Mix Examples - Electricity Questions in English

(A) When a wire of resistance $R$ is cut into $n$ equal pieces,the resistance of each piece becomes $R' = R/n$.
Here,the original resistance $R = 5\, \Omega$ and the number of pieces $n = 5$.
Therefore,the resistance of each piece $R' = 5\, \Omega / 5 = 1\, \Omega$.
When $n$ resistors of equal resistance $R'$ are connected in parallel,the equivalent resistance $R_{eq}$ is given by $R_{eq} = R'/n$.
Substituting the values,$R_{eq} = 1\, \Omega / 5 = 0.2\, \Omega$.

(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.
Rearranging the formula for resistivity,we get $\rho = R \cdot \frac{A}{L}$.
The $SI$ unit of resistance $R$ is Ohm $(\Omega)$,the unit of area $A$ is square meter $(m^2)$,and the unit of length $L$ is meter $(m)$.
Substituting these units into the formula: $\text{Unit of } \rho = \Omega \cdot \frac{m^2}{m} = \Omega \cdot m$.
Therefore,the $SI$ unit of resistivity is $\Omega \cdot m$.

(D) In the given circuit,all the resistors are connected in series.
When resistors are connected in series,the equivalent resistance $R_{eq}$ is the sum of individual resistances.
$R_{eq} = R_1 + R_2 + R_3 + R_4 + R_5$
Here,there are $5$ resistors,each of $2 \, \Omega$.
$R_{eq} = 2 \, \Omega + 2 \, \Omega + 2 \, \Omega + 2 \, \Omega + 2 \, \Omega = 10 \, \Omega$.
Therefore,the equivalent resistance between points $A$ and $B$ is $10 \, \Omega$.

(A) In the given circuit,there are two branches connected between points $A$ and $B$.
Branch $1$ consists of two resistors of $4 \, \Omega$ each,connected in series. The equivalent resistance of this branch is $R_1 = 4 \, \Omega + 4 \, \Omega = 8 \, \Omega$.
Branch $2$ consists of a single resistor of $8 \, \Omega$ connected directly between $A$ and $B$.
Now,these two branches ($R_1 = 8 \, \Omega$ and $R_2 = 8 \, \Omega$) are connected in parallel.
The equivalent resistance $R_{eq}$ is given by the formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.
Substituting the values: $\frac{1}{R_{eq}} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.
Therefore,$R_{eq} = 4 \, \Omega$.

(C) $1$. Identify the parallel combination: Two resistors of $5 \, \Omega$ each are connected in parallel. The equivalent resistance $(R_p)$ of this parallel part is given by $\frac{1}{R_p} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$,so $R_p = 2.5 \, \Omega$.
$2$. Identify the series combination: The circuit now consists of three resistors in series: the equivalent resistance $R_p = 2.5 \, \Omega$,and two other resistors of $5 \, \Omega$ each.
$3$. Calculate total resistance: The total equivalent resistance $(R_{eq})$ is the sum of these resistances in series: $R_{eq} = 2.5 \, \Omega + 5 \, \Omega + 5 \, \Omega = 12.5 \, \Omega$.

(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) An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent,such as water.
When an electric potential is applied across an electrolyte,the dissolved solute dissociates into charged particles called ions.
The positive ions (cations) move towards the cathode (negative electrode),and the negative ions (anions) move towards the anode (positive electrode).
Therefore,the flow of electric current through an electrolyte is due to the movement of both positive and negative ions.

(B) Distilled water is pure water $(H_2O)$ and does not contain dissolved salts or ions.
Since electricity requires free ions or electrons to flow through a medium,the absence of these charge carriers makes distilled water a poor conductor of electricity.
Therefore,distilled water acts as an insulator.

(A) proton is a subatomic particle found in the nucleus of an atom.
It carries a fundamental positive electric charge of approximately $+1.602 \times 10^{-19} \ C$.
Therefore,the correct option is $A$.

(C) An electron is a subatomic particle that carries a fundamental negative electric charge. By convention,the charge of an electron is defined as $-1.602 \times 10^{-19} \ C$.

(A) The elementary charge,denoted by $e$,is the electric charge carried by a single proton or,equivalently,the magnitude of the negative electric charge carried by a single electron.
It is a fundamental physical constant.
The value of the charge on an electron is approximately $1.602 \times 10^{-19} \ C$.
Therefore,the correct magnitude is $1.6 \times 10^{-19} \ C$.

(B) 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: $Q = 1 \ C$ and $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for $n$: $n = Q / e$.
Substituting the values: $n = 1 / (1.6 \times 10^{-19})$.
$n = 10^{19} / 1.6$.
$n = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,$6.25 \times 10^{18}$ electrons constitute a total charge of $1 \ C$.

(C) The charge of a single proton is $e = 1.6 \times 10^{-19} \ C$.
To find the total charge $(Q)$ on $n = 100$ protons,we use the formula $Q = n \times e$.
Substituting the values: $Q = 100 \times (1.6 \times 10^{-19} \ C)$.
$Q = 10^2 \times 1.6 \times 10^{-19} \ C$.
$Q = 1.6 \times 10^{-17} \ C$.
Therefore,the total electric charge is $1.6 \times 10^{-17} \ C$.

(D) Free electrons are the charge carriers responsible for the flow of electric current in conductors.
Metals like $Copper$,$Silver$,and $Aluminum$ are good conductors of electricity because they possess a large number of free electrons in their atomic structure.
$Plastic$ is an insulator (non-conductor) of electricity.
In insulators,electrons are tightly bound to their respective atoms and are not free to move,which prevents the flow of electric current.
Therefore,$Plastic$ does not contain free electrons.

(A) The electric force between two stationary point charges is determined by Coulomb's law.
According to Coulomb's law,the magnitude of the electrostatic force $(F)$ between two point charges ($q_1$ and $q_2$) separated by a distance $(r)$ is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The formula is given by: $F = k \cdot \frac{|q_1 \cdot q_2|}{r^2}$,where $k$ is Coulomb's constant.

(B) The electron was discovered by $J.J. Thomson$ in $1897$ during his experiments with cathode ray tubes. He identified these particles as fundamental constituents of atoms.

(B) Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor.
Mathematically,it is expressed as $I = Q / t$,where $I$ is the electric current,$Q$ is the net charge,and $t$ is the time taken.
The $SI$ unit of electric current is the Ampere $(A)$.

(D) The $SI$ unit of electric current is the $Ampere$ $(A)$.
Electric current is defined as the rate of flow of electric charge through a conductor.
One $Ampere$ is equivalent to one $Coulomb$ of charge flowing per second $(1 \ A = 1 \ C/s)$.

(A) Electric current $(I)$ is defined as the rate of flow of electric charge $(Q)$ through a conductor.
Mathematically,$I = Q / t$.
The $SI$ unit of charge $(Q)$ is the coulomb $(C)$ and the $SI$ unit of time $(t)$ is the second $(s)$.
Therefore,the unit of electric current is $Coulomb/second$ $(C/s)$,which is defined as an ampere $(A)$.

(D) The prefix $m$ stands for milli,which represents a factor of $10^{-3}$.
Therefore,$1\, mA = 1 \times 10^{-3}\, A$.
Thus,$1\, mA = 10^{-3}\, A$.

(D) The prefix $\mu$ (micro) represents a factor of $10^{-6}$.
Therefore,$1 \, \mu A$ is equal to $10^{-6} \, A$.

(D) We know that $1 \text{ milliampere } (mA) = 10^{-3} \text{ amperes } (A)$.
We also know that $1 \text{ microampere } (\mu A) = 10^{-6} \text{ amperes } (A)$.
To find the value of $1 \text{ mA}$ in terms of $\mu A$,we set up the equation:
$1 \text{ mA} = x \times 1 \mu A$
$10^{-3} \text{ A} = x \times 10^{-6} \text{ A}$
$x = \frac{10^{-3}}{10^{-6}}$
$x = 10^{-3 - (-6)} = 10^{-3 + 6} = 10^{3}$
Therefore,$1 \text{ mA} = 10^{3} \mu A$.

(B) The formula for electric current is $I = Q / t$,where $I$ is current,$Q$ is total charge,and $t$ is time.
Given $I = 1 \text{ A}$ and $t = 1 \text{ s}$,the total charge $Q = I \times t = 1 \text{ A} \times 1 \text{ s} = 1 \text{ C}$.
The quantization of charge is given by $Q = n \times e$,where $n$ is the number of electrons and $e$ is the elementary charge $(e \approx 1.6 \times 10^{-19} \text{ C})$.
To find $n$,we rearrange the formula: $n = Q / e$.
Substituting the values: $n = 1 \text{ C} / (1.6 \times 10^{-19} \text{ C}) = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,$6.25 \times 10^{18}$ electrons pass through the conductor per second.

(C) An $Ammeter$ is an instrument used to measure the electric current in a circuit. It is always connected in series with the component through which the current is to be measured. $A$ $Voltmeter$ measures potential difference,a $Galvanometer$ detects small amounts of current,and a $Wattmeter$ measures electric power.

(C) By convention,the direction of electric current is taken as the direction of flow of positive charges.
Since electrons are negatively charged particles,they flow from the negative terminal to the positive terminal of a source.
Therefore,the direction of conventional current is considered to be opposite to the direction of the flow of electrons.

(C) The electric potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit charge from one point to the other.
Its $SI$ unit is the volt $(V)$,named after the Italian physicist Alessandro Volta.
$1 \text{ volt} = 1 \text{ joule} / 1 \text{ coulomb}$.
Therefore,the correct unit is the volt.

(B) The electric potential difference $(V)$ between two points is defined as the amount of work done $(W)$ in moving a unit positive charge $(Q)$ 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 $Joule/Coulomb$ $(J/C)$,which is also known as Volt $(V)$.

(C) The $SI$ unit of electric potential (or potential difference) is $volt$ $(V)$.
Electric current is measured in $ampere$ $(A)$.
Resistance is measured in $ohm$ ($\Omega$).
Electric energy is measured in $joule$ $(J)$ or $kilowatt-hour$ $(kWh)$.
Therefore, the correct option is $C$.

(B) The device used to measure the electric potential difference between two points in an electric circuit is called a $Voltmeter$.
It is always connected in parallel across the points where the potential difference is to be measured.
An $Ammeter$ is used to measure electric current.
$A$ $Galvanometer$ is used to detect the presence of small electric currents.
$A$ $Voltameter$ is an electrolytic cell used for measuring the quantity of electricity by the amount of chemical action produced.

(A) The simple battery,known as the voltaic pile,was invented by the Italian physicist Alessandro Volta in $1800$.
It was the first electrical battery that could provide a continuous current to a circuit.
Therefore,the correct option is $A$.

(D) Voltaic cell (or simple galvanic cell) consists of two electrodes immersed in an electrolyte.
In this cell,the zinc plate acts as the negative electrode (anode) and the copper plate acts as the positive electrode (cathode).
Therefore,the positive electrode is made of copper.

(A) Voltaic cell (or simple galvanic cell) consists of two different metal electrodes,typically zinc and copper,immersed in an electrolyte solution.
In the original design by Alessandro Volta,the electrolyte used is dilute sulfuric acid $(H_2SO_4)$.
The zinc electrode acts as the anode (negative terminal) and the copper electrode acts as the cathode (positive terminal).

(C) Voltaic cell (also known as a Galvanic cell) is an electrochemical device that generates electrical energy from spontaneous chemical reactions occurring within it.
In this cell,redox reactions (oxidation and reduction) take place at the electrodes.
The chemical potential energy stored in the reactants is converted into electrical energy,which can then be used to power an external circuit.

(B) In a Voltaic cell (or any electrochemical cell),the chemical reactions occur at the electrodes.
In the external circuit,electrons flow from the anode (negative electrode) to the cathode (positive electrode).
This flow of electrons constitutes the electric current,which is conventionally defined as flowing from the positive terminal to the negative terminal (opposite to the direction of electron flow).
Therefore,the correct direction of electron flow is from the negative electrode to the positive electrode.

(A) In an electric circuit,the symbol for an electric cell consists of two parallel vertical lines of unequal length. The longer line represents the positive terminal $(+)$,and the shorter,thicker line represents the negative terminal $(-)$.
Option $A$ correctly represents this symbol.

(C) In electrical circuit diagrams,different components are represented by standard symbols.
- The symbol shown in option $A$ represents an open switch.
- The symbol shown in option $B$ represents a cell.
- The symbol shown in option $C$ represents a resistor,which is a component used to provide resistance in a circuit.
- The symbol shown in option $D$ represents a galvanometer,used to detect small electric currents.
Therefore,the correct symbol for a resistor is given in option $C$.

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

(B) 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 remains constant.
Mathematically,this is expressed as $V = IR$ or $I = V/R$.
Since $V = IR$,the relationship between $V$ and $I$ is linear,meaning the graph of $I$ versus $V$ (or $V$ versus $I$) is a straight line passing through the origin.
Resistance $(R)$ is a constant property of the conductor at a given temperature and does not change with changes in $V$ or $I$.

(C) Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across its ends,provided the temperature remains constant.
Mathematically,this is expressed as $V \propto I$,which leads to $V = IR$,where $R$ is the resistance of the conductor.
From this formula,we can also derive $R = \frac{V}{I}$ and $I = \frac{V}{R}$.
Option $A$ $(R = \frac{V}{I})$ is a common representation of the definition of resistance derived from Ohm's law,but $V = IR$ is the standard form of the law.

(C) The unit of electrical resistance is the $ohm$ $(\Omega)$.
According to Ohm's Law, $V = IR$, where $V$ is the potential difference in $volts$, $I$ is the current in $amperes$, and $R$ is the resistance in $ohms$.
Therefore, $R = V/I$, which gives the unit as $volt/ampere$, defined as $ohm$.

(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.

(D) The unit of electrical resistance is the ohm.
It is denoted by the Greek letter omega,represented by the symbol $\Omega$.
$R$ is the symbol for resistance itself.
$I$ is the symbol for electric current.
$\rho$ (rho) is the symbol for electrical resistivity.

(D) Insulators are materials that do not allow the flow of electric current through them.
This is because they have a very high electrical resistance,which prevents the movement of free electrons.
Therefore,the resistance of insulating materials is very large compared to conductors.

(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.

(A) Resistors used in electronic devices like radios and televisions are typically carbon resistors.
These are made from a mixture of carbon and graphite,which acts as the resistive material,often bound with a ceramic or plastic material.
Carbon resistors are preferred in these applications due to their small size,low cost,and ability to handle high-frequency signals effectively.

(D) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where:
$1$. $\rho$ (rho) is the resistivity,which depends on the nature of the material.
$2$. $l$ is the length of the conductor.
$3$. $A$ is the area of cross-section.
Since the resistance depends on all these three factors,the correct option is $D$.

(C) The resistance $R$ of a conducting 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 of the wire.
From this relation,it is clear that resistance is inversely proportional to the cross-sectional area $(R \propto \frac{1}{A})$.
If the area $A$ is doubled $(A' = 2A)$,the new resistance $R'$ becomes $R' = \rho \frac{L}{2A} = \frac{1}{2} R$.
Therefore,the resistance becomes half of its original value.

(D) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.
Since the wire is cylindrical,the cross-sectional area $A = \pi r^2$,where $r$ is the radius.
Since the diameter $D = 2r$,we can write $A = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Substituting this into the resistance formula: $R = \rho \frac{L}{\pi D^2 / 4} = \frac{4 \rho L}{\pi D^2}$.
This shows that $R \propto \frac{1}{D^2}$.
If the diameter $D$ is doubled $(D' = 2D)$,the new resistance $R'$ becomes $R' = \frac{4 \rho L}{\pi (2D)^2} = \frac{4 \rho L}{4 \pi D^2} = \frac{1}{4} R$.
Therefore,the resistance becomes one-fourth of its original value.