Mix Examples - Electricity Questions in English

(A) When $R_{1}$ and $R_{2}$ are connected in parallel,the net resistance $(R_{p})$ is given by:
$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
Therefore,we have:
$R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = 3 \, \Omega$ $...(1)$
When $R_{1}$ and $R_{2}$ are connected in series,the net resistance $(R_{s})$ is given by:
$R_{s} = R_{1} + R_{2} = 16 \, \Omega$ $...(2)$
Substituting equation $(2)$ into equation $(1)$:
$\frac{R_{1} R_{2}}{16} = 3 \implies R_{1} R_{2} = 48$ $...(3)$
From equation $(2)$,$R_{2} = 16 - R_{1}$. Substituting this into equation $(3)$:
$R_{1}(16 - R_{1}) = 48$
$16R_{1} - R_{1}^{2} = 48$
$R_{1}^{2} - 16R_{1} + 48 = 0$
$(R_{1} - 12)(R_{1} - 4) = 0$
Thus,$R_{1} = 12 \, \Omega$ or $R_{1} = 4 \, \Omega$.
If $R_{1} = 12 \, \Omega$,then $R_{2} = 4 \, \Omega$. If $R_{1} = 4 \, \Omega$,then $R_{2} = 12 \, \Omega$.
Therefore,the resistances are $4 \, \Omega$ and $12 \, \Omega$.

(N/A) Given: Voltage $V = 200 \ V$, Power $P = 100 \ W$, Number of bulbs $n = 5$, Time $t = 4 \ hours$.
$(a)$ We know that $P = \frac{V^2}{R}$, therefore, the resistance of each bulb is:
$R = \frac{V^2}{P} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \ \Omega$.
$(b)$ Electrical energy consumed by one bulb in $4 \ hours$ is:
$E_1 = P \times t = 100 \ W \times 4 \ h = 400 \ Wh$.
Total energy consumed by $5$ bulbs in $4 \ hours$ is:
$E_{total} = 5 \times 400 \ Wh = 2000 \ Wh = 2 \ kWh$.
$(c)$ Cost of consumed electricity:
Since $1 \ unit = 1 \ kWh$ and the rate is $50 \ paise = ₹ 0.50$ per unit:
$Cost = 2 \ units \times ₹ 0.50/unit = ₹ 1.00$.

(A) Given: Voltage $V = 5\,V$,Current $I = 500\,mA = 0.5\,A$,Time $t = 4\,\text{hours} = 4 \times 3600\,s = 14400\,s$.
$(i)$ Power $P = V \times I = 5\,V \times 0.5\,A = 2.5\,W$.
$(ii)$ Resistance $R = V / I = 5\,V / 0.5\,A = 10\,\Omega$.
$(iii)$ Energy consumed $E = P \times t = 2.5\,W \times 14400\,s = 36000\,J$.

(N/A) $(i)$ Resistors of value $10 \ \Omega$ and $20 \ \Omega$ are connected in series,therefore,their equivalent resistance is
$R_{S} = 10 + 20 = 30 \ \Omega$
Similarly,resistors of value $5 \ \Omega$ and $25 \ \Omega$ are in series,therefore,their equivalent resistance is
$R_{S_{1}} = 5 + 25 = 30 \ \Omega$
Now,$R_{S}$ and $R_{S_{1}}$ are in parallel,therefore,we have
$R_{P} = \frac{R_{S} \times R_{S_{1}}}{R_{S} + R_{S_{1}}} = \frac{30 \times 30}{30 + 30} = \frac{900}{60} = 15 \ \Omega$
$(ii)$ The total current is given by Ohm's law:
$I = \frac{V}{R_{P}} = \frac{12}{15} = 0.8 \ A$

(B) When electric lamps are connected in parallel,the total power consumed by the circuit is the sum of the power of individual lamps.
Total power,$P = P_{1} + P_{2} = 100\, W + 25\, W = 125\, W$.
The relationship between power $(P)$,voltage $(V)$,and current $(I)$ is given by the formula $P = V \times I$.
To find the total current $(I)$,we rearrange the formula: $I = \frac{P}{V}$.
Substituting the given values: $I = \frac{125\, W}{200\, V} = 0.625\, A$.
Therefore,the total current flowing through the circuit is $0.625\, A$.

(B) Given: $R_{1} = 2 \Omega, R_{2} = 2 \Omega, V = 12 \text{ V}$.
$(i)$ When resistors are connected in series,the equivalent resistance is:
$R_{S} = R_{1} + R_{2} = 2 + 2 = 4 \Omega$.
Power consumed in series is:
$P_{S} = \frac{V^{2}}{R_{S}} = \frac{12^{2}}{4} = \frac{144}{4} = 36 \text{ W}$.
$(ii)$ When resistors are connected in parallel,the equivalent resistance is:
$R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1 \Omega$.
Power consumed in parallel is:
$P_{P} = \frac{V^{2}}{R_{P}} = \frac{12^{2}}{1} = 144 \text{ W}$.
The ratio of power consumed in the two cases (parallel to series) is:
$\frac{P_{P}}{P_{S}} = \frac{144}{36} = 4:1$.

(N/A) $(i)$ Since $R_{1}$ and $R_{2}$ are in series,their resultant resistance,
$R_{S} = R_{1} + R_{2} = 3 + 2 = 5 \Omega$
Further,$R_{S}$ and $R_{3}$ are in parallel,their resultant resistance is given by
$R_{P} = \frac{R_{S} \times R_{3}}{R_{S} + R_{3}} = \frac{5 \times 5}{5 + 5} = 2.5 \Omega$
(ii) Also,the current $I$ shown by the ammeter is given by Ohm's law:
$I = \frac{V}{R_{P}} = \frac{2.5 \text{ V}}{2.5 \Omega} = 1 \text{ A}$

$1$. Energy consumed by $5$ tube lights of $40 \text{ W}$ each for $5 \text{ hours}$ per day:
$E_1 = 5 \times 40 \text{ W} \times 5 \text{ h} = 1000 \text{ Wh} = 1 \text{ kWh}$.
$2$. Energy consumed by an electric press of $500 \text{ W}$ for $4 \text{ hours}$ per day:
$E_2 = 500 \text{ W} \times 4 \text{ h} = 2000 \text{ Wh} = 2 \text{ kWh}$.
$3$. Total energy consumed per day:
$E_{\text{total/day}} = 1 \text{ kWh} + 2 \text{ kWh} = 3 \text{ kWh}$.
$4$. Total energy consumed in a month of $30 \text{ days}$:
$E_{\text{month}} = 3 \text{ kWh/day} \times 30 \text{ days} = 90 \text{ kWh}$.

(N/A) In arm $CE$, the resistors of value $2 \, \Omega$ and $4 \, \Omega$ are connected in series. Therefore, the total resistance in arm $CE = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega$.
$(b)$ The circuit consists of two parallel branches, arm $CE$ and arm $AB$, each having a resistance of $6 \, \Omega$. The total resistance of the circuit $R_P$ is given by:
$R_P = \frac{6 \, \Omega \times 6 \, \Omega}{6 \, \Omega + 6 \, \Omega} = 3 \, \Omega$.
Therefore, the total current $I$ drawn from the battery is:
$I = \frac{V}{R_P} = \frac{3 \, V}{3 \, \Omega} = 1 \, A$.
$(c)$ Since both arms $CE$ and $AB$ have the same resistance $(6 \, \Omega)$, the total current is split equally between them. Hence, the current in each arm ($AB$ and $CE$) is $0.5 \, A$.

(N/A) Since the wire is divided into four equal parts,the new length of each part becomes one-fourth of the original length. Since resistance is directly proportional to length $(R \propto l)$,the resistance of each part becomes one-fourth of the original resistance,$i.e., R' = R/4$.
$(b)$ When these four parts are connected in parallel,the equivalent resistance $R_p$ is given by the formula:
$\frac{1}{R_p} = \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'}$
Substituting $R' = R/4$:
$\frac{1}{R_p} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R} = \frac{16}{R}$
Therefore,$R_p = R/16$.
Thus,the joint resistance is $1/16$ times the resistance of the original wire.

(N/A) $(i)$ The resistance $R$ of the lamp is calculated using the formula $R = \frac{V^2}{P}$. Substituting the given values: $R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega$.
$(ii)$ The energy consumed $E$ is calculated using the formula $E = P \times t$. Given $P = 100 \, W$ and $t = 5 \, h$,we get $E = 100 \, W \times 5 \, h = 500 \, Wh$. Converting to $kWh$,$E = \frac{500}{1000} \, kWh = 0.5 \, kWh$.

(A) Given: Length $L = 1.0 \, m$,Diameter $D = 0.2 \, mm = 0.2 \times 10^{-3} \, m$.
Radius $r = D/2 = 0.1 \times 10^{-3} \, m = 10^{-4} \, m$.
Resistance $R = 10 \, \Omega$.
Formula for resistivity: $\rho = \frac{R \cdot A}{L}$,where $A = \pi r^2$.
Substituting the values: $A = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8} \, m^2$.
$\rho = \frac{10 \times 3.14 \times 10^{-8}}{1.0} = 3.14 \times 10^{-7} \, \Omega \cdot m$.

(A) Given: Current $I = 10 \ A$,Potential difference $V = 220 \ V$.
According to Ohm's law,the relationship between voltage,current,and resistance is given by $V = I \times R$.
To find the resistance $R$,we rearrange the formula: $R = V / I$.
Substituting the given values: $R = 220 \ V / 10 \ A$.
Therefore,$R = 22 \ \Omega$.

(N/A) Given: Power $P = 1100 \ W$,Voltage $V = 220 \ V$.
$1$. Calculation of Resistance:
Using the formula $R = \frac{V^2}{P}$,we have:
$R = \frac{(220)^2}{1100} = \frac{48400}{1100} = 44 \ \Omega$.
$2$. Calculation of Energy Consumed:
The month of November has $30$ days.
Total time $t = 4 \ \text{hours/day} \times 30 \ \text{days} = 120 \ \text{hours}$.
Energy $E = P \times t = 1100 \ \text{W} \times 120 \ \text{h} = 132000 \ \text{Wh}$.
To convert into $kWh$,divide by $1000$:
$E = \frac{132000}{1000} = 132 \ kWh$.

(N/A) $(i)$ According to Ohm's law,the current $I$ through a resistor is given by $I = V / R$. Since the resistors are connected in parallel,the potential difference across each resistor is the same,i.e.,$V = 6\, V$.
For the $30\, \Omega$ resistor:
$I = 6\, V / 30\, \Omega = 0.2\, A$.
$(ii)$ For a parallel combination of resistors $R_1 = 5\, \Omega$,$R_2 = 10\, \Omega$,and $R_3 = 30\, \Omega$,the equivalent resistance $R_P$ is given by:
$1 / R_P = 1 / R_1 + 1 / R_2 + 1 / R_3$
$1 / R_P = 1 / 5 + 1 / 10 + 1 / 30$
$1 / R_P = (6 + 3 + 1) / 30 = 10 / 30 = 1 / 3$
Therefore,$R_P = 3\, \Omega$.
$(b)$ Two advantages of connecting electrical devices in parallel are:
$(i)$ If one device fails or is switched off,the others continue to work.
$(ii)$ Each device gets the full voltage of the power supply.

(3 A) The power dissipated by a resistor is given by the formula $P = I^2 R$.
Given that the maximum power each resistor can withstand is $P_{max} = 18 \,W$ and the resistance is $R = 2 \,\Omega$.
First,calculate the maximum current $I_{max}$ that any single resistor can handle:
$I_{max}^2 = P_{max} / R = 18 / 2 = 9$
$I_{max} = \sqrt{9} = 3 \,A$.
In the given circuit,resistor $A$ is in series with the parallel combination of resistors $B$ and $C$.
The total current $I$ flows through resistor $A$. Therefore,the maximum current $I$ that can flow through the circuit is limited by the maximum current capacity of resistor $A$,which is $3 \,A$.
If $I = 3 \,A$ flows through the circuit,then at the parallel junction,the current splits equally between $B$ and $C$ (since $R_B = R_C = 2 \,\Omega$).
Thus,current through $B$ is $I_B = 1.5 \,A$ and current through $C$ is $I_C = 1.5 \,A$.
Since $1.5 \,A < 3 \,A$,both $B$ and $C$ are safe.
Therefore,the maximum current that can flow through the circuit is $3 \,A$.

(N/A) The two $8\, \Omega$ resistors are connected in parallel. Their effective resistance $R_P$ is given by:
$R_P = \frac{R_1 R_2}{R_1 + R_2} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\, \Omega$
$(b)$ The total resistance of the circuit $R_{eq} = R_{series} + R_P = 4\, \Omega + 4\, \Omega = 8\, \Omega$. The total current $I$ in the circuit is:
$I = \frac{V}{R_{eq}} = \frac{8\, V}{8\, \Omega} = 1\, A$. Since the $4\, \Omega$ resistor is in series with the rest of the circuit,the same current flows through it.
$(c)$ The potential difference $V$ across the $4\, \Omega$ resistor is:
$V = I \times R = 1\, A \times 4\, \Omega = 4\, V$
$(d)$ The power $P$ dissipated in the $4\, \Omega$ resistor is:
$P = I^2 R = (1\, A)^2 \times 4\, \Omega = 4\, W$
$(e)$ There is no difference in the ammeter readings. In a series circuit,the current remains the same at all points,so both ammeters $A_1$ and $A_2$ will show the same reading of $1\, A$.

(B) Given:
Power $(P)$ = $600 \ W = 0.6 \ kW$
Time $(t)$ = $6 \ hours/day \times 30 \ days = 180 \ hours$
Energy consumed $(E)$ = $P \times t = 0.6 \ kW \times 180 \ h = 108 \ kWh$
Since $1 \ unit = 1 \ kWh$, the total energy consumed is $108 \ units$.
Cost = Energy consumed $\times$ Rate per unit
Cost = $108 \times 3 = ₹ \ 324$.

(N/A) The circuit diagram is as shown in the image.
Given:
$R_1 = 5 \, \Omega$,$R_2 = 10 \, \Omega$,$R_3 = 15 \, \Omega$
$V = 30 \, V$
Since the resistors are connected in series,the total resistance $R$ is:
$R = R_1 + R_2 + R_3$
$R = 5 + 10 + 15 = 30 \, \Omega$
The current $I$ flowing through the circuit is given by Ohm's Law:
$I = V / R = 30 \, V / 30 \, \Omega = 1 \, A$
The potential difference $V'$ across the $10 \, \Omega$ resistor is:
$V' = I \times R_2 = 1 \, A \times 10 \, \Omega = 10 \, V$

(1) From the diagram,we observe that the following pairs of resistors are in parallel:
$R_1$ and $R_2$,$R_3$ and $R_4$,$R_5$ and $R_6$,and $R_7$ and $R_8$.
Since each resistor has a value of $2 \, \Omega$,the equivalent resistance for each parallel pair is:
$\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{2} = 1 \, \Omega \implies R_{12} = 1 \, \Omega$.
Similarly,$R_{34} = 1 \, \Omega$,$R_{56} = 1 \, \Omega$,and $R_{78} = 1 \, \Omega$.
Now,$R_{12}$ and $R_{34}$ are in series,so their equivalent resistance is:
$R_{1234} = R_{12} + R_{34} = 1 + 1 = 2 \, \Omega$.
Also,$R_{56}$ and $R_{78}$ are in series,so their equivalent resistance is:
$R_{5678} = R_{56} + R_{78} = 1 + 1 = 2 \, \Omega$.
Finally,$R_{1234}$ and $R_{5678}$ are in parallel between points $A$ and $B$. The net resistance $R_{eq}$ is:
$\frac{1}{R_{eq}} = \frac{1}{R_{1234}} + \frac{1}{R_{5678}} = \frac{1}{2} + \frac{1}{2} = 1 \, \Omega^{-1}$.
Therefore,the equivalent resistance is $1 \, \Omega$.

(N/A) Given: Voltage $V = 200 \, V$,Resistance of coil $A$ $(R_A)$ = $30 \, \Omega$,Resistance of coil $B$ $(R_B)$ = $30 \, \Omega$.
$(i)$ When coil $A$ or $B$ is used separately:
$I = V / R = 200 / 30 = 6.67 \, A$.
(ii) When the coils are connected in series:
Total resistance $R_S = R_A + R_B = 30 + 30 = 60 \, \Omega$.
$I_S = V / R_S = 200 / 60 = 3.33 \, A$.
(iii) When the coils are connected in parallel:
Total resistance $1 / R_P = 1 / R_A + 1 / R_B = 1 / 30 + 1 / 30 = 2 / 30 = 1 / 15$.
$R_P = 15 \, \Omega$.
$I_P = V / R_P = 200 / 15 = 13.33 \, A$.

(B) Power of the electric heater $= 1200 \ W = 1.2 \ kW$.
Time used per day $= 2 \ h$.
Energy consumed per day $= \text{Power} \times \text{Time} = 1.2 \ kW \times 2 \ h = 2.4 \ kWh$.
Total energy consumed in $30$ days $= 2.4 \ kWh/day \times 30 \ days = 72 \ kWh$.
Since $1 \ kWh = 1 \ \text{unit}$, total energy $= 72 \ \text{units}$.
Cost of energy $= \text{Total units} \times \text{Rate per unit} = 72 \times 5 = ₹ \ 360$.

(N/A) An electric circuit is a continuous and closed path in which an electric current flows.
The circuit diagram consists of a battery of two cells,a resistor,an ammeter,and a closed plug key connected in series. The ammeter is connected in series with the resistor,and the positive terminal of the battery is connected to the positive terminal of the ammeter.

(N/A) The bulb will glow only when the piece of silver is placed between points $A$ and $B$.
Silver is a metal and acts as a good conductor of electricity,which completes the electrical circuit and allows current to flow through the bulb.
In contrast,wood is an insulator (a bad conductor of electricity). When a piece of wood is placed between points $A$ and $B$,it breaks the electrical circuit,preventing the flow of current,and therefore the bulb does not glow.

(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 temperature of the conductor does not change significantly. Therefore, the resistance remains constant.
$(ii)$ When a heavy current passes through a conductor for about $30$ seconds, the conductor heats up due to the heating effect of electric current $(H = I^2Rt)$. Since the resistance of a metallic conductor increases with an increase in temperature, the resistance of the conductor will increase.

(N/A) The circuit diagram is as shown in the image.
Activity to find the resistance $R$:
$1$. Set up a circuit using the wire of resistance $R$,an ammeter,a voltmeter,a plug key,and four cells of $1.5 \, V$ each.
$2$. First,use one cell as the source and note the readings of the ammeter $(I)$ and voltmeter $(V)$.
$3$. Increase the number of cells one by one and note the corresponding readings of $I$ and $V$.
$4$. Calculate the ratio $V/I$ for each set of readings. The average of these ratios gives the resistance $R$ of the wire.
Numerical calculation:
Given: Current $(I) = 0.2 \, A$,Resistance $(R) = 20 \, \Omega$.
Using Ohm's Law: $V = I \times R$
$V = 0.2 \, A \times 20 \, \Omega = 4 \, V$.
Thus,the potential difference across the ends of the resistor is $4 \, V$.

$(A)$ Sample $A$ is a good conductor, and sample $B$ is an insulator.
Metals and alloys typically have low resistivity in the range of $10^{-8} \ \Omega \cdot m$ to $10^{-6} \ \Omega \cdot m$, making them good conductors of electricity. In contrast, insulators possess very high resistivity, typically in the order of $10^{12} \ \Omega \cdot m$ to $10^{17} \ \Omega \cdot m$.
$(b)$ When a wire of resistance $R = 4 \ \Omega$ is folded on itself, its length $l$ becomes $l' = l/2$ and its cross-sectional area $A$ becomes $A' = 2A$.
The resistance formula is $R = \rho (l/A)$.
The new resistance $R'$ is given by:
$R' = \rho (l'/A') = \rho ((l/2) / (2A)) = (1/4) \times \rho (l/A) = R/4$.
Substituting $R = 4 \ \Omega$:
$R' = 4 \ \Omega / 4 = 1 \ \Omega$.
Therefore, the new resistance is $1 \ \Omega$.

(B) The heat produced per unit time is given by the power formula $P = \frac{V^2}{R}$, where $V$ is the potential difference and $R$ is the resistance.
Since both wires are connected to the same battery, $V$ is constant for both.
Therefore, $P \propto \frac{1}{R}$, meaning the wire with lower resistance will produce more heat.
In a $V-I$ graph, the slope represents the resistance $(R = \frac{V}{I})$. The slope of wire $A$ is greater than the slope of wire $B$, which means $R_A > R_B$.
Since $R_B < R_A$, wire $B$ will have a lower resistance and thus will produce more heat per unit time.

(A) $(i)$ The current in the parallel branches is inversely proportional to the resistance of the branch. The branches containing ammeters $B$ and $C$ have resistances of $6\,\Omega$ and $3\,\Omega$ respectively.
$\frac{\text{Reading of ammeter } C}{\text{Reading of ammeter } B} = \frac{6\,\Omega}{3\,\Omega} = 2$
Therefore,the reading of ammeter $C = 2 \times 0.5\, A = 1.0\, A$.
The reading of ammeter $A$ is the total current in the circuit,which is the sum of currents in the parallel branches:
Reading of ammeter $A = (0.5 + 1.0)\, A = 1.5\, A$.
$(ii)$ Let the total resistance of the circuit be $R$. The circuit consists of a $2\,\Omega$ resistor in series with a parallel combination of $3\,\Omega$ and $6\,\Omega$ resistors.
First,calculate the equivalent resistance of the parallel part $(R_p)$:
$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_p = 2\,\Omega$
Now,the total resistance $R = 2\,\Omega + R_p = 2\,\Omega + 2\,\Omega = 4\,\Omega$.

(A) In a series combination,the equivalent resistance $R_{s} = R_{1} + R_{2}$ is greater than the individual resistances. In a parallel combination,the equivalent resistance $R_{p} = (R_{1}R_{2}) / (R_{1} + R_{2})$ is smaller than the individual resistances.
$1$. For the $V-I$ graph (top): The slope of the graph represents resistance $(R = V/I)$. Since $R_{s} > R_{p}$,the line for 'series' must have a steeper slope than the line for 'parallel'. Thus,the top graph is correctly labelled.
$2$. For the $I-V$ graph (bottom): The slope of the graph represents conductance $(1/R = I/V)$. Since $R_{s} > R_{p}$,the conductance for parallel $(1/R_{p})$ is greater than for series $(1/R_{s})$. Therefore,the line for 'parallel' must have a steeper slope than the line for 'series'. Thus,the bottom graph is also correctly labelled.
Conclusion: Both graphs are correctly labelled.

(D) Reasons for parallel connection:
$(i)$ Each appliance receives the same potential difference (voltage) as the supply source.
$(ii)$ If one appliance fails or is switched off,the others continue to operate independently.
$(b)$ Calculation:
Given: Fuse rating $= 5 \, A$,Power $(P) = 1.5 \, kW = 1500 \, W$,Voltage $(V) = 220 \, V$.
Using the formula $I = \frac{P}{V}$:
$I = \frac{1500}{220} \approx 6.82 \, A$.
Since the current required by the heater $(6.82 \, A)$ is greater than the fuse rating $(5 \, A)$,the fuse will melt (blow off) and the circuit will break.
$(c)$ Required change:
$A$ fuse with a higher rating,typically $10 \, A$,should be installed to accommodate the current drawn by the heater.

(A) The light will increase.
In a series circuit,the total resistance $R$ is the sum of the resistances of individual bulbs $(R = R_1 + R_2 + ... + R_{20})$.
The power consumed by the circuit is given by the formula $P = \frac{V^2}{R}$,where $V$ is the supply voltage and $R$ is the total resistance.
When one bulb is removed (or fused),the total resistance of the circuit decreases because there are now only $19$ bulbs in series.
Since the power $P$ is inversely proportional to the resistance $R$ $(P \propto \frac{1}{R})$,a decrease in total resistance leads to an increase in the total power consumed by the remaining bulbs.
Therefore,the brightness of the light in the room will increase.

(N/A) To get a reading of $2.0\, V$ across the $20\, \Omega$ resistor,we need to arrange the circuit such that the voltage divider rule provides the desired potential difference.
$1$. Connect the two $10\, \Omega$ resistors in parallel. Their equivalent resistance is $R_p = \frac{10 \times 10}{10 + 10} = 5\, \Omega$.
$2$. Connect this parallel combination $(5\, \Omega)$ in series with the $20\, \Omega$ resistor.
$3$. The total resistance of the circuit is $R_{eq} = 5\, \Omega + 20\, \Omega = 25\, \Omega$.
$4$. The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{2.5\, V}{25\, \Omega} = 0.1\, A$.
$5$. The voltage across the $20\, \Omega$ resistor is $V = I \times R = 0.1\, A \times 20\, \Omega = 2.0\, V$.
$6$. Thus,the voltmeter should be connected in parallel across the $20\, \Omega$ resistor as shown in the diagram.

(C) Heating elements in devices like toasters, heaters, and irons require materials with high electrical resistivity so that they can generate significant heat when current passes through them (Joule's heating effect, $H = I^2Rt$).
Comparing the given values:
$A$: $6.84 \times 10^{-8} \, \Omega m$ (Conductor)
$B$: $1.60 \times 10^{-8} \, \Omega m$ (Conductor)
$C$: $1.00 \times 10^{-4} \, \Omega m$ (Alloy/Resistive material)
$D$: $2.50 \times 10^{12} \, \Omega m$ (Insulator)
$E$: $4.40 \times 10^{-5} \, \Omega m$ (Alloy/Resistive material)
$F$: $2.30 \times 10^{17} \, \Omega m$ (Insulator)
Materials $C$ and $E$ have significantly higher resistivity compared to conductors like $A$ and $B$, making them suitable for heating elements. Among the options provided, $C$ is a standard choice for such applications.

(N/A) $(i)$ The net resistance of a series combination of resistors is greater than the net resistance of a parallel combination of resistors. Therefore,the resistance of circuit $I$ is greater than the resistance of circuit $II$.
$(ii)$ Since current is inversely proportional to resistance for the same potential difference $(I = V/R)$,the current in circuit $II$ is greater than the current in circuit $I$.
$(iii)$ In a parallel circuit,the potential difference across each resistor is equal. Therefore,the potential difference across each resistor in circuit $II$ is equal.

(A) The resistance $R$ of an electric bulb is calculated using the formula $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power.
For bulb $A$: $R_A = \frac{220^2}{60} \approx 806.67\, \Omega$.
For bulb $B$: $R_B = \frac{220^2}{100} = 484\, \Omega$.
Since the voltage $V$ is constant for both bulbs,the resistance is inversely proportional to the power $(R \propto \frac{1}{P})$.
Therefore,the bulb with the lower power rating $(60\, W)$ will have a higher resistance.
Thus,bulb $A$ has greater resistance.

(B) The heat produced per second in a resistor is given by the power formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Since the resistance $R$ remains constant,the heat produced per second is directly proportional to the square of the potential difference,i.e.,$P \propto V^2$.
If the heat produced per second increases by a factor of $16$,then $P' = 16P$.
Substituting this into the proportionality,we get $16P \propto (V')^2$.
Since $P \propto V^2$,we have $16V^2 = (V')^2$.
Taking the square root of both sides,we get $V' = 4V$.
Therefore,the applied potential difference changes by a factor of $4$.

(B) When bulbs are connected in parallel,the potential difference across each bulb remains the same,which is equal to the supply voltage $(220\, V)$.
The brightness of a bulb depends on the power consumed by it,which is given by the formula $P = V^2 / R$.
Since the voltage $V$ is constant for both bulbs,the power consumed is directly proportional to the rated power of the bulb.
The bulb with the higher wattage rating $(100\, W)$ will consume more power and therefore glow brighter than the $60\, W$ bulb.

(C) In a parallel circuit,the voltage across each component remains the same as the supply voltage $(V)$.
The rate of heat produced by a heater is given by the formula $P = V^2 / R$,where $V$ is the voltage and $R$ is the resistance of the heater.
Since the heater is connected in parallel to the mains,its voltage $V$ remains constant regardless of the bulb connected in parallel with it.
Therefore,replacing the $60\, W$ bulb with a $100\, W$ bulb does not change the voltage across the heater.
Consequently,the rate of heat produced by the heater remains the same.

(B) The rate of heat produced in a conductor is given by the formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Since both conductors are connected across the same $dc$ source,the potential difference $V$ remains constant for both.
Therefore,the rate of heat produced is inversely proportional to the resistance: $Q \propto \frac{1}{R}$.
For the first conductor with resistance $R_1 = R$,the rate of heat is $Q_1 = \frac{V^2}{R}$.
For the second conductor with resistance $R_2 = 2R$,the rate of heat is $Q_2 = \frac{V^2}{2R}$.
Calculating the ratio $Q_1 / Q_2$:
$\frac{Q_1}{Q_2} = \frac{V^2 / R}{V^2 / 2R} = \frac{V^2}{R} \times \frac{2R}{V^2} = 2$.
Thus,the value of $Q_1 / Q_2$ is $2$.

(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.
Since both wires are of the same metal,their resistivity $\rho$ is the same. Given that $A$ is the same for both,$R \propto L$.
Therefore,the ratio of resistances is $\frac{R_A}{R_B} = \frac{L_A}{L_B} = \frac{2}{1}$.
According to Ohm's Law,$I = \frac{V}{R}$. Since the potential difference $V$ is the same for both,$I \propto \frac{1}{R}$.
Thus,the ratio of currents is $\frac{I_A}{I_B} = \frac{R_B}{R_A} = \frac{1}{2}$.

(N/A) Given: Rated voltage $V = 220 \, V$,Rated power $P = 60 \, W$.
$(i)$ Using the formula $P = \frac{V^2}{R}$,the resistance $R$ of the bulb is:
$R = \frac{V^2}{P} = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega$.
$(ii)$ When two identical bulbs are connected in series,the total resistance $R_T = R + R = 2R = 2 \times 806.67 = 1613.34 \, \Omega$.
The current $I$ flowing through the circuit is $I = \frac{V}{R_T} = \frac{220}{1613.34} \approx 0.136 \, A$.
The rate of energy conversion (power consumed) in each bulb is $P' = I^2 R = (0.136)^2 \times 806.67 \approx 15 \, W$.
$(iii)$ Total power consumed in series connection is $P_{total} = I^2 R_T = (0.136)^2 \times 1613.34 \approx 30 \, W$.
$(iv)$ If the bulbs are connected in parallel,the total power is $P_{total} = P_1 + P_2 = 60 + 60 = 120 \, W$.

(A) For a metallic wire,the slope of the $I-V$ graph represents the conductance,which is the reciprocal of resistance $(R = V/I)$.
From the graph,the slope is given by $\Delta I / \Delta V = 1/R$.
As the temperature of a metallic conductor increases,its resistance $(R)$ increases.
Since $R$ increases with temperature,the slope $(1/R)$ decreases as the temperature increases.
Therefore,the line with the smallest slope corresponds to the highest temperature.
In the given graph,line $1$ has the smallest slope,so it represents the highest temperature.

(A) 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. The area $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
$(i)$ When $V$ is halved: The resistance $R$ depends only on the physical dimensions and material properties of the conductor. Therefore,$R$ remains unchanged.
$(ii)$ When $L$ is halved: Since $R \propto L$,if the length $L$ is halved,the resistance $R$ also becomes halved.
$(iii)$ When $D$ is doubled: Since $R \propto \frac{1}{A}$ and $A \propto D^2$,we have $R \propto \frac{1}{D^2}$. If $D$ is doubled,$R$ becomes $\frac{1}{(2)^2} = \frac{1}{4}$ of its original value.

(A) In an $I-V$ graph,the slope represents the conductance,which is the reciprocal of resistance $(1/R)$.
To determine the series combination,draw a vertical line from any point on the $V$ axis such that it intersects the graphs $1, 2,$ and $3$ at points $P, Q,$ and $R$ respectively,as shown in the provided solution image.
For a constant potential $V$,the currents corresponding to the three graphs are $I_1, I_2,$ and $I_3$.
From the graph,we observe that $I_1 < I_2 < I_3$.
Since $R = V/I$,for a fixed $V$,a smaller current implies a larger resistance. Therefore,$R_1 > R_2 > R_3$.
In a series combination,the equivalent resistance $(R_s = R_a + R_b)$ is always greater than the individual resistances ($R_a$ and $R_b$).
Since graph $1$ has the lowest current for a given potential,it corresponds to the highest resistance. Thus,graph $1$ represents the series combination of the other two resistors (graphs $2$ and $3$).

(B) The graph shows the variation of current $(I)$ on the $y$-axis with respect to voltage $(V)$ on the $x$-axis.
According to Ohm's law, $V = IR$, which implies $I = (1/R)V$.
The slope of the $I-V$ graph is given by $\text{slope} = I/V = 1/R$.
From the graph, the slope of line $X$ is greater than the slope of line $Y$ $(\text{slope}_X > \text{slope}_Y)$.
Since $\text{slope} = 1/R$, this means $1/R_X > 1/R_Y$, which implies $R_X < R_Y$.
Resistance $(R)$ is related to resistivity $(\rho)$ by the formula $R = \rho(L/A)$.
Given that both wires have the same length $(L)$ and the same diameter (and thus the same cross-sectional area $A$), the resistance is directly proportional to the resistivity $(R \propto \rho)$.
Since $R_Y > R_X$, it follows that the resistivity of wire $Y$ is greater than the resistivity of wire $X$ $(\rho_Y > \rho_X)$.

(B) The resistance $R$ of a conductor is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area.
Since both resistors are made of the same material ($\rho$ is constant) and have the same radii ($A$ is constant), the resistance $R$ is directly proportional to the length $L$ $(R \propto L)$.
Given $L_{1} > L_{2}$, it follows that $R_{1} > R_{2}$.
In a $V-I$ graph, the slope represents the resistance $(R = \frac{V}{I} = \text{slope})$. A steeper slope indicates a higher resistance.
Comparing the two graphs, graph $B$ has a greater slope than graph $A$, meaning the resistance of graph $B$ is greater than that of graph $A$.
Since $R_{1} > R_{2}$, graph $B$ corresponds to the longer length $L_{1}$.

(D) Electric current is defined as the flow of electric charge from a region of higher potential to a region of lower potential.
When two charged bodies are connected,the flow of charge occurs due to a potential difference between them.
If both bodies have the same potential,there is no potential difference $(V_1 - V_2 = 0)$.
Consequently,no current will flow between the two bodies.

(A) 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.
When the wire is melted and recast to half its length $(L' = L / 2)$,its volume remains constant.
Since $V = A \times L$,we have $A \times L = A' \times L'$.
Substituting $L' = L / 2$,we get $A \times L = A' \times (L / 2)$,which implies $A' = 2A$.
The new resistance $R'$ is given by $R' = \rho \frac{L'}{A'}$.
Substituting the new values: $R' = \rho \frac{L / 2}{2A} = \frac{1}{4} \times \rho \frac{L}{A}$.
Therefore,$R' = R / 4$.

(B) Let the resistance of each resistor be $R$.
When $n$ resistors are connected in series,the equivalent resistance is $R_{s} = nR$.
When $n$ resistors are connected in parallel,the equivalent resistance is $R_{p} = R / n$.
The ratio of the maximum resistance (series) to the minimum resistance (parallel) is given by:
Ratio $= R_{s} / R_{p} = (nR) / (R / n) = n^{2}$.