Mix Examples - Electricity Questions in English

(A) According to Ohm's Law,the electric current flowing through a metallic conductor is directly proportional to the potential difference $(V)$ applied across its ends,provided its temperature remains constant.
Mathematically,this is expressed as $V \propto I$,which leads to $V = IR$ or $\frac{V}{I} = R$.
Here,$R$ is a constant for the given metallic conductor at a given temperature and is known as its resistance.
Resistance is the property of a conductor to resist the flow of charges through it.

(B) The resistivity of a material is a fundamental property that quantifies how strongly a given material opposes the flow of electric current.
It is denoted by the Greek letter $\rho$ (rho).
The $SI$ unit of resistivity is the ohm-meter $(\Omega \cdot m)$.
In contrast,$R$ represents resistance,and $\Omega$ is the symbol for the unit ohm.

(A) Silicon $(Si)$ is a chemical element with atomic number $14$.
It belongs to group $14$ of the periodic table.
Its electrical conductivity lies between that of a conductor and an insulator.
Materials with such properties are classified as semiconductors.
Therefore,silicon is a semiconductor.

(C) semiconductor is a material that has electrical conductivity between that of a conductor (like copper) and an insulator (like glass).
Germanium $(Ge)$ and Silicon $(Si)$ are the most common examples of semiconductor elements used in electronic devices.
Zinc $(Zn)$,Copper $(Cu)$,and Silver $(Ag)$ are all metals and act as good conductors of electricity.
Therefore,the correct option is $C$.

(D) The resistivity $(\rho)$ of materials is a measure of how strongly a material opposes the flow of electric current.
Materials are classified based on their resistivity as follows:
$1$. Conductors: Have very low resistivity (e.g., metals like copper, silver).
$2$. Insulators: Have very high resistivity (e.g., glass, rubber).
$3$. Semiconductors: Have resistivity values that lie between those of conductors and insulators (e.g., silicon, germanium).
Therefore, the resistivity of semiconductors is greater than that of conductors but less than that of insulators.

(C) When resistors are connected in parallel,the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.
For three resistors $R_{1}$,$R_{2}$,and $R_{3}$ connected in parallel,the formula is:
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
To find $R$,we take the reciprocal of the sum:
$R = \frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}}$
$R = \frac{R_{1}R_{2}R_{3}}{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}$
Since the provided options were incomplete or incorrect regarding the final expression for $R$,the correct relationship is defined by the sum of reciprocals.

(D) When resistors are connected in series,the equivalent resistance $R$ is the sum of the individual resistances.
The formula for equivalent resistance in series is $R = R_1 + R_2 + R_3$.
Given values are $R_1 = 5 \Omega$,$R_2 = 10 \Omega$,and $R_3 = 15 \Omega$.
Substituting these values into the formula: $R = 5 \Omega + 10 \Omega + 15 \Omega = 30 \Omega$.
Therefore,the equivalent resistance is $30 \Omega$.

(C) When resistors $R_{1}$ and $R_{2}$ are connected in parallel,the reciprocal of the equivalent resistance $R_{eq}$ is equal to the sum of the reciprocals of the individual resistances.
The formula is given by: $\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$.
To simplify this,we find a common denominator,which is $R_{1} R_{2}$.
So,$\frac{1}{R_{eq}} = \frac{R_{2} + R_{1}}{R_{1} R_{2}}$.
Taking the reciprocal of both sides,we get the equivalent resistance: $R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$.

(C) When resistors are connected in parallel,the equivalent resistance $R$ is given by the formula: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$.
Given $R_1 = 3 \Omega$ and $R_2 = 6 \Omega$.
Substituting the values: $\frac{1}{R} = \frac{1}{3} + \frac{1}{6}$.
Finding a common denominator: $\frac{1}{R} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6}$.
Therefore,$\frac{1}{R} = \frac{1}{2}$.
Thus,$R = 2 \Omega$.

(A) In a series combination of resistors,the resistors are connected end-to-end such that the same amount of electric current $(I)$ flows through each resistor.
According to Ohm's Law $(V = IR)$,since the current $(I)$ is constant and the resistance $(R)$ of each resistor may differ,the voltage drop $(V)$ across each resistor will be different.
The equivalent resistance $(R_{eq})$ in a series circuit is the sum of individual resistances $(R_{eq} = R_1 + R_2 + R_3)$,which is always greater than the largest individual resistor.
Therefore,the statement that the current flowing through each resistor is the same is correct.

(C) In a parallel combination of resistors,the potential difference $(V)$ across each resistor remains the same.
However,the current $(I)$ divides among the resistors based on their resistance values.
The formula for equivalent resistance $(R_{eq})$ in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$.
Due to this reciprocal relationship,the equivalent resistance $(R_{eq})$ is always smaller than the smallest individual resistance in the circuit.
Therefore,option $C$ is correct.

(D) In a series combination of resistors:
$1$. The current $(I)$ flowing through each resistor remains the same.
$2$. The total voltage $(V)$ across the combination is the sum of the individual voltage drops: $V = V_1 + V_2 + ... + V_n$.
$3$. The equivalent resistance $(R_{eq})$ is the sum of the individual resistances: $R_{eq} = R_1 + R_2 + ... + R_n$.
$4$. Because $R_{eq}$ is the sum of all resistances,it is always greater than the value of the largest individual resistor in the series.
Therefore,the statement that the equivalent resistance is smaller than the smallest resistor is false.

(B) The circuit consists of two parallel branches connected between points $A$ and $B$.
Branch $1$ consists of resistors $R_1$ and $R_2$ in series: $R_{s1} = R_1 + R_2 = 3 \ \Omega + 3 \ \Omega = 6 \ \Omega$.
Branch $2$ consists of resistors $R_3$ and $R_4$ in series: $R_{s2} = R_3 + R_4 = 3 \ \Omega + 3 \ \Omega = 6 \ \Omega$.
Branch $3$ consists of resistor $R_5$ connected directly between $A$ and $B$: $R_5 = 6 \ \Omega$.
Now,these three branches ($R_{s1}$,$R_{s2}$,and $R_5$) are connected in parallel.
The equivalent resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R_{s1}} + \frac{1}{R_{s2}} + \frac{1}{R_5}$.
Substituting the values: $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.
Therefore,$R_{eq} = 2 \ \Omega$.

(A) In domestic electrical circuits,all appliances are connected in parallel to the main line.
This is because,in a parallel connection,each appliance receives the same voltage as the source.
Furthermore,if one appliance stops working or is switched off,it does not affect the operation of other appliances.
Additionally,each appliance can be operated independently with its own switch.

(D) In a series circuit,there is only one path for the flow of electric current.
If any one component (like a bulb) in the series circuit blows out or is disconnected,the circuit becomes open (broken).
Because the current cannot flow through an open circuit,the flow of electricity stops completely.
Therefore,if the third bulb blows out,the entire circuit is interrupted,and all the remaining bulbs will turn off.

(A) In a parallel circuit,each electrical appliance (or bulb) is connected across the same potential difference independently.
Since each bulb has its own separate path for the flow of current,the failure or blowing out of one bulb does not affect the operation of the other bulbs.
Therefore,if the second bulb blows out,the remaining four bulbs will continue to receive the same voltage and will remain lit.

(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.

(C) 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.
Therefore,the correct option is $C$.

(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) Solutions that conduct electricity are known as electrolytes. These substances dissociate into ions when dissolved in water,allowing the flow of electric current through the solution.

(C) The process of depositing a layer of any desired metal on another material by means of electricity is known as $Electroplating$.
In this process,an electric current is passed through an electrolyte solution containing the metal to be deposited (in this case,gold) to coat the base metal (silver).
Therefore,the correct option is $C$.

(A) In the process of electroplating,the object to be coated (in this case,the steel spoon) is connected to the negative terminal (cathode) of the battery.
The metal to be deposited (in this case,the copper rod) is connected to the positive terminal (anode) of the battery.
When the current flows,copper ions from the copper rod dissolve into the electrolyte and get deposited on the steel spoon connected to the negative terminal.
Therefore,the copper rod must be connected to the positive terminal.

(A) In the process of electroplating,the object to be coated (in this case,the iron spoon) is connected to the negative terminal (cathode) of the battery. The metal to be deposited (in this case,the chromium rod) is connected to the positive terminal (anode). When the current flows,chromium ions from the rod dissolve into the electrolyte and get deposited on the iron spoon.

(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.

(C) Given:
Electric current $(I)$ = $0.5\;A$
Time $(t)$ = $1\;hour = 60 \times 60\;seconds = 3600\;s$
Formula:
Electric charge $(Q)$ = $I \times t$
Calculation:
$Q = 0.5\;A \times 3600\;s$
$Q = 1800\;C$
Therefore, the total electric charge passing through the bulb is $1800\;C$.

(A) The work done $(W)$ in moving a charge $(Q)$ through an electric potential difference $(V)$ is given by the formula: $W = Q \times V$.
Given:
Charge $(Q)$ = $2 \, C$
Potential difference $(V)$ = $6 \, V$
Substituting the values into the formula:
$W = 2 \, C \times 6 \, V = 12 \, J$.
Therefore,the work done is $12 \, J$.

(B) According to Ohm's law,the relationship between voltage $(V)$,current $(I)$,and resistance $(R)$ is given by the formula: $V = I \times R$.
Given values are:
Current $(I)$ = $0.5\, A$
Resistance $(R)$ = $440\, \Omega$
Substituting these values into the formula:
$V = 0.5\, A \times 440\, \Omega$
$V = 220\, V$
Therefore,the electric bulb should be connected to a $220\, V$ line.

(A) According to Ohm's Law,the relationship between voltage $(V)$,current $(I)$,and resistance $(R)$ is given by the formula $V = I \times R$.
Given values are:
Voltage $(V)$ = $120\;V$
Current $(I)$ = $2\;A$
To find the resistance $(R)$,we rearrange the formula: $R = V / I$.
Substituting the values: $R = 120\;V / 2\;A = 60\;\Omega$.
Therefore,the resistance of the heater coil is $60\;\Omega$.

(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, $E = 12,000 / 1,000 = 12 \; \text{units}$.

(D) When resistors are connected in parallel,the equivalent resistance $(R_{eq})$ is given by the formula:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
Given $R_1 = 5 \Omega$,$R_2 = 10 \Omega$,and $R_3 = 30 \Omega$.
Substituting these values into the formula:
$\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30}$
To add these fractions,find the least common multiple $(LCM)$ of $5, 10,$ and $30$,which is $30$.
$\frac{1}{R_{eq}} = \frac{6}{30} + \frac{3}{30} + \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$
Therefore,$R_{eq} = 3 \Omega$.

(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 $11 \; \Omega$.

(A) In a parallel circuit,the potential difference $(V)$ across each resistor is the same.
According to Ohm's Law,$I = V / R$.
Since $V$ is constant,the current $(I)$ is inversely proportional to the resistance $(R)$: $I \propto 1 / R$.
Therefore,the resistor with the smallest resistance will have the highest current flowing through it.
Given $R_{1} = 5 \Omega$,$R_{2} = 10 \Omega$,and $R_{3} = 30 \Omega$,the smallest resistance is $R_{1} = 5 \Omega$.
Thus,the highest current flows through $R_{1}$.

(C) The formula for electric current $(I)$ is given by $I = \frac{Q}{t}$, where $Q$ is the charge and $t$ is the time in seconds.
Given: Charge $(Q) = 1800\,C$, Time $(t) = 1\,hour = 60 \times 60\,s = 3600\,s$.
Substituting the values into the formula: $I = \frac{1800}{3600} = 0.5\,A$.
Therefore, the current flowing through the bulb is $0.5\,A$.

(B) In a parallel combination of resistors,the equivalent resistance $(R_{eq})$ is given by the formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
Substituting the given values: $\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30}$.
Finding the common denominator $(30)$: $\frac{1}{R_{eq}} = \frac{6 + 3 + 1}{30} = \frac{10}{30} = \frac{1}{3}$.
Therefore,$R_{eq} = 3 \Omega$.
In a parallel circuit,the equivalent resistance is always less than the smallest individual resistance present in the circuit. Since the smallest resistance is $5 \Omega$,the equivalent resistance must be less than $5 \Omega$.

(C) The current $I$ drawn by an appliance is given by the formula $P = V \times I$,which implies $I = P / V$.
We calculate the current for each option:
$A) I = 40 \ W / 110 \ V \approx 0.36 \ A$
$B) I = 75 \ W / 220 \ V \approx 0.34 \ A$
$C) I = 60 \ W / 110 \ V \approx 0.54 \ A$
$D) I = 100 \ W / 220 \ V \approx 0.45 \ A$
Comparing these values,the appliance with the rating $110 \ V / 60 \ W$ draws the maximum current of approximately $0.54 \ A$.

(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) $1$. Analyze the observations: The bulb glows $(ON)$ only if both materials connected at $X$ and $Y$ are conductors. If the bulb is $OFF$,it means at least one of the materials is an insulator.
$2$. From observation $2$,$I$ and $IV$ result in $ON$,so both $I$ and $IV$ are conductors.
$3$. From observation $4$,$III$ and $V$ result in $ON$,so both $III$ and $V$ are conductors.
$4$. From observation $1$,$I$ and $II$ result in $OFF$. Since $I$ is a conductor,$II$ must be an insulator.
$5$. For the third row,the bulb is $OFF$. This means at least one of the materials must be an insulator. Since $II$ is the only identified insulator,the pair must contain $II$.
$6$. Looking at the options,option $A$ ($II$ and $III$) contains $II$,which is an insulator. Therefore,the bulb will remain $OFF$.