Mix Examples - Electricity Questions in English

(N/A) Based on standard circuit symbols:
$A$ represents an electric bulb,which is a load in the circuit.
$B$ represents a rheostat (variable resistor),used to control the current in the circuit.
$C$ represents an electric cell,which acts as the source of potential difference.
$D$ represents an ammeter,which is used to measure the electric current flowing through the circuit.

(N/A) The heat produced $(H)$ in a current-carrying conductor is given by Joule's Law of Heating: $H = I^2Rt$.
$(i)$ The factors on which the heat produced depends are:
$(a)$ The square of the current flowing through the conductor $(I^2)$.
$(b)$ The resistance of the conductor $(R)$.
$(c)$ The time for which the current flows through the conductor $(t)$.
$(ii)$ $A$ practical application of this heating effect is the electric iron,which converts electrical energy into heat energy to press clothes.

(N/A) In metals,free electrons are in a state of random thermal motion. In the absence of an external electric field,the average velocity of these electrons in any given direction is zero,meaning there is no net flow of charge. Current is defined as the rate of flow of charge. For a net flow to occur,an external electric potential difference must be applied across the conductor. This potential difference creates an electric field that exerts a force on the electrons,causing them to drift in a specific direction,thereby establishing an electric current.

(0.2 A, 2.2 A) According to Ohm's law,the current $I$ is given by the formula $I = \frac{V}{R}$.
For the electric bulb:
Given,$V = 220 \ V$ and $R = 1100 \ \Omega$.
$I = \frac{220}{1100} = 0.2 \ A$.
For the electric heater:
Given,$V = 220 \ V$ and $R = 100 \ \Omega$.
$I = \frac{220}{100} = 2.2 \ A$.

(B) Wire $Y$ has greater resistance.
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.
Since both wires have the same cross-sectional area $(A)$ and are made of the same material (same $\rho$),the resistance is directly proportional to the length $(l)$ of the wire $(R \propto l)$.
As the length of wire $Y$ $(2l)$ is greater than the length of wire $X$ $(l)$,wire $Y$ has greater resistance.

(A) In a parallel combination of resistors, the equivalent resistance is always less than the smallest individual resistance in the circuit.
Since the smallest resistor is $2 \, \Omega$, the total resistance must be less than $2 \, \Omega$.
This can be calculated using the formula for parallel resistors:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
$\frac{1}{R_p} = \frac{15 + 10 + 6}{30} = \frac{31}{30}$
$R_p = \frac{30}{31} \approx 0.967 \, \Omega$
Thus, the total resistance is approximately $0.967 \, \Omega$, which is less than $2 \, \Omega$.

(A) In a series circuit,the current $(I)$ flowing through all components is the same.
According to Ohm's Law,the potential difference $(V)$ across a resistor is given by $V = I \times R$.
Since $I$ is constant,$V$ is directly proportional to the resistance $(R)$: $V \propto R$.
The power rating of a bulb is given by $P = \frac{V^2}{R}$,which implies $R = \frac{V^2}{P}$.
For a fixed voltage rating,the resistance is inversely proportional to the power rating $(R \propto \frac{1}{P})$.
Therefore,the bulb with the lowest wattage $(40\, W)$ has the highest resistance,and the bulb with the highest wattage $(100\, W)$ has the lowest resistance.
Consequently,the potential difference is highest across the $40\, W$ bulb and lowest across the $100\, W$ bulb.

(D) The resistance $R$ of a cylindrical conductor depends on the following factors:
$(i)$ Length $(l)$: Resistance is directly proportional to the length $(R \propto l)$.
$(ii)$ Area of cross-section $(A)$: Resistance is inversely proportional to the area of cross-section $(R \propto 1/A)$.
$(iii)$ Nature of the material (resistivity,$\rho$).
$(iv)$ Temperature.
When a conductor is stretched to double its length,its volume remains constant. If the original length is $l$ and the new length is $l' = 2l$,then the new area of cross-section $A'$ will be $A/2$ because $V = A \times l = A' \times l'$.
The new resistance $R'$ is given by:
$R' = \rho \times (l' / A') = \rho \times (2l / (A/2)) = 4 \times (\rho \times l / A) = 4R$.
Thus,the resistance becomes $4$ times the original resistance.

(N/A) The filament of an electric bulb is made of $Tungsten$.
$(b)$ The characteristics that make $Tungsten$ suitable for this purpose are:
$1$. It has a very high melting point (approximately $3380^{\circ}C$),which prevents it from melting even when it becomes white-hot during the emission of light.
$2$. It has high resistivity,which allows it to produce a large amount of heat when an electric current passes through it,enabling it to glow brightly.

(N/A) An ammeter is always connected in series with the circuit components. It measures the electric current flowing through the circuit.
$A$ voltmeter is always connected in parallel across the points between which the potential difference is to be measured. It measures the potential difference across a conductor.

(C) Sample $C$ is suitable for heating elements of electrical appliances.
Reason: Heating elements require materials with high resistivity so that they can produce a significant amount of heat when an electric current passes through them, according to Joule's Law of heating $(H = I^2Rt)$. Among the given samples, Sample $C$ has the highest resistivity $(100 \times 10^{-6} \Omega \cdot m)$, making it the most efficient choice for generating heat.

(N/A) Tungsten is used for making the filament of electric lamps because:
$(i)$ It has a very high melting point (approximately $3380^{\circ}C$),which allows it to glow at high temperatures without melting.
$(ii)$ It has high resistivity,which allows it to produce a large amount of heat when current passes through it.
$(b)$ Heating elements are made of alloys because:
$(i)$ Alloys have higher resistivity than pure metals,which generates more heat.
$(ii)$ Alloys do not oxidize (burn) easily even at high temperatures,which increases the longevity of the appliance.

(N/A) $(i)$ Since the bulbs are connected in parallel,the potential difference across each bulb remains the same $(4.5 \, V)$. Therefore,if bulb $B_{1}$ fuses,the other two bulbs ($B_{2}$ and $B_{3}$) will continue to glow with the same intensity.
$(ii)$ Initially,the total current $I = 3 \, A$ is divided equally among the three identical bulbs in parallel,so each bulb draws $I_{1} = I_{2} = I_{3} = 1 \, A$.
When bulb $B_{2}$ fuses,it creates an open circuit in its branch,so the current through it becomes zero. Thus,the reading of $A_{2}$ becomes $0 \, A$.
Since $B_{1}$ and $B_{3}$ remain connected in parallel,they still experience the same potential difference and draw $1 \, A$ each. Therefore,$A_{1}$ shows $1 \, A$ and $A_{3}$ shows $1 \, A$.
The total current $A$ recorded by the main ammeter will be the sum of the currents in the remaining branches: $A = 1 \, A + 1 \, A = 2 \, A$.

(C) The wires of the connecting cord of an electric heater are typically made of copper $(Cu)$,which has extremely low electrical resistance. According to Joule's law of heating,$H = I^2Rt$,the heat produced is directly proportional to the resistance $(R)$. Since the resistance of copper wires is negligible,very little heat is produced,and they do not glow.
Conversely,the heating element is made of an alloy like nichrome,which has a very high electrical resistance. Due to this high resistance,a large amount of heat is generated when current flows through it,causing it to glow red-hot.

(B) Ammeter $A_{2}$ will show a higher reading.
According to the formula for resistance,$R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the area of cross-section.
Since both wires have the same material ($\rho$ is constant) and the same area of cross-section,the resistance is directly proportional to the length $(R \propto l)$.
Given that wire $A$ is longer than wire $B$,the resistance of wire $A$ $(R_{A})$ is greater than the resistance of wire $B$ $(R_{B})$.
According to Ohm's law,$I = \frac{V}{R}$. Since both resistors are connected in parallel,the potential difference $(V)$ across them is the same.
Therefore,current is inversely proportional to resistance $(I \propto \frac{1}{R})$.
Since $R_{A} > R_{B}$,the current flowing through wire $A$ $(I_{A})$ will be less than the current flowing through wire $B$ $(I_{B})$.
Thus,ammeter $A_{2}$ (connected in series with wire $B$) will show a higher reading than ammeter $A_{1}$ (connected in series with wire $A$).

(N/A) The total resistance of the circuit is $R = R_1 + R_2 + R_3 = 1 \Omega + 3 \Omega + 2 \Omega = 6 \Omega$.
The total current in the circuit is $I = \frac{V}{R} = \frac{3 \text{ V}}{6 \Omega} = 0.5 \text{ A}$.
When the voltmeter is connected between $B$ and $C$,it measures the potential difference across the $3 \Omega$ resistor.
The potential difference across the $3 \Omega$ resistor is $V_{BC} = I \times R_{BC} = 0.5 \text{ A} \times 3 \Omega = 1.5 \text{ V}$.
Initially,the voltmeter was connected across the $1 \Omega$ resistor,measuring $V_{AB} = I \times R_{AB} = 0.5 \text{ A} \times 1 \Omega = 0.5 \text{ V}$.
Therefore,the reading of the voltmeter will increase from $0.5 \text{ V}$ to $1.5 \text{ V}$.

(A) Ammeter $A_{1}$ will show a higher reading.
Resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A_{c}}$,where $\rho$ is resistivity,$l$ is length,and $A_{c}$ is the cross-sectional area.
Since wires $A$ and $B$ are of the same material and length,$\rho$ and $l$ are constant.
Therefore,$R \propto \frac{1}{A_{c}}$.
Since wire $A$ is thicker than wire $B$,it has a larger cross-sectional area $(A_{c})$,which means it has lower resistance.
According to Ohm's law,$I = \frac{V}{R}$. Since both wires are connected in parallel,the potential difference $V$ across them is the same.
Because wire $A$ has lower resistance,the current $I$ flowing through it will be higher.
Thus,ammeter $A_{1}$ will show a higher reading.

(N/A) $1$. Tungsten has a very high melting point (approximately $3380 ^\circ C$), which allows it to glow at high temperatures without melting.
$2$. It has a high resistance, which allows it to produce a large amount of heat and light when an electric current passes through it.

(C) We know that for a $V-I$ graph,the slope represents the resistance $(R = V/I)$.
In a series combination,the equivalent resistance $(R_s = R_1 + R_2)$ is always greater than the individual resistances ($R_1$ and $R_2$).
Therefore,the graph with the highest slope represents the series combination.
In the given graph,line $C$ has the maximum slope compared to lines $A$ and $B$.
Thus,$C$ represents the series combination of the other two resistors.

(N/A) Charges can move in a conductor only if there is a potential difference across its two ends. This potential difference is necessary to create an electric field that exerts a force on the charges,causing them to flow. The device used to maintain this potential difference in an electric circuit is an electric cell or a battery (a combination of one or more cells).

(N/A) The electrical fuse works on the heating effect of electric current. When the current exceeds the safe limit,the heat generated melts the fuse wire,breaking the circuit.
$(b)$ An electrical fuse is always connected in series with the live wire of the domestic circuit to protect appliances from high current surges.

(A) We know that $V = IR$,which implies $R = V/I$. The slope of the $V-I$ graph represents the resistance $R$ of the wire.
From the given graph,the slope of line $A$ is greater than the slope of line $B$,therefore,the resistance of wire $A$ $(R_A)$ is greater than the resistance of wire $B$ $(R_B)$.
We know that 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 made of the same material,their resistivity $\rho$ is the same. Since they are of equal thickness,their cross-sectional area $A$ is also the same.
Thus,$R \propto l$. Since $R_A > R_B$,it follows that the length of wire $A$ is greater than the length of wire $B$.

(N/A) Electrical resistance is the property of a conductor to oppose the flow of electric current through it.
$(i)$ When a low current passes through a conductor for a short duration, the heat generated $(H = I^2Rt)$ is negligible. Therefore, there is no significant rise in temperature, and the resistance remains effectively unchanged.
$(ii)$ When a heavy current passes through a conductor for about $30$ seconds, a significant amount of heat is produced. This leads to a considerable rise in the temperature of the conductor. Since the resistance of most metallic conductors increases with an increase in temperature, the resistance of the conductor will increase.

(A) The $SI$ unit of electric current is the ampere $(A)$.
One ampere is defined as the flow of $1$ coulomb of charge per second through a conductor $(1 \text{ A} = 1 \text{ C/s})$.
To calculate the number of electrons $(n)$:
Given: Current $(I)$ = $1 \text{ A}$,Time $(t)$ = $1 \text{ s}$,Charge on an electron $(e)$ = $1.6 \times 10^{-19} \text{ C}$.
Using the formula $Q = I \times t$ and $Q = n \times e$,we get $n = \frac{I \times t}{e}$.
$n = \frac{1 \text{ A} \times 1 \text{ s}}{1.6 \times 10^{-19} \text{ C}} = \frac{1}{1.6} \times 10^{19} = 6.25 \times 10^{18}$.
Thus,$6.25 \times 10^{18}$ electrons flow through the conductor in $1$ second.

(A) In a series combination,all components are connected in a single path. If the filament of one lamp breaks,the circuit becomes open,and the flow of current is interrupted throughout the entire circuit. Therefore,the other two lamps will stop glowing.
$(b)$ In a parallel combination,each lamp is connected in a separate branch across the power source. If the filament of one lamp breaks,it only affects that specific branch. The other two lamps remain connected to the voltage source and will continue to glow normally.

(N/A) Based on the provided $V-I$ graph,we can observe the relationship between voltage $(V)$ and current $(I)$. The table below shows the values of $I$ corresponding to four different values of $V$:

$V$ (volts)	$2$	$4$	$6$	$8$
$I$ (amperes)	$1$	$2$	$3$	$4$

From the graph,when $V = 10 \text{ volts}$,the corresponding value of current $I$ is $5 \text{ amperes}$.
To determine the resistance $(R)$ of the resistor,we use Ohm's Law,which states $R = \frac{V}{I}$. The resistance is equal to the reciprocal of the slope of the $I-V$ graph,or the slope of the $V-I$ graph itself. Using the values from the graph:
$R = \frac{V}{I} = \frac{10 \text{ V}}{5 \text{ A}} = 2 \ \Omega$.

(A) Given: Potential difference $(V) = 20 \ V$,Work $(W) = 40 \ J$.
We know that the work done in moving a charge $(Q)$ between two points is given by the formula:
$W = Q \times V$
Substituting the given values:
$40 \ J = Q \times 20 \ V$
$Q = \frac{40 \ J}{20 \ V} = 2 \ C$
We also know that the total charge $(Q)$ is related to the number of electrons $(n)$ by the relation:
$Q = n \times e$
where $e$ is the elementary charge,$e = 1.6 \times 10^{-19} \ C$.
Therefore,the number of electrons $(n)$ is:
$n = \frac{Q}{e} = \frac{2 \ C}{1.6 \times 10^{-19} \ C} = 1.25 \times 10^{19}$

(A) $(i)$ The $SI$ unit of resistance is $Ohm (\Omega)$.
$(ii)$ One $Ohm$ is defined as the resistance of a conductor such that when a potential difference of $1\, V$ is applied across its ends, a current of $1\, A$ flows through it.
$(iii)$ Given:
$I = 200\, mA = 0.2\, A$
$V = 0.8\, V$
Using Ohm's Law, $V = IR$:
$R = \frac{V}{I}$
$R = \frac{0.8\, V}{0.2\, A}$
$R = 4\, \Omega$

(N/A) The flow of electric current in a conductor is due to the motion of electrons. These electrons are not completely free to move; they collide with the atoms and ions of the conductor as they drift. These collisions and the electrostatic attraction exerted by the atoms hinder the motion of electrons,which manifests as resistance.
$(b)$ Conductor: $A$ material that allows electric current to flow through it easily due to the presence of free electrons.
Resistor: $A$ component or conductor that has a specific,significant amount of resistance,used to control or limit the flow of current in a circuit.
Resistance: It is the inherent property of a material by virtue of which it opposes the flow of electric current through it.

(D) Given: Resistance $(R) = 6 \, \Omega$,Potential difference $(V) = 12 \, V$.
Using Ohm's law: $V = IR$.
Therefore,current $(I) = \frac{V}{R} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.
For the second part,the original resistance is $R = \rho \frac{l}{A}$.
When the wire is stretched to double its length,the new length $l' = 2l$. Since the volume of the wire remains constant,$V = A \times l = A' \times l'$.
Thus,$A' = \frac{A \times l}{l'} = \frac{A \times l}{2l} = \frac{A}{2}$.
The new resistance $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right) = 4R$.
Substituting the original resistance value: $R' = 4 \times 6 \, \Omega = 24 \, \Omega$.

(A) $R = \rho \frac{L}{A}$
Where $R$ is the resistance and $\rho$ is the resistivity of the material.
Initial length of the wire $= L$. The new length is $L' = 2L$ and the new area of cross-section is $A' = A/2$.
$(a)$ The new resistance $R'$ is given by:
$R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left( \rho \frac{L}{A} \right) = 4R$.
Thus,the resistance becomes $4$ times the original resistance.
$(b)$ Resistivity $(\rho)$ is an intrinsic property of the material and depends only on the nature of the material and temperature. Therefore,it remains unchanged.

(A) The resistance $(R)$ 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 area of cross-section.
For the first wire,$R_1 = 10\, \Omega = \rho \frac{l}{A}$.
For the second wire,the length $l' = 3l$ and the area $A' = 4A$.
The resistance of the second wire is $R_2 = \rho \frac{l'}{A'} = \rho \frac{3l}{4A}$.
Substituting the value of $R_1$ into the equation,we get $R_2 = \frac{3}{4} \times R_1$.
$R_2 = \frac{3}{4} \times 10\, \Omega = 7.5\, \Omega$.

(A) Separately: Current $I = \frac{V}{R} = \frac{220 \ V}{22 \ \Omega} = 10 \ A$.
$(b)$ In series: Total resistance $R_s = R_1 + R_2 = 22 \ \Omega + 22 \ \Omega = 44 \ \Omega$.
Therefore,current $I = \frac{V}{R_s} = \frac{220 \ V}{44 \ \Omega} = 5 \ A$.
$(c)$ In parallel: Total resistance $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{22} + \frac{1}{22} = \frac{2}{22} = \frac{1}{11} \ \Omega^{-1}$.
Therefore,$R_p = 11 \ \Omega$.
Thus,current $I = \frac{V}{R_p} = \frac{220 \ V}{11 \ \Omega} = 20 \ A$.

$(7.5 \, \Omega)$ The circuit consists of resistors in series and parallel combinations.
First, consider the two $2 \, \Omega$ resistors connected in parallel. Let their combined resistance be $R_{P1}$.
$\frac{1}{R_{P1}} = \frac{1}{2} + \frac{1}{2} = 1 \implies R_{P1} = 1 \, \Omega$.
Next, consider the two $1 \, \Omega$ resistors connected in parallel. Let their combined resistance be $R_{P2}$.
$\frac{1}{R_{P2}} = \frac{1}{1} + \frac{1}{1} = 2 \implies R_{P2} = 0.5 \, \Omega$.
Finally, the total equivalent resistance $R$ of the circuit is the sum of the series components:
$R = 3 \, \Omega + 3 \, \Omega + R_{P1} + R_{P2}$
$R = 6 \, \Omega + 1 \, \Omega + 0.5 \, \Omega = 7.5 \, \Omega$.

(D) The power $P$ and voltage $V$ are given for both lamps. The current $I$ drawn by each lamp can be calculated using the formula $P = VI$,or $I = P/V$.
For the first lamp,$I_1 = P_1 / V = 100 \, W / 220 \, V = 10/22 \, A = 5/11 \, A$.
For the second lamp,$I_2 = P_2 / V = 60 \, W / 220 \, V = 6/22 \, A = 3/11 \, A$.
Since the lamps are connected in parallel,the total current $I$ drawn from the supply is the sum of the individual currents:
$I = I_1 + I_2 = 5/11 \, A + 3/11 \, A = 8/11 \, A$.
$I \approx 0.727 \, A$.

(N/A) Given: Power, $P = 2 kW = 2000 W$, Time, $t = 2 h$ daily.
Energy consumed per day, $E = P \times t = 2 kW \times 2 h = 4 kWh$.
In $SI$ units (Joules): $4 kWh = 4 \times 3.6 \times 10^6 J = 14.4 \times 10^6 J$.
In commercial units: $4 kWh$ (or $4$ units).
$(b)$ Total energy consumed in June ($30$ days):
Total Energy $= 4 kWh/day \times 30 days = 120 kWh$.
Cost of electricity $= 120 \text{ units} \times ₹ 3.00/unit = ₹ 360$.

(A) $(i)$ For bulb $A$:
Resistance,$R_{A} = \frac{V^{2}}{P_{A}} = \frac{(120)^{2}}{90} = 160 \, \Omega$.
Therefore,current $I_{A} = \frac{V}{R_{A}} = \frac{120}{160} = 0.75 \, A$.
For bulb $B$:
Resistance,$R_{B} = \frac{V^{2}}{P_{B}} = \frac{(120)^{2}}{60} = 240 \, \Omega$.
Therefore,current $I_{B} = \frac{V}{R_{B}} = \frac{120}{240} = 0.5 \, A$.
$(ii)$ Since the bulbs are connected in parallel to a $120 \, V$ source,each bulb operates at its rated voltage. Bulb $A$ has a higher power rating $(90 \, W)$ compared to bulb $B$ $(60 \, W)$,therefore bulb $A$ will consume more energy.

(B) The filament of the bulb and the connecting wires are connected in series,so the same current flows through both.
According to Joule's law of heating,the heat produced is given by $H = I^2Rt$.
Since the resistance $(R)$ of the connecting wires is negligibly small compared to the resistance of the filament,the heat produced in the wires is negligible.
Conversely,the filament has a high resistance,which leads to a significant amount of heat production,causing it to glow.
Therefore,the bulb glows while the connecting wires remain cool.

(N/A) Consider three resistors of resistances $R_{1}, R_{2}$ and $R_{3}$ connected in series to a cell of potential $V$ as shown in the figure. Since the three resistors are connected in series,the current $I$ flowing through each of them is the same.
By Ohm's law,the potential drop across each resistor is given by:
$V_{1} = IR_{1}, V_{2} = IR_{2}$ and $V_{3} = IR_{3}$
Since $V$ is the total potential in the circuit,by the conservation of energy,we have:
$V = V_{1} + V_{2} + V_{3}$ $...(1)$
Substituting the values of $V_{1}, V_{2}$ and $V_{3}$ in equation $(1)$,we get:
$V = IR_{1} + IR_{2} + IR_{3}$ $...(2)$
If $R_{S}$ is the equivalent resistance of the series combination,then by Ohm's law:
$V = IR_{S}$ $...(3)$
Comparing equations $(2)$ and $(3)$,we have:
$IR_{S} = IR_{1} + IR_{2} + IR_{3}$
Dividing both sides by $I$,we get:
$R_{S} = R_{1} + R_{2} + R_{3}$ $...(4)$
Thus,in a series combination,the equivalent resistance is the sum of the individual resistances.

(N/A) Consider three resistors $R_{1}, R_{2}$ and $R_{3}$ connected in parallel as shown in the figure.
When the total current $I$ reaches point $a$,it splits into three parts: $I_{1}$ flowing through $R_{1}$,$I_{2}$ flowing through $R_{2}$,and $I_{3}$ flowing through $R_{3}$.
Since charge must be conserved,the total current $I$ entering point $a$ must be equal to the sum of the currents leaving that point. Therefore,we have:
$I = I_{1} + I_{2} + I_{3}$ $...(1)$
Since the resistors are connected in parallel,the potential difference $V$ across each resistor is the same. According to Ohm's law,the current through each resistor is:
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}, I_{3} = \frac{V}{R_{3}}$
Substituting these values into equation $(1)$,we get:
$I = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}$ $...(2)$
If $R_{P}$ is the equivalent resistance of the parallel combination,then the total current $I$ can be written as:
$I = \frac{V}{R_{P}}$ $...(3)$
Comparing equations $(2)$ and $(3)$,we have:
$\frac{V}{R_{P}} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}$
Dividing both sides by $V$,we get the expression for equivalent resistance:
$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

(N/A) The relationship is defined by Ohm's Law, which states that: "Provided the physical conditions (like temperature) remain constant, the electric current flowing through a conductor is directly proportional to the potential difference applied across its two ends."
The formula is $V = I \times R$, where $V$ is potential difference, $I$ is current, and $R$ is resistance.
The graph of potential difference $(V)$ versus current $(I)$ is a straight line passing through the origin.
Given:
Potential difference $(V)$ = $1.4 \ V$
Current $(I)$ = $0.35 \ A$
Using Ohm's Law, the resistance $(R)$ is calculated as:
$R = \frac{V}{I}$
$R = \frac{1.4}{0.35}$
$R = 4 \ \Omega$
Thus, the resistance of the conductor is $4 \ \Omega$.

(N/A) One volt is defined as the potential difference between two points in a current-carrying conductor when $1 \ J$ of work is done to move a charge of $1 \ C$ from one point to the other.
$(b)$ The relation between potential difference $(V)$,work $(W)$,and charge $(Q)$ is given by the formula: $V = W / Q$.
Given:
Work $(W)$ = $100 \ J$
Charge $(Q)$ = $20 \ C$
Using the formula $V = W / Q$:
$V = 100 \ J / 20 \ C = 5 \ V$.
Therefore,the potential difference between the two terminals of the battery is $5 \ V$.

(N/A) The direction of electric current is conventionally defined as opposite to the direction of the flow of electrons in a wire.
$(b)$ Given:
Charge $(Q) = 500 \, C$
Time $(t) = 10 \, \text{minutes} = 10 \times 60 \, \text{s} = 600 \, \text{s}$
Using the formula for electric current:
$I = \frac{Q}{t}$
$I = \frac{500 \, C}{600 \, \text{s}}$
$I = 0.833 \, \text{A}$
Therefore,the current in the circuit is approximately $0.833 \, \text{A}$.

(N/A) Electric current is defined as the rate of flow of electric charge through a conductor.
The $SI$ unit of electric current is the ampere $(A)$.
According to $Ohm's$ law,$V = I \times R$,where $V$ is the potential difference,$I$ is the current,and $R$ is the resistance.
From this,$R = V / I$.
If the potential difference across a conductor is $1$ volt and the current flowing through it is $1$ ampere,then the resistance $R$ is given by $R = 1 \text{ volt} / 1 \text{ ampere} = 1 \text{ ohm } (\Omega)$.
Thus,$1$ ohm is defined as the resistance of a conductor when a potential difference of $1$ volt applied across its ends causes a current of $1$ ampere to flow through it.

(N/A) In a series combination,the total resistance is the sum of individual resistances:
$R_{S} = R + R + \dots + R \ (n \ \text{times}) = nR$
In a parallel combination,the reciprocal of the total resistance is the sum of the reciprocals of individual resistances:
$\frac{1}{R_{P}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} \ (n \ \text{times}) = \frac{n}{R}$
Therefore,$R_{P} = \frac{R}{n}$
$(b)$ For resistors connected in parallel,the resultant resistance $R_{P}$ is given by:
$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{P}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}$
Taking the least common multiple $(LCM)$ of $3, 4$,and $12$,which is $12$:
$\frac{1}{R_{P}} = \frac{4 + 3 + 1}{12} = \frac{8}{12} = \frac{2}{3}$
$R_{P} = \frac{3}{2} = 1.5 \ \Omega$

(N/A) $Ohm's$ 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.
Mathematically: $V = IR$,where $V$ is potential difference,$I$ is current,and $R$ is resistance.
$(b)$ $(i)$ Variable resistance: Represented by a resistor symbol with an arrow or a sliding contact.
$(ii)$ Voltmeter: Represented by a circle with a '$V$' inside,having positive and negative terminals.
$(c)$ Given: Rated power $P = 100 \, W$,Rated voltage $V = 220 \, V$.
First,calculate the resistance of the bulb:
$R = \frac{V^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega$.
Now,calculate the power consumed at $V_1 = 110 \, V$:
$P_{new} = \frac{V_1^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25 \, W$.
The power consumed will be $25 \, W$.

(B) Let the resistance of each wire be $R$.
When connected in series,the total resistance is $R_{S} = R + R = 2R$.
The heat produced by the series combination is given by $H = \frac{V^{2}}{R_{S}} t = \frac{V^{2}}{2R} t$.
The heat produced by an individual resistor (if connected alone to the same voltage source) is $H' = \frac{V^{2}}{R} t$.
Comparing the two,we find $H = \frac{1}{2} \left( \frac{V^{2}}{R} t \right) = \frac{H'}{2}$.
Thus,the heat produced by the series combination is half the heat produced by a single wire.

(N/A) $(i)$ The two resistors are connected in a parallel combination because both ends of the resistors are connected to the same two points in the circuit.
$(ii)$ According to Ohm's law,$I = \frac{V}{R}$. Since the resistors are in parallel,the potential difference $(V)$ across each resistor is $3 \, V$.
$(a)$ Current through the $10 \, \Omega$ resistor: $I_1 = \frac{V}{R_1} = \frac{3 \, V}{10 \, \Omega} = 0.3 \, A$.
$(b)$ Current through the $15 \, \Omega$ resistor: $I_2 = \frac{V}{R_2} = \frac{3 \, V}{15 \, \Omega} = 0.2 \, A$.
$(iii)$ The ammeter measures the total current $(I)$ flowing through the circuit,which is the sum of the currents through the individual branches:
$I = I_1 + I_2 = 0.3 \, A + 0.2 \, A = 0.5 \, A$.
Therefore,the ammeter reading is $0.5 \, A$.

(B) Copper and aluminium are used because they have low electrical resistivity,which results in low resistance and minimal energy loss during transmission. Additionally,they are highly ductile,allowing them to be easily drawn into thin wires.
$(b)$ The resistance of a wire is given by the formula $R = \frac{\rho L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.
Since $R$ and $L$ are equal for both wires,we have $A = \frac{\rho L}{R}$,which implies $A \propto \rho$.
Given that the resistivity of manganin is higher than that of copper $(\rho_{\text{manganin}} > \rho_{\text{copper}})$,it follows that the cross-sectional area of the manganin wire must be greater than that of the copper wire $(A_{\text{manganin}} > A_{\text{copper}})$.
Therefore,the manganin wire is thicker.

(N/A) $(i)$ The resistance $R$ of a wire is directly proportional to its length $L$ $(R \propto L)$. When the length is doubled,the resistance also becomes double. Therefore,the new resistance is $2 \times 10 \, \Omega = 20 \, \Omega$.
$(ii)$ The resistance $R$ is inversely proportional to the square of the diameter $D$ $(R \propto 1/D^2)$. When the diameter is doubled,the resistance becomes $1/4$ of its original value. Therefore,the new resistance is $10/4 \, \Omega = 2.5 \, \Omega$.
$(b)$ The heating elements of electrical devices are made of alloys because alloys have a much higher resistivity than their constituent metals. Furthermore,alloys do not oxidize (burn) readily even at high temperatures,which increases the durability and efficiency of the heating element.