Mix Examples - Electricity Questions in English

(A-B, B-A) $(i)$ Element $B$: It is used for the filament of an electric bulb because it has a high resistivity $\left(520 \times 10^{-8} \, \Omega m\right)$. High resistivity allows the filament to produce a large amount of heat when current passes through it,enabling it to glow.
$(ii)$ Element $A$: It is used for electrical transmission lines because it has a very low resistivity $\left(1.62 \times 10^{-8} \, \Omega m\right)$. Low resistivity ensures minimal energy loss in the form of heat during the transmission of electric power over long distances.

(A) $Ohm's$ law states that the electric current flowing through a metallic conductor is directly proportional to the potential difference across its ends,provided that the physical conditions (like temperature) remain constant.
Given:
Current $(I)$ = $0.2 \, A$
Potential difference $(V)$ = $0.8 \, V$
Using the formula $R = V / I$:
$R = 0.8 \, V / 0.2 \, A$
$R = 4 \, \Omega$
Therefore,the resistance of the conductor is $4 \, \Omega$.

(N/A) Tungsten is a metal that has a very high melting point and high resistivity,which allows it to glow at high temperatures without melting,making it ideal for bulb filaments.
$(b)$ Copper and aluminium are excellent conductors of electricity because they possess very low resistivity,which minimizes energy loss during the transmission of electric current.
$(c)$ Alloys are used in electrical heating devices because their resistivity is significantly higher than that of their constituent pure metals,and they do not oxidize (burn) readily at high temperatures.

(N/A) The $V-I$ graph for a nichrome wire is a straight line passing through the origin,as nichrome follows Ohm's law.
Given:
Resistivity $\rho = 4.8 \times 10^{-7} \, \Omega \cdot m$
Length $L = 625 \, mm = 0.625 \, m$
Resistance $R = 4 \, \Omega$
Using the formula for resistance: $R = \frac{\rho L}{A}$
Rearranging for area $A$: $A = \frac{\rho L}{R}$
Substituting the values:
$A = \frac{4.8 \times 10^{-7} \times 0.625}{4}$
$A = 1.2 \times 10^{-7} \times 0.625$
$A = 0.75 \times 10^{-7} \, m^2$ or $7.5 \times 10^{-8} \, m^2$

(N/A) The circuit diagram is as shown in the image.
Join three resistors of different values in series. Connect them with a battery,an ammeter,a rheostat,and a plug key,as shown in the figure. Close the plug key. Note the ammeter reading. Change the position of the ammeter to anywhere in between the resistors. Note the ammeter reading each time. We observe that the value of the current in the ammeter is the same,independent of its position in the electric circuit. It means that in a series combination of resistors,the current is the same in every part of the circuit or the same current flows through each resistor.
Precautions:
$(i)$ The current should not be passed for a long time to avoid heating.
$(ii)$ All the connections should be tight.

(N/A) Suppose a current $I$ flows through a conductor of resistance $R$ for time $t$ under a potential difference of $V$ volts.
The charge flowing through the conductor is $Q = I \times t$.
The work done $(W)$ in moving a charge $Q$ across a potential difference $V$ is given by:
$W = V \times Q$
Substituting $Q = I \times t$ into the equation:
$W = V \times I \times t$
$(b)$ Given:
For bulb $1$: $P_1 = 40 \ W$,$V_1 = 240 \ V$
For bulb $2$: $P_2 = 25 \ W$,$V_2 = 240 \ V$
We know that $P = \frac{V^2}{R}$,so $R = \frac{V^2}{P}$.
$R_1 = \frac{240^2}{40} = \frac{57600}{40} = 1440 \ \Omega$
$R_2 = \frac{240^2}{25} = \frac{57600}{25} = 2304 \ \Omega$
Comparing $R_1$ and $R_2$,$R_2 > R_1$.
Ratio: $\frac{R_2}{R_1} = \frac{2304}{1440} = 1.6$
Thus,the $25 \ W$ bulb has higher resistance,and it is $1.6$ times the resistance of the $40 \ W$ bulb.

(N/A) The instrument used to measure electric current in a circuit is an ammeter.
The unit of electric current is the ampere $(A)$. One ampere is defined as the flow of $1$ coulomb of electric charge through a cross-section of a conductor in $1$ second.
$(b)$ $(i)$ The symbol represents a rheostat or variable resistance.
$(ii)$ The symbol represents a closed plug key or a closed switch.
$(c)$ The circuit diagram consists of a battery of four cells (total $6\, V$),a plug key (closed),an ammeter connected in series with the $0.5\, m$ nichrome wire $XY$,and a voltmeter connected in parallel across the wire $XY$.

(N/A) Resistance is the property of a conductor to oppose the flow of electric current through it.
The $SI$ unit of resistance is $Ohm$ ($\Omega$). One $Ohm$ is defined as the resistance of a conductor through which a current of $1$ $Ampere$ flows when a potential difference of $1$ $Volt$ is applied across its ends.
The factors on which the resistance of a conductor depends are:
$1$. Length of the conductor $(l)$
$2$. Area of cross-section $(A)$
$3$. Nature of the material (resistivity, $\rho$)
$4$. Temperature of the conductor
$(i)$ Since resistance $R \propto l$, if the length is doubled, the resistance becomes $2$ times the original value.
$(ii)$ Since resistance $R \propto \frac{1}{A}$ and $A = \pi r^2$, if the radius $r$ is doubled, the area $A$ becomes $4$ times $(A' = \pi(2r)^2 = 4\pi r^2)$. Consequently, the resistance becomes $\frac{1}{4}$th of its original value.

(N/A) The heating effect of electric current is the phenomenon where heat is generated in a conductor when an electric current passes through it.
When an electric current flows through a resistor,the electrons move through the conductor and collide with the atoms or ions of the resistor. During these collisions,the kinetic energy of the electrons is transferred to the atoms of the resistor,causing them to vibrate more vigorously. This increase in the vibrational energy of the atoms manifests as an increase in the temperature of the resistor,resulting in the production of heat.
Two devices based on the heating effect of electric current are:
$(i)$ Electric iron
$(ii)$ Electric geyser

(A) The instrument used to measure the potential difference between two points in a circuit is a Voltmeter.
The $SI$ unit of potential difference is the Volt $(V)$. It is defined as the amount of work done $(W)$ in moving a unit positive charge $(q)$ from one point to another,i.e.,$V = \frac{W}{q}$. Thus,$1 \text{ Volt} = 1 \text{ Joule} / 1 \text{ Coulomb}$.
The circuit symbols are:
$(i)$ Variable resistor: $A$ resistor symbol with an arrow pointing across it.
$(ii)$ Closed plug key: $A$ circle with a dot inside it.
$(b)$ $(i)$ Circuit $(I)$ has more resistance because the resistors are connected in series,and the equivalent resistance is the sum of individual resistances $(R_{eq} = R_{1} + R_{2} + R_{3})$.
$(ii)$ More current passes through circuit $(II)$ because the equivalent resistance in parallel is less than the individual resistances,leading to a higher total current from the source.
$(iii)$ In circuit $(II)$,the potential difference across each resistor is equal because they are connected in parallel.
$(iv)$ In circuit $(I)$,more heat is produced in $R_{1}$ because in a series circuit,the current $(I)$ is the same through all resistors,and heat produced $H = I^{2}Rt$. Since $R_{1}$ is the largest resistance,it will produce the most heat.

(B) The circuit diagram consists of an electric bulb,a voltmeter connected in parallel to the bulb,an ammeter connected in series,a cell,and a key,as shown in the provided image.
$(b)$ Given:
Current $(I) = 0.4 \ A$
Potential difference $(V) = 2 \ V$
According to Ohm's law,$V = IR$
Therefore,$R = \frac{V}{I} = \frac{2 \ V}{0.4 \ A} = 5 \ \Omega$
So,the resistance of the electric bulb is $5 \ \Omega$.
$(c)$ The law applied is Ohm's law. The $V-I$ graph for an ohmic conductor is a straight line passing through the origin,as shown in the provided image.

(N/A) The potential difference between two points is said to be $1$ volt if $1$ joule of work is done in moving a charge of $1$ coulomb from one point to another.
$(b)$ The heating element of an electrical heater is made of an alloy (like nichrome) which has high resistivity. Due to high resistivity, it produces a large amount of heat when current passes through it, causing it to glow. In contrast, the connecting cord is made of copper, which has very low resistivity, so it does not produce significant heat and does not glow.
$(c)$ $(i)$ Silver is a better conductor than copper because it has lower electrical resistivity ($1.60 \times 10^{-8} \, \Omega m$ for silver vs $1.62 \times 10^{-8} \, \Omega m$ for copper).
$(ii)$ Nichrome is advised for electrical heating devices because it has a very high resistivity compared to other materials listed, which allows it to generate more heat according to Joule's law of heating $(H = I^2Rt)$.

$(a)$ Total resistance $R = 5\, \Omega + 8\, \Omega + 12\, \Omega = 25\, \Omega$.
Total voltage $V = 5 \times 2\, V = 10\, V$.
Using Ohm's law, the current $I = \frac{V}{R} = \frac{10\, V}{25\, \Omega} = 0.4\, A$.
Therefore, the ammeter reading is $0.4\, A$.
$(b)$ The potential difference across the $12\, \Omega$ resistor is $V' = I \times R_{12} = 0.4\, A \times 12\, \Omega = 4.8\, V$.
Therefore, the voltmeter reading is $4.8\, V$.

(N/A) The chemical action within a cell generates a potential difference across its terminals. This potential difference drives and maintains the flow of electric current in the circuit.
$(b)$ $(i)$ The paths $xTy$ and $xZy$ are connected in parallel between points $x$ and $y$. The equivalent resistance $R_e$ is given by:
$\frac{1}{R_e} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3} \, \Omega^{-1}$
$R_e = 1.5 \, \Omega$
$(ii)$ The total resistance of the circuit is the sum of the equivalent resistance $R_e$ and the series resistor of $1.5 \, \Omega$:
$R_{total} = 1.5 \, \Omega + 1.5 \, \Omega = 3 \, \Omega$
Using Ohm's law,the main current $I = \frac{V}{R_{total}} = \frac{6 \, V}{3 \, \Omega} = 2 \, A$
$(iii)$ The potential difference across the parallel combination of $xTy$ and $xZy$ is $V_{xy} = I \times R_e = 2 \, A \times 1.5 \, \Omega = 3 \, V$.
Current through path $xTy$ $(2 \, \Omega)$: $I_1 = \frac{V_{xy}}{R_1} = \frac{3 \, V}{2 \, \Omega} = 1.5 \, A$
Current through path $xZy$ $(6 \, \Omega)$: $I_2 = \frac{V_{xy}}{R_2} = \frac{3 \, V}{6 \, \Omega} = 0.5 \, A$

(N/A) The commercial unit of electric energy is kilowatt-hour $(kWh)$.
The $SI$ unit of electrical energy is Joule $(J)$.
$1 \, kWh = 1000 \, W \times 3600 \, s = 3.6 \times 10^6 \, J$.
$(b)$ According to Joule's law of heating,$H = I^2Rt$.
If the current $I$ becomes $3I$,the new heat $H' = (3I)^2Rt = 9I^2Rt = 9H$.
Thus,the heat produced will become nine times the initial value.
$(c)$ Initial heat $H = I^2Rt$.
If the resistance $R$ is replaced by a conductor of double resistance $(2R)$,the new heat $H' = I^2(2R)t = 2I^2Rt = 2H$.
Thus,the heat produced will become twice the initial value.

(N/A) Ohm's law states that the current flowing through a metallic conductor is directly proportional to the potential difference applied across its ends,provided its physical conditions (like temperature) remain constant.
Mathematically,this is expressed as:
$V \propto I$
$V = IR$
Here,$V$ is the potential difference,$I$ is the current,and $R$ is the constant of proportionality known as the resistance of the conductor.
Verification:
To verify Ohm's law,set up a circuit as shown in the figure,consisting of a nichrome wire $R$,an ammeter,a voltmeter,a battery,and a rheostat $(Rh)$.
$1$. Close the key $K$ and note the current $I$ in the ammeter and potential difference $V$ in the voltmeter for a specific position of the rheostat.
$2$. Change the position of the rheostat to vary the current in the circuit and record the corresponding values of $V$ and $I$.
$3$. Calculate the ratio $V/I$ for each set of readings. You will observe that the ratio $V/I$ is constant.
$4$. Plot a graph of $V$ (on the y-axis) against $I$ (on the x-axis). The graph will be a straight line passing through the origin,which confirms that $V$ is directly proportional to $I$,thus verifying Ohm's law.

(N/A) An ammeter is used to measure electric current in an electric circuit.
One ampere is defined as the flow of one coulomb of electric charge through a circuit in one second.
$(b)$ $(i)$ Variable resistor (Rheostat) and $(ii)$ Closed plug key (switch).
$(c)$ $(i)$ The circuit diagram consists of a battery of four cells,an ammeter in series,a nichrome wire $XY$ in series,and a voltmeter connected in parallel across $XY$.
$(ii)$ Since the graph is a straight line passing through the origin,the ratio $V/I$ is constant,which represents the resistance $R$ of the nichrome wire.
From the graph,at $V = 1.6 \ V$,$I = 0.6 \ A$.
$R = V/I = 1.6 / 0.6 = 2.67 \ \Omega$.
Since the ratio $V/I$ is constant for all values $(0.8 \ V, 1.2 \ V, 1.6 \ V)$,it confirms Ohm's Law.

(A) $(i)$ It means that the work done in moving a charge of $1 \, C$ between the two points is $1 \, J$.
$(ii)$ Using the formula $W = qV$,where $q = 5 \, C$ and $V = 12 \, V$,we get $W = 5 \times 12 = 60 \, J$.
$(b)$ To study the factors affecting resistance:
$(i)$ Set up a circuit with a cell,an ammeter,a nichrome wire of length $l$,and a plug key.
$(ii)$ Close the key and note the current reading in the ammeter.
$(iii)$ Replace the wire with another nichrome wire of the same thickness but double the length $(2l)$. Note the current.
$(iv)$ Replace the wire with a thicker nichrome wire of the same length $l$. Note the current.
$(v)$ Replace the wire with a copper wire of the same length and thickness as the first nichrome wire. Note the current.
$(vi)$ Observe that the current varies in each case,demonstrating that resistance depends on length,cross-sectional area,and the material of the conductor.

(N/A) $(i)$ Variable resistor or rheostat: It is used to change the resistance in a circuit,thereby controlling the current flowing through it.
$(ii)$ Wires crossing without joining: It represents two wires that cross each other without making any electrical contact.
$(b)$ The circuit diagram consists of a $12 \, V$ battery,a switch $(K)$,an ammeter $(A)$ in series with the combination,and a voltmeter $(V)$ connected in parallel across the three resistors ($5 \, \Omega$,$10 \, \Omega$,and $20 \, \Omega$) which are themselves connected in parallel.
$(c)$ Advantage of parallel connection: In a parallel circuit,if one electrical device stops working due to a defect or switch-off,the other devices continue to operate independently because each device has its own separate circuit path.

(N/A) In a series arrangement,if any one component stops working,the circuit breaks,and none of the other devices will function.
$(b)$ Tungsten is used because it has a very high melting point $(3380 ^\circ C)$ and high resistivity,allowing it to emit light without melting.
$(c)$ Alloys have higher resistivity than pure metals and do not oxidize (burn) easily at high temperatures,making them ideal for heating elements.
$(d)$ Copper and aluminium have very low resistivity,which minimizes energy loss during the transmission of electricity.
$(e)$ The cord has very low resistance,so it produces negligible heat,whereas the heating element has high resistance,causing it to glow due to the heating effect of current.

(N/A) The resistance $R$ is given by Ohm's law: $R = V / I$.
From the graph,at $V = 2 \, V$,the current $I = 0.1 \, A$.
Therefore,$R = 2 \, V / 0.1 \, A = 20 \, \Omega$.
$(b)$ Given: Total current $I = 5 \, A$,Voltage $V = 220 \, V$.
The required total resistance $R_{eq}$ of the circuit is $R_{eq} = V / I = 220 \, V / 5 \, A = 44 \, \Omega$.
Let $n$ resistors of $176 \, \Omega$ each be connected in parallel.
The equivalent resistance for $n$ resistors in parallel is $1 / R_{eq} = n / R$.
$1 / 44 = n / 176$.
$n = 176 / 44 = 4$.
Thus,$4$ resistors are required.
$(c)$ Electric power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.
We know that power $P = W / t$ and work done $W = V \cdot I \cdot t$.
So,$P = (V \cdot I \cdot t) / t = V \cdot I$.
From Ohm's law,$I = V / R$.
Substituting $I$ in the power equation: $P = V \cdot (V / R) = V^2 / R$.

(N/A) The circuit diagram shows three resistors $R_{1}, R_{2}, R_{3}$ connected in parallel across a potential difference $V$. The total current $I$ splits into $I_{1}, I_{2}, I_{3}$ through the resistors.
Since they are in parallel,the potential difference $V$ across each resistor is the same.
Total current $I = I_{1} + I_{2} + I_{3}$.
Using Ohm's law,$I_{1} = V/R_{1}$,$I_{2} = V/R_{2}$,and $I_{3} = V/R_{3}$.
If $R_{p}$ is the equivalent resistance,$I = V/R_{p}$.
Substituting these,$V/R_{p} = V/R_{1} + V/R_{2} + V/R_{3}$.
Thus,$1/R_{p} = 1/R_{1} + 1/R_{2} + 1/R_{3}$.
$(b)$ Electric bulbs are filled with chemically inactive gases like nitrogen or argon to prevent the oxidation of the tungsten filament at high temperatures,thereby prolonging its life.
$(c)$ The statement means that the fuse is designed to allow a maximum current of $5 \ A$ to flow through the circuit. If the current exceeds this value,the fuse wire melts due to excessive heating and breaks the circuit,protecting the appliances.

(N/A) $(i)$ The ammeter reading will decrease (become half). This is because resistance $R$ is directly proportional to length $L$ $(R \propto L)$. When length is doubled, resistance doubles, and according to Ohm's law $(I = V/R)$, the current $I$ becomes half.
$(ii)$ The ammeter reading will increase (become two times). This is because resistance $R$ is inversely proportional to the area of cross-section $A$ $(R \propto 1/A)$. When the area is doubled, resistance becomes half, and consequently, the current $I$ becomes double.
$(b)$ It means that $1\, J$ of work is being done to move a charge of $1\, C$ from point $A$ to point $B$.

(A) The circuit diagram is as shown in the provided image.
Given:
Total voltage $V = 5 \times 2 \text{ V} = 10 \text{ V}$.
Resistors in series: $R_{1} = 5 \, \Omega$, $R_{2} = 10 \, \Omega$, $R_{3} = 15 \, \Omega$.
The equivalent resistance $R$ of the series combination is:
$R = R_{1} + R_{2} + R_{3} = 5 + 10 + 15 = 30 \, \Omega$.
$(i)$ The electric current $I$ passing through the circuit is:
$I = V / R = 10 \text{ V} / 30 \, \Omega = 1/3 \text{ A} \approx 0.33 \text{ A}$.
$(ii)$ The potential difference $V_{1}$ across the $5 \, \Omega$ resistor is:
$V_{1} = I \times R_{1} = (1/3) \text{ A} \times 5 \, \Omega = 5/3 \text{ V} \approx 1.67 \text{ V}$.

(N/A) An electric circuit is a continuous and closed path in which an electric current can flow.
$(b)$ We know that $I = Q / t$ and $Q = n e$,where $n$ is the number of electrons and $e$ is the charge on an electron. Therefore,we have:
$I = n e / t$
$n = (I \times t) / e = (1 \ \text{A} \times 1 \ \text{s}) / (1.6 \times 10^{-19} \ \text{C})$
$n = 6.25 \times 10^{18}$ electrons.
$(c)$ The circuit diagram used to study Ohm's law is shown below.
Electric current is measured by an Ammeter $(A)$.
Potential difference is measured by a Voltmeter $(V)$.

$(a)$ Electric potential difference between two points in an electric circuit carrying current is defined as the work done to move a unit positive charge from one point to the other.
$(b)$ The instrument used to measure potential difference is a Voltmeter. It is always connected in parallel across the points between which the potential difference is to be measured.
$(c)$ Given: Current $(I) = 2 \, A$, Time $(t) = 1 \, \text{minute} = 60 \, s$, Potential difference $(V) = 3 \, V$.
First, calculate the total charge $(Q)$ transferred: $Q = I \times t = 2 \, A \times 60 \, s = 120 \, C$.
Now, calculate the work done $(W)$ using the formula $W = V \times Q$: $W = 3 \, V \times 120 \, C = 360 \, J$.
Thus, the work done is $360 \, J$.

(N/A) $Ohm's$ law: The potential difference across the ends of a resistor is directly proportional to the current flowing through it, provided its temperature remains constant.
Mathematical representation: $V = IR$ or $R = V / I$, where $V$ is the potential difference, $I$ is the current, and $R$ is the resistance.
Definition of $1\, ohm$: If the potential difference across the two ends of a conductor is $1\, V$ and the current flowing through it is $1\, A$, then the resistance $R$ of the conductor is $1\, ohm$ $(1\, \Omega = 1\, V / 1\, A)$.
Disadvantages of series connection:
$(i)$ Different electrical appliances require different values of current to operate properly, which is not possible in a series circuit because the current remains the same throughout.
$(ii)$ If one appliance fails or is switched off, the entire circuit is broken, and none of the other appliances will work.

(N/A) Power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.
$1$ watt is the power consumed by a device that carries a current of $1 \ A$ when operated at a potential difference of $1 \ V$, whereas $1$ watt hour is the amount of energy consumed when $1$ watt of power is used for $1$ hour.
The commercial unit of electric energy is $kWh$.
$1 \ kWh = 1000 \ W \times 3600 \ s = 3.6 \times 10^6 \ J$.
Total energy consumed by the heater in $30$ days:
$E = \text{Power} \times \text{Time} \times \text{Days} = 1000 \ W \times 2 \ h \times 30 = 60000 \ Wh = 60 \ kWh$.
Cost of energy = $60 \ kWh \times ₹ 5.00/kWh = ₹ 300$.

(N/A) When three resistors $R_1, R_2,$ and $R_3$ are connected in series,the same current $I$ flows through each resistor. The potential difference across each resistor is $V_1 = IR_1, V_2 = IR_2,$ and $V_3 = IR_3$. The total potential difference $V$ is $V = V_1 + V_2 + V_3 = I(R_1 + R_2 + R_3)$. If $R_s$ is the equivalent resistance,then $V = IR_s$. Thus,$R_s = R_1 + R_2 + R_3$.
$(b)$ The given circuit shows three resistors connected in parallel with values $R_1 = 6 \Omega, R_2 = 10 \Omega,$ and $R_3 = 15 \Omega$.
For parallel connection,the equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{10} + \frac{1}{15}$
$\frac{1}{R_p} = \frac{5 + 3 + 2}{30} = \frac{10}{30} = \frac{1}{3}$
Therefore,$R_p = 3 \Omega$.

(N/A) $(i)$ Ohm's law.
It states that,provided the physical conditions remain the same,the current flowing through a conductor is directly proportional to the potential difference across its two ends.
Mathematically,we have $V \propto I$ or $V = IR$,where $R$ is the resistance of the conductor.
$(ii)$ The $V-I$ graph is a straight line passing through the origin. This justifies the law because a straight-line graph indicates that the potential difference $V$ is directly proportional to the current $I$ $(V \propto I)$.
$(iii)$ $(a)$ Variable resistor (Rheostat): It is used to vary the resistance in the circuit to control the current.
$(b)$ Closed switch (or closed key): It allows the current to flow through the circuit by completing the path.

(D) Resistance is defined as the opposition offered by a conductor to the flow of electric current.
It depends on the following factors:
$(i)$ Nature of the material of the conductor.
$(ii)$ Length of the conductor $(L)$.
$(iii)$ Area of cross-section of the conductor $(A)$.
$A$ rheostat or variable resistor is the device used to change the resistance in an electric circuit without changing the voltage source.
Calculation:
Given: $L = 50 \, m$,$A = 0.01 \, mm^2 = 0.01 \times 10^{-6} \, m^2$,$\rho = 5 \times 10^{-8} \, \Omega \, m$.
Using the formula $R = \frac{\rho L}{A}$:
$R = \frac{5 \times 10^{-8} \times 50}{0.01 \times 10^{-6}}$
$R = \frac{250 \times 10^{-8}}{10^{-8}}$
$R = 250 \, \Omega$.

(N/A) $(i)$ Current $(I)$ is calculated using the formula $P = V \times I$,so $I = P / V = 750\, W / 200\, V = 3.75\, A$.
$(ii)$ Resistance $(R)$ is calculated using Ohm's Law $V = I \times R$,so $R = V / I = 200\, V / 3.75\, A \approx 53.33\, \Omega$.
$(iii)$ Energy consumed $(E)$ is calculated as $E = P \times t = 750\, W \times 2\, h = 1500\, Wh = 1.5\, kWh$.

(D) Let the length of the original wire be $l$ and its radius be $r$.
The original resistance is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
The original volume is $V_1 = \pi r^2 l$.
When the radius is halved,the new radius becomes $r' = \frac{r}{2}$. Let the new length be $l'$.
Since the volume remains constant,$V_1 = V_2$,so $\pi r^2 l = \pi (\frac{r}{2})^2 l'$.
$\pi r^2 l = \pi \frac{r^2}{4} l'$,which simplifies to $l' = 4l$.
The new resistance is $R' = \rho \frac{l'}{\pi (r')^2} = \rho \frac{4l}{\pi (r/2)^2}$.
$R' = \rho \frac{4l}{\pi (r^2/4)} = 16 \left( \rho \frac{l}{\pi r^2} \right)$.
Therefore,$R' = 16R$.
The new resistance is $16$ times the original resistance.

(A) Given: Potential difference $(V)$ = $1.5 \ V$,Resistance $(R)$ = $30 \ \Omega$.
According to Ohm's Law,$V = IR$.
Therefore,the current $(I)$ is given by the formula: $I = \frac{V}{R}$.
Substituting the values: $I = \frac{1.5}{30} = 0.05 \ A$.
Thus,the current flowing through the circuit is $0.05 \ A$.

(A) Given: Current $I = 0.5\, A$, Time $t = 10\, \text{min} = 10 \times 60 = 600\, \text{s}$.
The formula for electric charge $Q$ is $Q = I \times t$.
Substituting the values: $Q = 0.5\, A \times 600\, \text{s} = 300\, C$.
Therefore, the amount of electric charge that flows through the circuit is $300\, C$.

(B) Given: Charge $(Q) = 2 \, C$,Potential difference $(V) = 12 \, V$.
The formula for work done $(W)$ in moving a charge between two points is given by $W = Q \times V$.
Substituting the given values into the formula:
$W = 2 \, C \times 12 \, V = 24 \, J$.
Therefore,the work done is $24 \, J$.

(N/A) Given: Potential difference $V = 220 \ V$,Resistance $R = 1200 \ \Omega$.
According to Ohm's law,$V = IR$,so $I = V / R$.
$I = 220 \ V / 1200 \ \Omega = 0.1833 \ A \approx 0.18 \ A$.
$(b)$ Given: Potential difference $V = 220 \ V$,Resistance $R = 100 \ \Omega$.
According to Ohm's law,$I = V / R$.
$I = 220 \ V / 100 \ \Omega = 2.2 \ A$.

(B) Given: Initial potential difference $V_1 = 60\, V$,Initial current $I_1 = 4\, A$.
According to Ohm's law,$V = IR$,where $R$ is the resistance of the heater. Since the heater remains the same,its resistance $R$ is constant.
First,calculate the resistance: $R = V_1 / I_1 = 60\, V / 4\, A = 15\, \Omega$.
Now,for the new potential difference $V_2 = 120\, V$,the new current $I_2$ is calculated as:
$I_2 = V_2 / R = 120\, V / 15\, \Omega = 8\, A$.
Alternatively,since $I \propto V$,doubling the potential difference from $60\, V$ to $120\, V$ will double the current from $4\, A$ to $8\, A$.

(A) Given:
Resistance $R = 26 \, \Omega$
Length $L = 1 \, m$
Diameter $D = 0.3 \, mm = 3 \times 10^{-4} \, m$
Radius $r = D/2 = 1.5 \times 10^{-4} \, m$
The formula for resistance is $R = \rho \frac{L}{A}$,where $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Rearranging for resistivity $\rho$:
$\rho = \frac{R A}{L} = \frac{R \pi D^2}{4 L}$
Substituting the values:
$\rho = \frac{26 \times 3.14 \times (3 \times 10^{-4})^2}{4 \times 1}$
$\rho = \frac{26 \times 3.14 \times 9 \times 10^{-8}}{4}$
$\rho = 1.8369 \times 10^{-6} \, \Omega \, m \approx 1.84 \times 10^{-6} \, \Omega \, m$.

(A) Given initial resistance $R = 4 \, \Omega$.
When a wire is doubled on itself,its length $L$ becomes half $(L_n = L/2)$ and its area of cross-section $A$ becomes double $(A_n = 2A)$.
The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity of the material.
The new resistance $R_n$ is given by:
$R_n = \rho \frac{L_n}{A_n} = \rho \frac{L/2}{2A} = \frac{1}{4} \left( \rho \frac{L}{A} \right) = \frac{R}{4}$.
Substituting the value of $R = 4 \, \Omega$:
$R_n = \frac{4 \, \Omega}{4} = 1 \, \Omega$.
Thus,the new resistance of the wire is $1 \, \Omega$.

(N/A) We know that the power input is $P = V \times I$.
Thus,the current is $I = P / V$.
$(a)$ When heating is at the maximum rate $(P = 840\, W)$:
$I = 840 / 220 = 3.82\, A$.
The resistance is $R = V / I = 220 / 3.82 = 57.60\, \Omega$.
$(b)$ When heating is at the minimum rate $(P = 360\, W)$:
$I = 360 / 220 = 1.64\, A$.
The resistance is $R = V / I = 220 / 1.64 = 134.15\, \Omega$.

(B) Given: Heat produced $H = 100 \, J$,Resistance $R = 4 \, \Omega$,Time $t = 1 \, s$.
We know that the formula for heat produced is $H = I^2 R t$.
Rearranging to find current $I$:
$I = \sqrt{\frac{H}{R t}} = \sqrt{\frac{100}{4 \times 1}} = \sqrt{25} = 5 \, A$.
Now,using Ohm's Law to find the potential difference $V$:
$V = I \times R$
$V = 5 \, A \times 4 \, \Omega = 20 \, V$.
Therefore,the potential difference across the resistor is $20 \, V$.

(A) Given: Potential difference $V = 220 \ V$ and current $I = 0.50 \ A$.
The formula for electric power is $P = V \times I$.
Substituting the given values: $P = 220 \ V \times 0.50 \ A$.
Therefore,the power of the bulb is $P = 110 \ W$.

(A) Given: Power $P = 400 \ W$,Time $t = 8 \ h$ per day for $30$ days,Rate $= ₹ 3.00$ per $kWh$.
Total time in hours $= 8 \times 30 = 240 \ h$.
Total energy consumed in $kWh = \frac{P \times t}{1000} = \frac{400 \times 240}{1000} = 96 \ kWh$.
Cost of energy $= \text{Total energy} \times \text{Rate} = 96 \times 3 = ₹ 288.00$.
Therefore,the cost to operate the refrigerator for $30$ days is ₹ $288.00$.

(N/A) The resistors $6 \, \Omega$ and $3 \, \Omega$ are connected in parallel.
Therefore,the equivalent resistance $R_P$ is given by:
$\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2}$
$\frac{1}{R_P} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
Thus,$R_P = 2 \, \Omega$.
Now,the $2 \, \Omega$ resistor and the parallel combination $R_P$ are in series.
Therefore,the total equivalent resistance $R$ is:
$R = 2 \, \Omega + R_P = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega$.
The equivalent resistance of the circuit is $4 \, \Omega$.

(A) Given:
Resistance $R = 480 \Omega$
Voltage $V = 220 \text{ V}$
Time $t = 10 \text{ s}$
The formula for electrical energy consumed is $E = \frac{V^2 t}{R}$.
Substituting the given values:
$E = \frac{(220)^2 \times 10}{480}$
$E = \frac{48400 \times 10}{480}$
$E = \frac{484000}{480}$
$E \approx 1008.33 \text{ J}$.

(B) Given: Charge $q = 2 \ C$,Initial potential $V_1 = 118 \ V$,Final potential $V_2 = 128 \ V$.
We know that the potential difference $V$ is defined as the work done $W$ per unit charge $q$,i.e.,$V = W / q$.
Therefore,the work done $W = q \times (V_2 - V_1)$.
Substituting the values: $W = 2 \times (128 - 118) = 2 \times 10 = 20 \ J$.
Thus,the total work done is $20 \ J$.

(A) Given:
Potential difference across $5 \, \Omega$ resistor $(V_1)$ = $10 \, V$
Potential difference across $R \, \Omega$ resistor $(V_2)$ = $6 \, V$
Resistance $R_1 = 5 \, \Omega$
$(i)$ By Ohm's law,the current $(I)$ through the $5 \, \Omega$ resistor is:
$I = \frac{V_1}{R_1} = \frac{10 \, V}{5 \, \Omega} = 2 \, A$
(ii) Since the two resistors are connected in series,the current remains the same throughout the circuit. Therefore,the current through resistor $R$ is also $2 \, A$.
(iii) Using Ohm's law for resistor $R$:
$R = \frac{V_2}{I} = \frac{6 \, V}{2 \, A} = 3 \, \Omega$
(iv) Since the resistors are in series,the total voltage $V$ is the sum of the potential differences across each resistor:
$V = V_1 + V_2 = 10 \, V + 6 \, V = 16 \, V$

(25.2 KWH) Energy consumed by one $100 \text{ W}$ bulb working for $2 \text{ hours}$ daily:
$= \frac{100 \text{ W}}{1000} \times 2 \text{ h} = 0.2 \text{ kWh}$
Energy consumed by four $40 \text{ W}$ bulbs working for $4 \text{ hours}$ daily:
$= \frac{4 \times 40 \text{ W}}{1000} \times 4 \text{ h} = 0.64 \text{ kWh}$
Total energy consumed everyday:
$= 0.2 \text{ kWh} + 0.64 \text{ kWh} = 0.84 \text{ kWh}$
Total energy consumed in $30 \text{ days}$:
$= 0.84 \text{ kWh} \times 30 = 25.2 \text{ kWh}$

$(11 \Omega)$ The given circuit can be redrawn by identifying the nodes connected by wires. Points $A$ and $C$ are at the same potential, and points $E$ and $F$ are connected to $D$ via a wire, effectively making them parallel.
It is clear that the resistances $20 \Omega$ (between $C$ and $D$), $10 \Omega$ (between $D$ and $E$), and $20 \Omega$ (between $E$ and $F$) are in a parallel combination between the effective nodes.
If $R_{1}$ is the effective resistance of these three parallel resistors, then:
$\frac{1}{R_{1}} = \frac{1}{20} + \frac{1}{10} + \frac{1}{20}$
$\frac{1}{R_{1}} = \frac{1 + 2 + 1}{20} = \frac{4}{20} = \frac{1}{5}$
Therefore, $R_{1} = 5 \Omega$.
Now, the resistor $R_{1}$ and the $6 \Omega$ resistor are in series. Therefore, the equivalent resistance $R$ between $A$ and $B$ is:
$R = R_{1} + 6 \Omega$
$R = 5 \Omega + 6 \Omega = 11 \Omega$.