Textbook - Electricity Questions in English

(A) An electric circuit is a continuous and closed path of an electric current.
It consists of electric devices,switching devices,and a source of electricity,all connected by conducting wires.

(A) The unit of electric current is the ampere $(A)$.
$1$ ampere is defined as the flow of $1$ coulomb $(C)$ of electric charge through a cross-section of a conductor in $1$ second $(s)$.
Mathematically,$I = Q / t$,where $I$ is current in amperes,$Q$ is charge in coulombs,and $t$ is time in seconds.

(N/A) One electron possesses a charge of $1.6 \times 10^{-19} \, C$,which means $1.6 \times 10^{-19} \, C$ of charge is contained in $1$ electron.
Using the formula $Q = ne$,where $Q = 1 \, C$ and $e = 1.6 \times 10^{-19} \, C$:
$n = \frac{Q}{e} = \frac{1}{1.6 \times 10^{-19}} = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Therefore,$6.25 \times 10^{18}$ electrons constitute one coulomb of charge.

(N/A) source of electricity,such as a cell,battery,or power supply,helps to maintain a potential difference across a conductor. This potential difference is necessary to drive the flow of electrons through the conductor,thereby creating an electric current.

(N/A) The potential difference between two points is said to be $1$ $V$ if $1$ $J$ of work is done in moving a charge of $1$ $C$ from one point to the other.
Mathematically,$V = W / Q$,where $V = 1$ $V$,$W = 1$ $J$,and $Q = 1$ $C$.

(B) The energy given to each coulomb of charge is equal to the amount of work required to move it.
The amount of work done is calculated using the formula:
Potential difference $(V)$ = Work done $(W)$ / Charge $(Q)$
Work done $(W)$ = Potential difference $(V)$ $\times$ Charge $(Q)$
Given:
Charge $(Q)$ = $1\, C$
Potential difference $(V)$ = $6\, V$
Calculation:
Work done $(W)$ = $6\, V \times 1\, C = 6\, J$
Therefore,$6\, J$ of energy is given to each coulomb of charge passing through a battery of $6\, V$.

(N/A) The resistance of a conductor depends upon the following factors:
$(a)$ Length of the conductor: Resistance is directly proportional to the length $(R \propto l)$.
$(b)$ Cross-sectional area of the conductor: Resistance is inversely proportional to the cross-sectional area $(R \propto 1/A)$.
$(c)$ Material of the conductor: Resistance depends on the nature of the material, represented by its resistivity $(\rho)$.
$(d)$ Temperature of the conductor: For most conductors, resistance increases with an increase in temperature.

(A) The resistance of a wire is given by the formula $R = \rho \frac{l}{A}$.
Where:
$\rho$ = Resistivity of the material of the wire.
$l$ = Length of the wire.
$A$ = Area of cross-section of the wire.
From the formula,it is clear that resistance is inversely proportional to the area of cross-section $(R \propto \frac{1}{A})$.
$A$ thicker wire has a larger area of cross-section,which results in lower resistance. Conversely,a thinner wire has a smaller area of cross-section,resulting in higher resistance.
Since current flows more easily through a path of lower resistance,current will flow more easily through a thick wire compared to a thin wire when connected to the same source.

(C) According to Ohm's law, the relationship between potential difference $(V)$, current $(I)$, and resistance $(R)$ is given by:
$V = IR$ or $I = \frac{V}{R}$
Given that the resistance $(R)$ remains constant, the current is directly proportional to the potential difference $(I \propto V)$.
If the potential difference is reduced to half of its former value, let the new potential difference be $V' = \frac{V}{2}$.
Substituting this into the formula for the new current $(I')$:
$I' = \frac{V'}{R} = \frac{\frac{V}{2}}{R} = \frac{1}{2} \left( \frac{V}{R} \right) = \frac{I}{2}$
Therefore, the current flowing through the electrical component is reduced to half of its original value.

(N/A) The resistivity of an alloy is generally much higher than that of its constituent pure metals.
Due to this high resistivity,alloys produce more heat when an electric current passes through them.
Furthermore,alloys do not oxidize (burn) or melt readily even at high temperatures.
Therefore,alloys are preferred for heating elements in appliances like electric toasters and electric irons.

(A) The electrical resistivity of iron is $10.0 \times 10^{-8} \,\Omega \,m$.
The electrical resistivity of mercury is $94.0 \times 10^{-8} \,\Omega \,m$.
Since the resistivity of mercury is higher than that of iron, mercury offers more resistance to the flow of electric current.
Therefore, iron is a better conductor than mercury.

(A) The electrical conductivity of a material is inversely proportional to its electrical resistivity.
Materials with lower resistivity are better conductors of electricity.
Comparing the values in the table,silver has the lowest resistivity of $1.6 \times 10^{-8} \, \Omega \, m$.
Therefore,silver is the best conductor among the given materials.

(N/A) Three cells of potential $2\, V$ each,connected in series,are equivalent to a battery of potential $2\, V + 2\, V + 2\, V = 6\, V$.
The following circuit diagram shows three resistors of resistances $5 \,\Omega$,$8 \,\Omega$,and $12 \,\Omega$ respectively,connected in series with a battery of potential $6\, V$ and a closed plug key.

(N/A) To measure the current flowing through the resistors,an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the $12\,\Omega$ resistor,a voltmeter should be connected in parallel to this resistor.
The total voltage of the battery is $V = 3 \times 2\, V = 6\, V$.
The resistances are connected in series,so the total resistance $R$ is:
$R = 5\,\Omega + 8\,\Omega + 12\,\Omega = 25\,\Omega$.
According to Ohm's law,$V = IR$,the current $I$ flowing through the circuit is:
$I = \frac{V}{R} = \frac{6\, V}{25\,\Omega} = 0.24\, A$.
Thus,the ammeter reading is $0.24\, A$.
The potential difference $V_1$ across the $12\,\Omega$ resistor is calculated using Ohm's law:
$V_1 = I \times R_{12\,\Omega} = 0.24\, A \times 12\,\Omega = 2.88\, V$.
Therefore,the reading of the voltmeter is $2.88\, V$.

(A) When two resistors $R_1$ and $R_2$ 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 $R_1 = 1\, \Omega$ and $R_2 = 10^6\, \Omega$.
Substituting the values: $\frac{1}{R} = \frac{1}{1} + \frac{1}{10^6} = 1 + 0.000001 = 1.000001\, \Omega^{-1}$.
Therefore,$R = \frac{1}{1.000001} \approx 0.999999\, \Omega$.
Since $0.999999\, \Omega < 1\, \Omega$,the equivalent resistance is slightly less than $1\, \Omega$.

(D) When $1 \, \Omega$,$10^3 \, \Omega$,and $10^6 \, \Omega$ are connected in parallel,the equivalent resistance $R$ is given by the formula:
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
Substituting the values:
$\frac{1}{R} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} = 1 + 0.001 + 0.000001 = 1.001001 \, \Omega^{-1}$
Therefore,$R = \frac{1}{1.001001} \approx 0.999 \, \Omega$.
In a parallel combination,the equivalent resistance is always less than the smallest individual resistance. Since $1 \, \Omega$ is the smallest,the result $0.999 \, \Omega$ is consistent with this principle.

(D) Resistance of electric lamp,$R_1 = 100 \,\Omega$.
Resistance of toaster,$R_2 = 50 \,\Omega$.
Resistance of water filter,$R_3 = 500 \,\Omega$.
Voltage of the source,$V = 220 \,V$.
These are connected in parallel.
Let $R$ be the equivalent resistance of the circuit.
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500}$.
$\frac{1}{R} = \frac{5 + 10 + 1}{500} = \frac{16}{500}$.
$R = \frac{500}{16} \,\Omega = 31.25 \,\Omega$.
According to Ohm's law,$V = IR$,so the total current $I = \frac{V}{R}$.
$I = \frac{220}{500/16} = \frac{220 \times 16}{500} = 7.04 \,A$.
Since the electric iron takes as much current as all three appliances,the current through the iron is $I' = 7.04 \,A$.
Using Ohm's law for the electric iron,$R' = \frac{V}{I'} = \frac{220}{7.04} = 31.25 \,\Omega$.
Therefore,the resistance of the electric iron is $31.25 \,\Omega$ and the current flowing through it is $7.04 \,A$.

(N/A) $1$. In a parallel circuit,the voltage remains the same across all electrical appliances,which is equal to the source voltage. This ensures that each device operates at its rated voltage.
$2$. If one electrical appliance stops working due to a defect or switch-off,the other appliances continue to function because each device has its own independent circuit path.
$3$. The total effective resistance of the circuit decreases when appliances are connected in parallel,allowing for a higher total current to be drawn from the source as needed.

(N/A) We have three resistors with resistances $R_1 = 2 \,\Omega$,$R_2 = 3 \,\Omega$,and $R_3 = 6 \,\Omega$.
To obtain a total resistance of $4 \,\Omega$,we connect the $3 \,\Omega$ and $6 \,\Omega$ resistors in parallel,and then connect this combination in series with the $2 \,\Omega$ resistor.
First,calculate the equivalent resistance $(R_p)$ of the $3 \,\Omega$ and $6 \,\Omega$ resistors connected in parallel:
$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_p = 2 \,\Omega$
Next,connect this equivalent resistance $(R_p)$ in series with the $2 \,\Omega$ resistor:
$R_{total} = R_p + 2 \,\Omega = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$
Thus,the total resistance of the circuit is $4 \,\Omega$.

(N/A) To obtain a total resistance of $1\, \Omega$ from resistors of $2\, \Omega$,$3\, \Omega$,and $6\, \Omega$,they must be connected in parallel.
The formula for the equivalent resistance $(R_p)$ of resistors connected in parallel is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
Substituting the given values:
$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
Taking the least common multiple $(LCM)$ of $2, 3,$ and $6$,which is $6$:
$\frac{1}{R_p} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\, \Omega^{-1}$
Therefore,$R_p = 1\, \Omega$.
Thus,connecting all three resistors in parallel results in a total resistance of $1\, \Omega$.

(A) Given four coils with resistances $R_1 = 4 \,\Omega, R_2 = 8 \,\Omega, R_3 = 12 \,\Omega$,and $R_4 = 24 \,\Omega$.
$(a)$ To obtain the highest total resistance,the coils must be connected in series. The equivalent resistance $R_s$ is the sum of individual resistances:
$R_s = R_1 + R_2 + R_3 + R_4 = 4 + 8 + 12 + 24 = 48 \,\Omega$.
$(b)$ To obtain the lowest total resistance,the coils must be connected in parallel. The equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}$.
Taking the least common multiple $(LCM)$ of $4, 8, 12, 24$,which is $24$:
$\frac{1}{R_p} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2}$.
Therefore,$R_p = 2 \,\Omega$.
Thus,the highest resistance is $48 \,\Omega$ and the lowest resistance is $2 \,\Omega$.

(B) The heating element of an electric heater is a resistor.
According to Joule's law of heating,the heat produced $(H)$ is given by $H = I^2Rt$,where $I$ is the current,$R$ is the resistance,and $t$ is the time.
The resistance of the heating element is very high,which causes it to produce a large amount of heat,making it glow red.
In contrast,the cord is made of good conductors like copper,which have very low resistance.
Therefore,the heat produced in the cord is negligible,and it does not glow.

(C) The amount of heat $(H)$ produced is given by the formula $H = V \times Q$,where $V$ is the potential difference and $Q$ is the total charge transferred.
Given:
Potential difference,$V = 50\, V$
Charge,$Q = 96000\, C$
Using the formula:
$H = V \times Q$
$H = 50\, V \times 96000\, C$
$H = 4800000\, J$
$H = 4.8 \times 10^{6}\, J$
Therefore,the heat generated is $4.8 \times 10^{6}\, J$.

(D) The amount of heat $(H)$ produced is given by Joule's law of heating as $H = I^2Rt$.
Given:
Resistance,$R = 20\, \Omega$
Current,$I = 5\, A$
Time,$t = 30\, s$
Using the formula:
$H = (5)^2 \times 20 \times 30$
$H = 25 \times 20 \times 30$
$H = 500 \times 30$
$H = 15000\, J$
$H = 1.5 \times 10^4\, J$
Therefore,the amount of heat developed in the electric iron is $1.5 \times 10^4\, J$.

(B) The rate of consumption of electric energy in an electric appliance is defined as electric power.
Mathematically,$P = \frac{W}{t} = VI$.
Therefore,the rate at which energy is delivered by a current is known as electric power.

(B) Power $(P)$ is given by the expression,
$P = V \times I$
Where,
Voltage,$V = 220\, V$
Current,$I = 5\, A$
$P = 220 \times 5 = 1100\, W$
Energy consumed $(E)$ by the motor is given by,
$E = P \times t$
Where,
Time,$t = 2\, h = 2 \times 60 \times 60 = 7200\, s$
$E = 1100 \times 7200 = 7,920,000\, J = 7.92 \times 10^6\, J$
Therefore,the power of the motor is $1100\, W$ and the energy consumed is $7.92 \times 10^6\, J$.

(C) The resistance of a wire is directly proportional to its length $(R \propto l)$.
When a wire of resistance $R$ is cut into five equal parts,the resistance of each part becomes $R/5$.
When these five parts are connected in parallel,the equivalent resistance $R'$ is given by the formula:
$1/R' = 1/R_1 + 1/R_2 + 1/R_3 + 1/R_4 + 1/R_5$
Substituting $R_1 = R_2 = R_3 = R_4 = R_5 = R/5$:
$1/R' = 5/R + 5/R + 5/R + 5/R + 5/R = 25/R$
Therefore,$R/R' = 25$.

(D) Electrical power is given by the expression,$P = VI$ $(i)$.
According to Ohm's law,$V = IR$ $(ii)$.
Where,
$V =$ Potential difference
$I =$ Current
$R =$ Resistance
From equation $(i)$,substituting $V = IR$,we get:
$P = (IR) \times I = I^2R$.
From equation $(ii)$,substituting $I = \frac{V}{R}$,we get:
$P = V \times \frac{V}{R} = \frac{V^2}{R}$.
Therefore,the expressions for power are $P = VI = I^2R = \frac{V^2}{R}$.
The term $IR^2$ does not represent electrical power.

(A) The power consumed by an electrical appliance is given by the formula $P = \frac{V^2}{R}$.
First,we calculate the resistance $R$ of the bulb using its rated values:
$R = \frac{V^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484\, \Omega$.
The resistance of the bulb remains constant regardless of the operating voltage.
Now,when the bulb is operated at a new voltage $V' = 110\, V$,the new power consumed $P'$ is:
$P' = \frac{(V')^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25\, W$.
Therefore,the power consumed will be $25\, W$.

(B) Heat produced in the circuit is inversely proportional to the resistance $R$ for a constant potential difference $V$,given by the formula $H = \frac{V^2}{R} t$.
Let $R$ be the resistance of each wire. If connected in series,the equivalent resistance is $R_S = R + R = 2R$.
If connected in parallel,the equivalent resistance is $R_P = \frac{R \times R}{R + R} = \frac{R}{2}$.
The ratio of heat produced in series $(H_S)$ to parallel $(H_P)$ is given by:
$\frac{H_S}{H_P} = \frac{V^2 / R_S \times t}{V^2 / R_P \times t} = \frac{R_P}{R_S}$.
Substituting the values of $R_S$ and $R_P$:
$\frac{H_S}{H_P} = \frac{R/2}{2R} = \frac{1}{4}$.
Thus,the ratio of heat produced is $1:4$.

(B) To measure the potential difference between two points in an electrical circuit,a voltmeter must be connected in parallel across those two points.
This is because a voltmeter has a very high resistance,which ensures that it does not draw significant current from the circuit,thereby maintaining the accuracy of the potential difference measurement.

(A) The resistance $(R)$ of a copper wire of length $l$ and cross-sectional area $A$ is given by the formula:
$R = \rho \frac{l}{A}$
Given:
Resistivity of copper,$\rho = 1.6 \times 10^{-8}\, \Omega \, m$
Diameter,$d = 0.5\, mm = 5 \times 10^{-4}\, m$
Resistance,$R = 10\, \Omega$
Area of cross-section,$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{5 \times 10^{-4}}{2}\right)^2 = \pi \times 6.25 \times 10^{-8}\, m^2$
Calculating length $(l)$:
$l = \frac{R \times A}{\rho} = \frac{10 \times 3.14 \times 6.25 \times 10^{-8}}{1.6 \times 10^{-8}} = \frac{196.25}{1.6} \approx 122.7\, m$
If the diameter is doubled,the new diameter $d' = 2d = 1.0\, mm = 10^{-3}\, m$.
Since $R \propto \frac{1}{d^2}$,if the diameter is doubled,the area $A$ becomes $4$ times larger.
Therefore,the new resistance $R' = \frac{R}{4} = \frac{10}{4} = 2.5\, \Omega$.

(N/A) The plot between voltage and current is called $IV$ characteristic. The voltage is plotted on the $x-$axis and current is plotted on the $y-$axis.
The $IV$ characteristic of the given resistor is plotted in the figure.
The slope of the line gives the value of $\frac{1}{R}$ as,
Slope $= \frac{1}{R} = \frac{\Delta I}{\Delta V} = \frac{BC}{AC} = \frac{3.0 - 1.0}{10.2 - 3.4} = \frac{2.0}{6.8}$
Therefore,the resistance $R$ is,
$R = \frac{6.8}{2.0} = 3.4\, \Omega$
Thus,the resistance of the resistor is $3.4\, \Omega$.

(B) Resistance $(R)$ of a resistor is given by Ohm's law as:
$V = IR$
$R = \frac{V}{I}$
Given:
Potential difference,$V = 12\, V$
Current in the circuit,$I = 2.5\, mA = 2.5 \times 10^{-3}\, A$
Substituting the values in the formula:
$R = \frac{12}{2.5 \times 10^{-3}} = \frac{12}{0.0025} = 4800\, \Omega$
$R = 4.8 \times 10^3\, \Omega = 4.8\, k\Omega$
Therefore,the resistance of the resistor is $4.8\, k\Omega$.

(C) In a series circuit,the current remains the same through all components. According to Ohm's law,the current $I$ is given by $I = \frac{V}{R_{eq}}$.
The equivalent resistance $R_{eq}$ for resistors connected in series is the sum of individual resistances:
$R_{eq} = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \, \Omega$.
Given the potential difference $V = 9 \, V$,the current is:
$I = \frac{9}{13.4} \approx 0.671 \, A$.
Thus,the current flowing through the $12 \, \Omega$ resistor is approximately $0.67 \, A$.

(D) According to Ohm's law,the equivalent resistance $R$ required for the circuit is given by $V = I R$,which implies $R = V / I$.
Given supply voltage $V = 220 \,V$ and current $I = 5 \,A$,the required equivalent resistance is $R = 220 / 5 = 44 \,\Omega$.
When $x$ resistors of resistance $r = 176 \,\Omega$ are connected in parallel,the equivalent resistance $R$ is given by $1 / R = x / r$.
Substituting the values,we get $1 / 44 = x / 176$.
Solving for $x$,we get $x = 176 / 44 = 4$.
Therefore,$4$ resistors of $176 \,\Omega$ are required.

(N/A) To obtain the desired equivalent resistance,we analyze different combinations of the three $6 \, \Omega$ resistors.
$(i)$ To get $9 \, \Omega$:
We connect two $6 \, \Omega$ resistors in parallel and the third $6 \, \Omega$ resistor in series with this combination.
Equivalent resistance of the parallel part: $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \implies R_p = 3 \, \Omega$.
Total resistance: $R_{eq} = R_p + 6 \, \Omega = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega$.
$(ii)$ To get $4 \, \Omega$:
We connect two $6 \, \Omega$ resistors in series and the third $6 \, \Omega$ resistor in parallel with this combination.
Equivalent resistance of the series part: $R_s = 6 \, \Omega + 6 \, \Omega = 12 \, \Omega$.
Total resistance: $\frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{6} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} = \frac{1}{4} \implies R_{eq} = 4 \, \Omega$.

(B) The power rating of each bulb is $P_1 = 10\, W$ at a voltage of $V = 220\, V$.
The current drawn by each bulb is $I_1 = \frac{P_1}{V} = \frac{10}{220} = \frac{1}{22}\, A$.
The maximum allowable current in the circuit is $I_{total} = 5\, A$.
Since the bulbs are connected in parallel,the total current is the sum of the currents through each bulb.
If $n$ is the number of bulbs,then $I_{total} = n \times I_1$.
$5 = n \times \frac{1}{22}$.
$n = 5 \times 22 = 110$.
Therefore,$110$ electric bulbs can be connected in parallel.

(N/A) Supply voltage,$V = 220\, V$.
Resistance of each coil,$R = 24\,\Omega$.
$(i)$ When coils are used separately:
According to Ohm's law,$I = V / R$.
$I_1 = 220 / 24 = 9.17\, A$.
$(ii)$ When coils are connected in series:
Total resistance,$R_s = R + R = 24 + 24 = 48\,\Omega$.
$I_2 = V / R_s = 220 / 48 = 4.58\, A$.
$(iii)$ When coils are connected in parallel:
Total resistance,$1 / R_p = 1 / R + 1 / R = 1 / 24 + 1 / 24 = 2 / 24 = 1 / 12$.
So,$R_p = 12\,\Omega$.
$I_3 = V / R_p = 220 / 12 = 18.33\, A$.

(C) Case $(i)$: Potential difference,$V = 6 \,V$. The $1 \,\Omega$ and $2 \,\Omega$ resistors are connected in series. The equivalent resistance $R = 1 + 2 = 3 \,\Omega$. According to Ohm's law,$I = V/R = 6/3 = 2 \,A$. In a series circuit,the current remains the same through all components. Thus,the current through the $2 \,\Omega$ resistor is $2 \,A$. Power $P = I^2 R = (2)^2 \times 2 = 8 \,W$.
Case $(ii)$: Potential difference,$V = 4 \,V$. The $12 \,\Omega$ and $2 \,\Omega$ resistors are connected in parallel. In a parallel circuit,the voltage across each component is the same as the source voltage. Thus,the voltage across the $2 \,\Omega$ resistor is $4 \,V$. Power $P = V^2/R = (4)^2/2 = 16/2 = 8 \,W$.
Conclusion: The power used in the $2 \,\Omega$ resistor is $8 \,W$ in both cases.

(A) Both the bulbs are connected in parallel. Therefore,the potential difference across each of them will be $220\, V$,because the voltage remains the same in a parallel circuit.
For the first bulb $(100\, W, 220\, V)$:
$I_1 = \frac{P_1}{V} = \frac{100}{220}\, A$
For the second bulb $(60\, W, 220\, V)$:
$I_2 = \frac{P_2}{V} = \frac{60}{220}\, A$
Total current drawn from the line $(I)$ is the sum of currents through both bulbs:
$I = I_1 + I_2 = \frac{100}{220} + \frac{60}{220} = \frac{160}{220}\, A$
$I \approx 0.727\, A$
Rounding to two decimal places,we get $0.73\, A$.

(A) Energy consumed by an electrical appliance is given by the formula, $H = P \times t$.
For the $TV$ set:
Power $(P)$ = $250 \,W$
Time $(t)$ = $1 \,hr = 3600 \,s$
Energy = $250 \,W \times 3600 \,s = 900,000 \,J = 9 \times 10^5 \,J$.
For the toaster:
Power $(P)$ = $1200 \,W$
Time $(t)$ = $10 \,minutes = 10 \times 60 \,s = 600 \,s$
Energy = $1200 \,W \times 600 \,s = 720,000 \,J = 7.2 \times 10^5 \,J$.
Comparing the two values, $9 \times 10^5 \,J > 7.2 \times 10^5 \,J$.
Therefore, the $250 \,W$ $TV$ set consumes more energy.

(C) The rate at which heat is developed in an electric heater is defined as its electric power $(P)$.
The formula for electric power is given by:
$P = I^2 R$
Given:
Resistance $(R)$ = $8 \, \Omega$
Current $(I)$ = $15 \, A$
Substituting the values into the formula:
$P = (15)^2 \times 8$
$P = 225 \times 8$
$P = 1800 \, W$
Since $1 \, W = 1 \, J/s$,the rate at which heat is developed is $1800 \, J/s$ or $1800 \, W$.

(N/A) The melting point and resistivity of tungsten are very high. It does not oxidize or burn readily at high temperatures. Electric lamps glow at very high temperatures,and tungsten can withstand these temperatures without melting. Hence,tungsten is used as the filament of electric bulbs.
$(b)$ The conductors of electric heating devices,such as bread-toasters and electric irons,are made of an alloy because the resistivity of an alloy is significantly higher than that of pure metals. Furthermore,alloys do not oxidize (burn) readily at high temperatures even when they are red hot. This property allows them to produce a large amount of heat efficiently.

(N/A) In a series circuit,the total voltage is divided among the components. If one component fails or is switched off,the entire circuit breaks,and no current flows. Furthermore,each device receives a lower voltage than required,and the total resistance increases,which reduces the current. Therefore,series arrangement is not used in domestic circuits.
$(b)$ The resistance $(R)$ of a wire is inversely proportional to its area of cross-section $(A)$. This is expressed as $R \propto \frac{1}{A}$. As the area of cross-section increases,the resistance decreases,and vice versa.

(N/A) Copper and aluminium are metals that possess very low electrical resistivity,which means they offer minimal resistance to the flow of electric current.
Because of this property,they act as excellent conductors of electricity.
Consequently,they are widely used in electrical wiring and transmission lines to ensure efficient power delivery with minimal energy loss as heat.