eq} = R + (R \times R) / (R + R) = R + R/2 = (3/2)R$.
Thus,there are $4$ distinct equivalent resistance values that can be obtained.

(A) The $SI$ unit of electric charge is the coulomb $(C)$. It is defined as the amount of charge transported by a constant current of one ampere in one second $(1 \ C = 1 \ A \cdot s)$.

(A) Metals are good conductors of electricity because they possess a large number of free electrons that can move freely throughout the material.
In contrast,non-metals are insulators because they have very few or no free electrons,which prevents the flow of electric current.

(A) In a metal,electrons in the outermost shells of atoms are loosely bound to the nucleus. These are called free electrons. They can move randomly throughout the entire volume of the metal conductor. However,they cannot escape the metal surface under normal conditions because there is a potential barrier at the surface that keeps them confined within the material.

(B) Electric potential and potential difference are scalar quantities.
They are defined as the amount of work done in moving a unit positive charge from one point to another.
Since they only have magnitude and no specific direction associated with them,they are classified as scalars.

(A) The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point.
If a body is positively charged,it exerts a repulsive force on a unit positive charge,meaning work must be done against the electric field to bring the charge closer. Thus,the potential is positive.
Conversely,if a body is negatively charged,it exerts an attractive force on a unit positive charge,meaning the field does the work. Thus,the potential is negative.
Therefore,the statement is True.

(A) By convention,the direction of electric current is taken as the direction of flow of positive charges. Positive charges always move from a region of higher electric potential to a region of lower electric potential. This flow continues until the potential difference between the two points becomes zero.

(A) battery or a cell is a device that converts chemical energy into electrical energy. It maintains a constant potential difference (voltage) across the ends of a conductor,which drives the flow of electrons (current) through the circuit. Therefore,the statement is True.

(B) The unit of resistance is the $ohm$ ($\Omega$), not the $ampere$.
The $ampere$ $(A)$ is the $SI$ unit of electric current.
Therefore, the statement is $False$.

(A) According to Ohm's Law,$V = IR$,where $V$ is potential difference,$I$ is current,and $R$ is resistance.
Rearranging for current,we get $I = V/R$.
If the resistance $R$ is doubled $(R' = 2R)$ while the potential difference $V$ remains constant,the new current $I'$ becomes $I' = V/(2R) = (1/2) \times (V/R) = I/2$.
Therefore,the current is halved.

(A) The resistance of a metallic conductor is given by the formula $R = \frac{ml}{ne^2 \tau A}$.
As the temperature increases,the thermal velocity of the free electrons increases,which leads to more frequent collisions between electrons and ions.
This results in a decrease in the relaxation time $(\tau)$.
Since $R \propto \frac{1}{\tau}$,a decrease in relaxation time leads to an increase in the resistance of the conductor.
Therefore,the statement is true.

(B) The statement is $False$.
Insulators are materials that do not allow electric current to flow through them easily.
This is because they possess a very high electrical resistance,which prevents the movement of electrons.

(A) In a series circuit,there is only one path for the electric current to flow. Therefore,the amount of charge passing through any point in the circuit per unit time is constant. As a result,the current $(I)$ flowing through each resistor connected in series is the same.

(A) In a parallel circuit,all components are connected across the same two points (nodes).
Therefore,the potential difference (voltage) across each resistor is equal to the total voltage supplied by the source.
This is a fundamental property of parallel circuits.

(A) The resistance of a material depends on its resistivity $(\rho)$.
Rubber is an insulator, meaning it has a very high resistivity, which prevents the flow of electric current.
Copper is a conductor, meaning it has a very low resistivity, which allows electric current to flow easily.
Since resistance $R = \rho \frac{l}{A}$, for pieces of the same dimensions (length $l$ and area $A$), the material with higher resistivity will have higher resistance.
Therefore, the resistance of rubber is significantly higher than that of copper.

(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 definition of electric current $(I)$ is the rate of flow of electric charge $(Q)$ through a conductor.
Mathematically,current is defined as the quantity of charge flowing past a point divided by the time $(t)$ taken for that flow.
Formula: $I = Q / t$.
Therefore,the statement that current is charge multiplied by time is incorrect.

(A) cell is a device that converts chemical energy into electrical energy. Inside the cell,chemical reactions occur that create a potential difference between the two terminals. This potential difference drives the flow of electrons (charge) through the connected conducting wire,resulting in an electric current.

(A) The resistivity of pure metals is given by the relation $\rho_T = \rho_0 [1 + \alpha(T - T_0)]$,where $\alpha$ is the temperature coefficient of resistivity.
For pure metals,$\alpha$ is positive,which means that as the temperature $T$ increases,the resistivity $\rho_T$ also increases.
This happens because,with an increase in temperature,the frequency of collisions between free electrons and the vibrating ions of the metal lattice increases,leading to greater resistance to the flow of current.

(B) $Ohm's$ law states that the current flowing through a conductor is directly proportional to the potential difference across its ends,provided the temperature and other physical conditions remain constant. The mathematical expression is $V = IR$. Power is a separate concept defined as $P = VI$ or $P = I^2R$. Therefore,the statement is False.

(A) In a series circuit,all components are connected end-to-end,forming a single continuous loop for the current to flow. Therefore,there is only one path for the electrons.
In a parallel circuit,components are connected across the same two points,creating multiple branches. This allows the current to divide and flow through different paths simultaneously.

(B) The statement is False.
Resistance in a conducting wire is primarily caused by the collisions of electrons with the atoms and ions of the conductor's lattice structure as they drift through it.
While electrons do exert repulsive forces on each other,this is not the primary cause of electrical resistance in a conductor.

(D) The $SI$ unit of electric charge is the Coulomb $(C)$.
$1$ Coulomb is defined as the quantity of charge that passes through a point in a circuit when a current of $1$ Ampere flows for $1$ second.
- Ampere $(A)$ is the $SI$ unit of electric current.
- Volt $(V)$ is the $SI$ unit of electric potential difference.
- Watt $(W)$ is the $SI$ unit of power.

(C) The quantization of electric charge is given by the formula $Q = ne$,where $Q$ is the total charge,$n$ is the number of electrons,and $e$ is the elementary charge of an electron.
Given: Total charge $Q = 1.6 \ C$ and the charge of one electron $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for $n$: $n = Q / e$.
Substituting the values: $n = 1.6 \ C / (1.6 \times 10^{-19} \ C) = 1 / 10^{-19} = 10^{19}$.
Therefore,there are $10^{19}$ electrons in a charge of $1.6 \ C$.

(B) The prefix $\mu$ (micro) represents $10^{-6}$ and the prefix $m$ (milli) represents $10^{-3}$.
To convert $1\, \mu A$ to $mA$,we use the following calculation:
$1\, \mu A = 10^{-6}\, A$
$1\, mA = 10^{-3}\, A$
Therefore,$1\, \mu A = \frac{10^{-6}}{10^{-3}}\, mA = 10^{-6 - (-3)}\, mA = 10^{-3}\, mA$.

(A) Free electrons are the charge carriers responsible for electrical conductivity in materials.
Materials are classified as conductors,insulators,or semiconductors based on the availability of free electrons.
$A$. Copper is a metal and an excellent conductor of electricity because it possesses a large number of free electrons.
$B$. Glass,$C$. Rubber,and $D$. Paper are insulators,meaning they have very few or negligible free electrons,which prevents the flow of electric current.
Therefore,Copper has the highest number of free electrons among the given options.

(C) Ohm's law states that the current $(I)$ flowing through a conductor is directly proportional to the potential difference $(V)$ applied across its ends,provided the temperature and other physical conditions remain constant.
Mathematically,this is expressed as $V = IR$ or $I = V/R$.
Here,$R$ is the resistance,which is a constant for a given conductor at a constant temperature.
Since $I \propto V$,as the voltage $(V)$ increases,the current $(I)$ also increases,assuming the resistance $(R)$ remains constant.
Therefore,option $C$ is the correct statement.

(B) Electric current $(I)$ is defined as the rate of flow of electric charge $(Q)$ through a conductor.
Mathematically,it is expressed as the amount of charge flowing per unit time $(t)$.
Therefore,the formula is $I = Q / t$,where $I$ is current in Amperes,$Q$ is charge in Coulombs,and $t$ is time in seconds.

(D) The formula for electric charge $(Q)$ is given by the product of electric current $(I)$ and time $(t)$: $Q = I \times t$.
Given:
Electric current $(I)$ = $2 \ A$
Time $(t)$ = $1 \ minute = 60 \ seconds$.
Substituting the values into the formula:
$Q = 2 \ A \times 60 \ s = 120 \ C$.
Therefore,the total electric charge passing through the wire is $120 \ C$.

(C) The formula for electric current $(I)$ is given by $I = \frac{Q}{t}$,where $Q$ is the total charge and $t$ is the time.
Since $Q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \text{ C})$,the formula becomes $I = \frac{ne}{t}$.
Given: $I = 4.8 \text{ A}$,$t = 1 \text{ s}$,and $e = 1.6 \times 10^{-19} \text{ C}$.
Rearranging for $n$: $n = \frac{I \times t}{e}$.
Substituting the values: $n = \frac{4.8 \times 1}{1.6 \times 10^{-19}}$.
$n = 3 \times 10^{19}$ electrons.
Therefore,the number of electrons passing through the appliance is $3 \times 10^{19}$.

(A) Voltage $(V)$ is defined as the work done $(W)$ per unit charge $(Q)$ to move it between two points in an electric circuit.
Mathematically,$V = W / Q$.
Since electric current $(I)$ is defined as the rate of flow of charge,$I = Q / t$,which implies $Q = I \times t$.
Substituting $Q = I \times t$ into the voltage formula,we get $V = W / (I \times t)$.
Therefore,the correct formula is $\text{Work} / (\text{Current} \times \text{Time})$.

(B) Electric potential difference $(V)$ is defined as the work done $(W)$ per unit charge $(Q)$ in moving a positive charge from one point to another.
Mathematically,$V = W / Q$.
The unit of work $(W)$ is Joule $(J)$ and the unit of charge $(Q)$ is Coulomb $(C)$.
Therefore,the unit of electric potential difference is $J/C$,which is also known as the Volt $(V)$.

(C) The potential difference $(V)$ between two points is defined as the work done $(W)$ per unit charge $(Q)$ to move the charge from one point to another.
Formula: $V = \frac{W}{Q}$
Given:
Work done $(W)$ = $15 \,J$
Charge $(Q)$ = $3 \,C$
Calculation:
$V = \frac{15 \,J}{3 \,C} = 5 \,V$
Therefore,the potential difference between the two points is $5 \,V$.

(A) According to Ohm's Law,the relationship between potential difference $(V)$,current $(I)$,and resistance $(R)$ is given by $V = I \times R$.
Given:
Resistance $(R)$ = $10\, \Omega$
Potential difference $(V)$ = $1.5\, V$
To find the current $(I)$,we rearrange the formula: $I = V / R$.
Substituting the values: $I = 1.5\, V / 10\, \Omega = 0.15\, A$.
To convert the current from Amperes $(A)$ to milliamperes $(mA)$,we multiply by $1000$: $0.15\, A \times 1000 = 150\, mA$.
Therefore,the electric current flowing through the wire is $150\, mA$.

(D) The resistivity or specific resistance of a material is an intrinsic property of the material.
It does not depend on the physical dimensions of the conductor,such as its length or area of cross-section.
It depends only on the nature of the material and the temperature of the conductor.
Therefore,the correct factor is the material of the wire.