Mix Examples-Current Electricity Questions in English

(A) The charge flowing through the resistor is given by $Q = 2t - 8t^2$.
The current $I$ is the rate of change of charge with respect to time:
$I = \frac{dQ}{dt} = \frac{d}{dt}(2t - 8t^2) = 2 - 16t$.
The heat produced $H$ in a resistor $R$ over a time interval is given by the integral:
$H = \int_{0}^{t} I^2 R \ dt$.
Substituting the limits $0$ to $\frac{1}{8}$:
$H = \int_{0}^{1/8} (2 - 16t)^2 R \ dt$
$H = R \int_{0}^{1/8} (4 - 64t + 256t^2) \ dt$
$H = R \left[ 4t - 32t^2 + \frac{256t^3}{3} \right]_{0}^{1/8}$
Evaluating at the limits:
$H = R \left[ 4(\frac{1}{8}) - 32(\frac{1}{8})^2 + \frac{256}{3}(\frac{1}{8})^3 \right]$
$H = R \left[ \frac{1}{2} - 32(\frac{1}{64}) + \frac{256}{3}(\frac{1}{512}) \right]$
$H = R \left[ \frac{1}{2} - \frac{1}{2} + \frac{1}{6} \right] = \frac{R}{6} \ J$.

(A) The rated power $P_r = 60\, W$ at rated voltage $V_r = 120\, V$. The resistance of each bulb is $R = \frac{V_r^2}{P_r} = \frac{120^2}{60} = 240\, \Omega$.
When three identical bulbs are connected in series to a $120\, V$ supply,the total resistance is $R_{eq} = 3R = 3 \times 240 = 720\, \Omega$.
The current flowing through the series circuit is $I = \frac{V}{R_{eq}} = \frac{120}{720} = \frac{1}{6}\, A$.
The power dissipated by each bulb is $P = I^2 R = (\frac{1}{6})^2 \times 240 = \frac{1}{36} \times 240 = \frac{240}{36} = 6.67\, W \approx 6.7\, W$.

(B) From the given figure $A$,we observe that the voltage $V$ increases non-linearly with current $I$,specifically $V$ is proportional to $I^n$ where $n > 1$ (as it is a concave-up curve).
Power $P$ is defined as $P = V \times I$.
Substituting $V \propto I^n$,we get $P \propto I^n \times I = I^{n+1}$.
Since $n > 1$,the exponent $n+1$ is greater than $2$. This implies that the graph of $P$ versus $I$ will also be a concave-up curve,similar to the original $V-I$ graph but with a steeper increase.
Comparing this with the given options,the graph in option $B$ represents an upward-curving (concave-up) relationship,which is consistent with $P \propto I^{n+1}$ where $n+1 > 2$.
Therefore,option $B$ is the correct representation.

(A) potentiometer is considered an ideal device that draws negligible current from the circuit,meaning its effective resistance is infinite. Thus,the initial circuit behaves as if $R$ is connected directly to the cell $C$ through the ammeter $A$.
When the potentiometer is replaced by a voltmeter of finite resistance $R_v$,the voltmeter acts as a resistor in parallel with $R$. The equivalent resistance of the parallel combination $(R || R_v)$ is less than $R$.
Since the total resistance of the circuit decreases,the total current drawn from the ideal cell $C$ increases. Therefore,the ammeter reading $I$ is greater than the initial reading $I_0$ $(I > I_0)$.
Because the ammeter has a finite resistance $R_A$,the voltage drop across the ammeter increases as the current increases. Since the cell $C$ is ideal (constant $EMF$ $E$),the voltage across the parallel combination $(R || R_v)$ is $V = E - I R_A$. Since $I > I_0$,it follows that $V < V_0$.

(C) Let the potential at node $F$ be $0 \ V$. Then the potential at node $C$ is $V_C$. Applying Kirchhoff's Current Law $(KCL)$ at node $C$:
$\frac{V_C - 30}{4 + 2} + \frac{V_C - 0}{3} + \frac{V_C - 15}{1 + 2} = 0$
$\frac{V_C - 30}{6} + \frac{V_C}{3} + \frac{V_C - 15}{3} = 0$
Multiplying by $6$:
$(V_C - 30) + 2V_C + 2(V_C - 15) = 0$
$V_C - 30 + 2V_C + 2V_C - 30 = 0$
$5V_C = 60 \implies V_C = 12 \ V$.
The current in the left branch is $I_1 = \frac{30 - 12}{6} = 3 \ A$.
The current in the middle branch is $I_2 = \frac{12 - 0}{3} = 4 \ A$.
The current in the right branch is $I_3 = \frac{12 - 15}{3} = -1 \ A$ (meaning current flows from $D$ to $C$).
The power consumed by the circuit is the sum of power dissipated in resistors: $P = I_1^2 R_1 + I_2^2 R_2 + I_3^2 R_3$
$P = (3^2 \times 6) + (4^2 \times 3) + ((-1)^2 \times 3) = 54 + 48 + 3 = 105 \ W$.

(B) Let the potential at node $F$ be $0 \ V$. Let the potential at node $C$ be $V_C$. Applying Kirchhoff's Current Law $(KCL)$ at node $C$:
$\frac{V_C - 30}{4 + 2} + \frac{V_C - 6}{0} + \frac{V_C - 15}{1 + 2} = 0$ (Note: The middle branch has only a battery,so $V_C = 6 \ V$).
If $V_C = 6 \ V$,the current in the left loop $(I_1)$ is: $I_1 = \frac{30 - 6}{4 + 2} = \frac{24}{6} = 4 \ A$.
The current in the right loop $(I_2)$ is: $I_2 = \frac{15 - 6}{1 + 2} = \frac{9}{3} = 3 \ A$.
The power dissipated in the resistances is $P = I_1^2 R_{left} + I_2^2 R_{right}$.
$P = (4)^2 \times (4 + 2) + (3)^2 \times (1 + 2) = 16 \times 6 + 9 \times 3 = 96 + 27 = 123 \ W$.

(D) When a battery of $emf$ $E$ and internal resistance $r$ is being charged,the current $I$ flows into the positive terminal of the battery.
The terminal potential difference $V$ across the battery is given by the equation: $V = E + Ir$.
Since $I > 0$ and $r > 0$,it follows that $V > E$. Thus,the potential difference across the charger terminals $A$ and $B$ must be greater than $E$.
Because the positive terminal of the battery is connected to $A$ and the negative terminal to $B$,and current flows from $A$ to $B$ through the battery,$A$ must be at a higher potential than $B$.
Inside the battery during charging,the current flows from the positive terminal to the negative terminal.
Therefore,all the given statements are correct.

(A) The potential difference $V$ across a battery is given by the equation $V = E - ir$.
Comparing this with the equation of a straight line $y = mx + c$,we get $V = -ri + E$.
From the given graph,the intercept on the $V$-axis is $E = 10\,V$.
The intercept on the $i$-axis is the short-circuit current when $V = 0$,which is $i_{max} = E/r = 2\,A$.
Substituting $E = 10\,V$ and $i_{max} = 2\,A$,we get $r = E / i_{max} = 10 / 2 = 5\,\Omega$.
Thus,the internal resistance $r = 5\,\Omega$,the $emf$ $E = 10\,V$,and the maximum current is $2\,A$.
Comparing these with the options,option $A$ is correct.

(D) Let the voltage of the battery be $V = 12\,V$. The resistor $R$ and the $8\,\Omega$ resistor are in parallel. Let their equivalent resistance be $R_p = \frac{8R}{8+R}$.
This combination is in series with the $2\,\Omega$ resistor.
The total resistance of the circuit is $R_{eq} = R_p + 2 = \frac{8R}{8+R} + 2 = \frac{8R + 16 + 2R}{8+R} = \frac{10R + 16}{8+R}$.
The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{12(8+R)}{10R+16} = \frac{6(8+R)}{5R+8}$.
The power dissipated in the $2\,\Omega$ resistor is $P = I^2 \times 2 = 2 \times \left[ \frac{6(8+R)}{5R+8} \right]^2 = 72 \times \frac{(8+R)^2}{(5R+8)^2}$.
To maximize $P$,we analyze the function $f(R) = \frac{8+R}{5R+8}$.
As $R \to 0$,$f(R) \to \frac{8}{8} = 1$. As $R \to \infty$,$f(R) \to \frac{1}{5} = 0.2$.
Since $f(R)$ is a decreasing function for $R \ge 0$,the maximum value occurs at $R = 0$.
At $R = 0$,$P = 72 \times \left( \frac{8+0}{0+8} \right)^2 = 72 \times 1^2 = 72\,W$.
Thus,both $(A)$ and $(C)$ are correct.

(D) $1$. Since the conductor is in a steady state,the current $I$ flowing through any cross-section must be the same. Therefore,$I_P = I_Q$.
$2$. Current density $J = I/A$. Since the area $A_P > A_Q$,the current density $J_P < J_Q$. From Ohm's law in microscopic form,$J = \sigma E$,where $\sigma$ is conductivity. Thus,$E_P < E_Q$.
$3$. The rate of heat generated per unit time is power $H = I^2 R$. For a small segment of length $dx$,$dR = \rho \frac{dx}{A}$. Since $A_P > A_Q$,the resistance $R_Q > R_P$. Therefore,$H_Q > H_P$.
$4$. All statements are correct.

(D) For a conductor of non-uniform cross-section, the current $I$ is constant throughout the length. The current density $J = I/A(x)$, where $A(x)$ is the cross-sectional area at distance $x$. Since the radius $r(x)$ increases with $x$, $A(x)$ increases, so $J$ decreases as $x$ increases. The electric field $E = \rho J = \rho I / A(x)$. Since $A(x)$ increases with $x$, $E$ decreases as $x$ increases. Thus, the graph of $E$ vs $x$ is a decreasing curve, not a straight line as shown in the provided image.
The rate of heat generation per unit length is $H = I^2 R_{unit} = I^2 (\rho / A(x))$. Since $A(x)$ increases with $x$, $H$ decreases as $x$ increases. The graph of $H$ vs $x$ is a decreasing curve.
Also, $H = I^2 \rho / A(x)$ and $E = I \rho / A(x)$. Therefore, $H = I E$. This implies that $H$ is directly proportional to $E$ $(H \propto E)$. Thus, the graph of $H$ vs $E$ is a straight line passing through the origin.
Comparing these with the provided options, graph $(C)$ is correct. Graph $(B)$ shows a decreasing curve for $H$ vs $x$, which is qualitatively correct for a truncated cone.

(D) Applying Kirchhoff's voltage law to the loops:
For loop $AFDBEA$ (assuming current $i_1$ clockwise):
$E_F - E_E = i_1(r_F + r_E + R) - R i_2$
$1 - 2 = i_1(1 + 2 + 2) - 2 i_2 \Rightarrow 5 i_1 - 2 i_2 = -1 \dots(1)$
For loop $CGDBHC$ (assuming current $i_2$ clockwise):
$E_H - E_G = i_2(r_H + r_G + R) - R i_1$
$1 - 3 = i_2(1 + 3 + 2) - 2 i_1 \Rightarrow -2 i_1 + 6 i_2 = -2 \dots(2)$
Multiplying $(1)$ by $3$: $15 i_1 - 6 i_2 = -3 \dots(3)$
Adding $(2)$ and $(3)$: $13 i_1 = -5 \Rightarrow i_1 = -5/13\, A$
Substituting $i_1$ in $(1)$: $5(-5/13) - 2 i_2 = -1 \Rightarrow -25/13 + 1 = 2 i_2 \Rightarrow i_2 = -6/13\, A$
Potential difference $V_D - V_B = R(i_2 - i_1) = 2(-6/13 - (-5/13)) = 2(-1/13) = -2/13\, V$.
Potential difference across $G$: $V_G = E_G + i_2 r_G = 3 + (-6/13)(3) = 3 - 18/13 = 21/13\, V$.
Potential difference across $H$: $V_H = E_H - i_2 r_H = 1 - (-6/13)(1) = 1 + 6/13 = 19/13\, V$.

(D) tetrahedron has $4$ vertices and $6$ edges. Let the vertices be $A, B, C, D$. Each edge has a resistance $r$.
To find the equivalent resistance between any two points,say $A$ and $B$,we can redraw the circuit.
The path $A-B$ is a direct edge with resistance $r$.
The path $A-C-B$ consists of two edges in series,each of resistance $r$,giving a total of $2r$.
The path $A-D-B$ also consists of two edges in series,each of resistance $r$,giving a total of $2r$.
Thus,the circuit between $A$ and $B$ consists of three parallel branches: one with resistance $r$,one with $2r$,and one with $2r$.
All the provided diagrams represent different ways to visualize or simplify this network,and they are all topologically equivalent representations of the same electrical circuit.
Therefore,all the diagrams are correct.

(A) In a thermoelectric series,the direction of current flow at the cold junction is from the metal that appears later in the series to the metal that appears earlier.
For an Antimony-Bismuth thermocouple,the thermoelectric series order is Bismuth followed by Antimony.
Therefore,at the cold junction,the current flows from Antimony to Bismuth.
This creates a continuous loop where the current flows from Bismuth to Antimony at the hot junction.

(B) Let $j$ be the current density.
Since the current $I$ spreads over a hemispherical surface,$j \times 2 \pi r^2 = I$,which gives $j = \frac{I}{2 \pi r^2}$.
Using Ohm's law,the electric field is $E = \rho j = \frac{\rho I}{2 \pi r^2}$.
The potential difference due to current $I$ entering at $A$ between points $B$ and $C$ is $\Delta V_{BC, A} = V_B - V_C = \int_{r_B}^{r_C} E dr = \int_{a}^{a+b} \frac{\rho I}{2 \pi r^2} dr = \frac{\rho I}{2 \pi} \left[ -\frac{1}{r} \right]_{a}^{a+b} = \frac{\rho I}{2 \pi} \left( \frac{1}{a} - \frac{1}{a+b} \right)$.
Similarly,for current $I$ leaving at $D$,the potential difference $\Delta V_{BC, D}$ is also $\frac{\rho I}{2 \pi} \left( \frac{1}{a} - \frac{1}{a+b} \right)$.
By the superposition principle,the total potential difference is $\Delta V = \Delta V_{BC, A} + \Delta V_{BC, D} = 2 \times \frac{\rho I}{2 \pi} \left( \frac{1}{a} - \frac{1}{a+b} \right) = \frac{\rho I}{\pi a} - \frac{\rho I}{\pi(a+b)}$.

(D) First,calculate the resistance of each bulb using $R = V^2 / P$:
$R_1 = (220)^2 / 25 = 1936\ \Omega$
$R_2 = (220)^2 / 100 = 484\ \Omega$
The total resistance in series is $R_{eff} = R_1 + R_2 = 1936 + 484 = 2420\ \Omega$.
The current flowing through the series circuit is $I = V_{supply} / R_{eff} = 440 / 2420 = 44 / 242 = 2 / 11\ A \approx 0.1818\ A$.
Now,calculate the maximum rated current for each bulb using $I_{rated} = P / V$:
For the $25\ W$ bulb: $I_{1,rated} = 25 / 220 = 5 / 44 \approx 0.1136\ A$.
For the $100\ W$ bulb: $I_{2,rated} = 100 / 220 = 5 / 11 \approx 0.4545\ A$.
Comparing the circuit current $I$ with the rated currents:
Since $I (0.1818\ A) > I_{1,rated} (0.1136\ A)$,the $25\ W$ bulb will fuse.
Since $I (0.1818\ A) < I_{2,rated} (0.4545\ A)$,the $100\ W$ bulb will not fuse.

(A) Power of bulb $P_b = 60\, W$. Resistance of bulb $R_b = \frac{V^2}{P_b} = \frac{120^2}{60} = 240\,\Omega$.
Resistance of lead wires $R_L = 6\,\Omega$.
Voltage across bulb before heater is switched on: $V_1 = \frac{R_b}{R_b + R_L} \times 120 = \frac{240}{240 + 6} \times 120 = \frac{240}{246} \times 120 \approx 117.07\, V$.
Power of heater $P_h = 240\, W$. Resistance of heater $R_h = \frac{V^2}{P_h} = \frac{120^2}{240} = 60\,\Omega$.
When heater is in parallel with bulb,equivalent resistance $R_p = \frac{R_b \times R_h}{R_b + R_h} = \frac{240 \times 60}{240 + 60} = \frac{14400}{300} = 48\,\Omega$.
Voltage across bulb after heater is switched on: $V_2 = \frac{R_p}{R_p + R_L} \times 120 = \frac{48}{48 + 6} \times 120 = \frac{48}{54} \times 120 \approx 106.67\, V$.
Decrease in voltage $= V_1 - V_2 = 117.07 - 106.67 = 10.4\, V$.

(A) For a metal like $Cu$,the resistance $R$ increases linearly with temperature $T$ in a limited range,given by $R = R_0(1 + \alpha \Delta T)$,where $\alpha$ is the temperature coefficient of resistance.
For an intrinsic (undoped) semiconductor like $Si$,the number of charge carriers increases exponentially with temperature,leading to an exponential decrease in resistance,given by $\rho = \rho_0 e^{E_g / k_B T}$,where $E_g$ is the band gap energy and $k_B$ is the Boltzmann constant.

(A) The circuit consists of a battery $V_0$ in series with a resistor $R$ and a parallel combination of resistors. Initially,all switches are open. When a switch $S_n$ is closed,it short-circuits the corresponding resistor $nR$.
As switches $S_1, S_2, \dots, S_6$ are closed sequentially at $1 \text{ minute}$ intervals,the total resistance of the circuit decreases.
According to Ohm's law,$I = V/R_{eq}$. Since the total resistance $R_{eq}$ decreases with each step,the current $I$ must increase at each step.
Comparing this with the given options,the graph that shows a step-wise increase in current over time is Graph $A$.

(B) The power $P$ dissipated in an external resistance $R$ connected to a cell of electromotive force $E$ and internal resistance $r$ is given by the formula:
$P = I^2 R$
Since the current $I = \frac{E}{R+r}$,we have:
$P = \left( \frac{E}{R+r} \right)^2 R = \frac{E^2 R}{(R+r)^2}$
To find the variation of $P$ with $R$,we analyze the function $P(R) = \frac{E^2 R}{(R+r)^2}$.
$1$. When $R = 0$,$P = 0$.
$2$. When $R \to \infty$,$P \to 0$.
$3$. The maximum power is dissipated when $R = r$,where $P_{max} = \frac{E^2}{4r}$.
This behavior corresponds to a curve that starts from the origin,increases to a maximum value at $R = r$,and then decreases asymptotically towards zero as $R$ increases further. This matches the shape of Graph $B$.

(A) When two elements $P$ and $Q$ are connected in series,the current $i$ flowing through both is the same,and the total potential difference $V$ across the combination is the sum of the potential differences across each element: $V = V_P + V_Q$.
From the given graphs,for element $P$,the current $i$ increases linearly with $V_P$ up to $10 \text{ V}$ (where $i = 1 \text{ A}$) and remains constant at $1 \text{ A}$ for $V_P > 10 \text{ V}$.
For element $Q$,the current $i$ increases with $V_Q$ and reaches $1 \text{ A}$ at $V_Q = 10 \text{ V}$.
When in series,for a given current $i$ (where $0 \le i \le 1 \text{ A}$),the total voltage $V$ is $V_P(i) + V_Q(i)$.
At $i = 0$,$V = 0 + 0 = 0 \text{ V}$.
At $i = 1 \text{ A}$,$V = 10 \text{ V} + 10 \text{ V} = 20 \text{ V}$.
Since $V_P$ is linear and $V_Q$ is non-linear (curved),the sum $V = V_P + V_Q$ will also be a curve that reaches $20 \text{ V}$ at $i = 1 \text{ A}$.
For $i > 1 \text{ A}$,element $P$ cannot conduct more than $1 \text{ A}$,so the series combination cannot conduct more than $1 \text{ A}$.
Thus,the graph must show a curve ending at $i = 1 \text{ A}$ and $V = 20 \text{ V}$,followed by a vertical line at $V = 20 \text{ V}$ for $i > 1 \text{ A}$. This corresponds to Graph $A$.

(D) When $S_2$ is open and $S_1$ is at position $2$,the circuit is a series combination of $a, b, c, d$. Since $P = I^2 R$,and $b$ is brightest,$R_b$ is the largest. Since $c$ and $d$ are dimmest and equal,$R_c = R_d$. Given the series circuit,$P_b > P_a > P_c = P_d$,which implies $R_b > R_a > R_c = R_d$.
When $S_2$ is closed and $S_1$ is at position $1$,the circuit configuration changes. Lamps $b$ and $c$ are now in parallel,and this combination is in series with $a$ and $d$. Let the voltage across the parallel combination be $V_0$. Then $P'_b = V_0^2 / R_b$ and $P'_c = V_0^2 / R_c$. Since $R_b > R_c$,we have $P'_c > P'_b$.
For lamps $a$ and $d$,they carry the total current $I$. Since $R_a > R_d$,$P'_a = I^2 R_a > P'_d = I^2 R_d$.
Comparing the parallel branch with the series components,the power in the parallel branch is $P'_c = I_2^2 R_c = [I (R_b / (R_b + R_c))]^2 R_c$. Since $R_b > R_c$,the effective resistance of the parallel part is less than $R_d$,leading to $P'_a > P'_d > P'_c > P'_b$.

(B) To find the maximum power delivered to the variable resistance $R$, we use the Maximum Power Transfer Theorem. According to this theorem, the power is maximum when $R = R_{th}$, where $R_{th}$ is the Thevenin equivalent resistance across the terminals of $R$.
$1$. Find the Thevenin voltage ($V_{th}$ or $V_{ab}$): Remove $R$ and calculate the open-circuit voltage across the terminals $a$ and $b$. The $6 \, \Omega$ and $3 \, \Omega$ resistors form a voltage divider across the $12 \, V$ source. The voltage across the $3 \, \Omega$ resistor is $V_{ab} = 12 \times \frac{3}{6+3} = 12 \times \frac{3}{9} = 4 \, V$.
$2$. Find the Thevenin resistance ($R_{th}$ or $R_{ab}$): Short the voltage source. The $6 \, \Omega$ and $3 \, \Omega$ resistors are now in parallel, and this combination is in series with the $8 \, \Omega$ resistor. Thus, $R_{ab} = 8 + \frac{6 \times 3}{6+3} = 8 + \frac{18}{9} = 8 + 2 = 10 \, \Omega$.
$3$. Calculate maximum power: $P_{\max} = \frac{V_{ab}^2}{4 R_{ab}} = \frac{4^2}{4 \times 10} = \frac{16}{40} = 0.4 \, W$.

(D) The phenomenon known as the $Skin \text{ } Effect$ causes the majority of $A.C.$ current to flow near the surface of a conductor.
For $A.C.$ transmission, using a single thick wire is inefficient because the interior of the wire carries very little current.
By using multiple thin strands, the total surface area is increased, which reduces the effective resistance for $A.C.$ and improves efficiency.
For $D.C.$ transmission, the current is distributed uniformly throughout the cross-section of the conductor, so either a single thick wire or multiple thin strands will function effectively.
Therefore, multiple strands are preferred for $A.C.$, while either type is suitable for $D.C.$

(D) In the steady state,the capacitors act as open circuits,so no current flows through the capacitor branch. The potential at point $a$ relative to the negative terminal of the battery is determined by the voltage divider rule for resistors: $V_a = \frac{\varepsilon R_2}{R_1 + R_2}$.
Similarly,the potential at point $b$ relative to the negative terminal is determined by the voltage divider rule for capacitors: $V_b = \frac{\varepsilon C_1}{C_1 + C_2}$.
For the potential difference between $a$ and $b$ to be zero,we must have $V_a = V_b$.
Therefore,$\frac{\varepsilon R_2}{R_1 + R_2} = \frac{\varepsilon C_1}{C_1 + C_2}$.
Simplifying this,$R_2(C_1 + C_2) = C_1(R_1 + R_2)$.
$R_2 C_1 + R_2 C_2 = C_1 R_1 + C_1 R_2$.
$R_2 C_2 = R_1 C_1$.

(D) The potential difference $V$ across the rheostat of resistance $R$ is given by the formula:
$V = IR = \left( \frac{E}{R+r} \right) R = \frac{E}{1 + \frac{r}{R}}$
Analyzing the behavior of $V$ with respect to $R$:
$1$. When $R = 0$,$V = 0$.
$2$. As $R$ increases,the term $\frac{r}{R}$ decreases,so $V$ increases.
$3$. As $R \to \infty$,the term $\frac{r}{R} \to 0$,so $V \to E$.
The graph starts from the origin $(0,0)$ and asymptotically approaches the value $E$ as $R$ increases. This corresponds to an upward convex curve. Thus,the correct graph is shown in option $D$.

(C) The current $I$ is related to the drift velocity $v_d$ by the formula $I = n e A v_d$,where $n$ is the charge carrier density,$e$ is the charge,and $A$ is the cross-sectional area.
Given that $v_d \propto \sqrt{E}$,we can substitute this into the current equation:
$I \propto v_d \propto \sqrt{E}$
Since the electric field $E$ is related to the potential difference $V$ across a wire of length $L$ by $E = V/L$,we have $E \propto V$.
Substituting this into the proportionality for current:
$I \propto \sqrt{V}$
Squaring both sides,we get:
$I^2 \propto V$
This represents a parabolic relationship where $V$ is proportional to the square of $I$ $(V = k I^2)$. This corresponds to a curve that is concave up,starting from the origin. Among the given options,the graph that shows $V$ increasing more rapidly than $I$ (a parabolic curve opening upwards) is the correct one.

(C) The resistance of a metallic filament (like tungsten) increases with temperature. When the bulb is switched $ON$,the filament is at room temperature,so its resistance is very low. According to Ohm's law,$I = V/R$,a very high surge current flows through the filament initially. This high current causes a sudden thermal shock and mechanical stress,which increases the probability of the filament breaking (fusing). As the bulb heats up,the resistance increases,and the current stabilizes to its rated value. Thus,both statements are true,and Statement $2$ is the correct explanation for Statement $1$.

(A) The circuit consists of a $12\,V$ voltage source in series with a parallel combination of an $8\,\Omega$ resistor and a variable resistor $R$,which is then in series with a $2\,\Omega$ resistor.
Let $R_p$ be the equivalent resistance of the parallel combination of $8\,\Omega$ and $R$,given by $R_p = \frac{8R}{8+R}$.
The total resistance of the circuit is $R_{eq} = R_p + 2 = \frac{8R}{8+R} + 2$.
The current $I$ flowing through the $2\,\Omega$ resistor is $I = \frac{V}{R_{eq}} = \frac{12}{\frac{8R}{8+R} + 2} = \frac{12(8+R)}{8R + 16 + 2R} = \frac{12(8+R)}{10R + 16}$.
The power dissipated in the $2\,\Omega$ resistor is $P = I^2 \times 2 = 2 \times \left[ \frac{12(8+R)}{10R + 16} \right]^2$.
To maximize $P$,we need to maximize the current $I$. The current $I$ is maximum when the total resistance $R_{eq}$ is minimum.
$R_{eq} = \frac{8R}{8+R} + 2$. Since $R$ is a resistance,$R \ge 0$. The function $f(R) = \frac{8R}{8+R}$ is an increasing function of $R$. Thus,$R_{eq}$ is minimum when $R = 0$.
When $R = 0$,the parallel combination becomes a short circuit,so $R_p = 0$.
The total resistance becomes $R_{eq} = 0 + 2 = 2\,\Omega$.
The current $I = \frac{12\,V}{2\,\Omega} = 6\,A$.
The power dissipated in the $2\,\Omega$ resistor is $P = I^2 R_{fixed} = (6)^2 \times 2 = 36 \times 2 = 72\,W$.

(C) The resistance of a bulb is given by $R = \frac{V^2}{P}$.
For the first bulb $(40\,W, 220\,V)$,$R_1 = \frac{220^2}{40} = 1210\,\Omega$.
For the second bulb $(60\,W, 220\,V)$,$R_2 = \frac{220^2}{60} = 806.67\,\Omega$.
When connected in series,the total resistance $R_s = R_1 + R_2 = \frac{220^2}{40} + \frac{220^2}{60} = 220^2 \left( \frac{3+2}{120} \right) = 220^2 \left( \frac{5}{120} \right) = \frac{220^2}{24}$.
The effective power in series is $P_1 = \frac{V^2}{R_s} = \frac{220^2}{220^2/24} = 24\,W$.
When connected in parallel,the effective power is $P_2 = P_1' + P_2' = 40\,W + 60\,W = 100\,W$.
Therefore,the ratio $\frac{P_1}{P_2} = \frac{24}{100} = 0.24$.

(A) Given the charge $Q = at - bt^2$.
The current $I$ is the rate of flow of charge: $I = \frac{dQ}{dt} = a - 2bt$.
The current becomes zero at time $t_0$ when $a - 2bt_0 = 0$,which gives $t_0 = \frac{a}{2b}$.
The heat produced $H$ in the resistance $R$ is given by $H = \int_0^{t_0} I^2 R \, dt$.
Substituting $I = a - 2bt$:
$H = \int_0^{a/2b} (a - 2bt)^2 R \, dt = R \int_0^{a/2b} (a^2 - 4abt + 4b^2t^2) \, dt$.
Integrating with respect to $t$:
$H = R \left[ a^2t - 2abt^2 + \frac{4b^2t^3}{3} \right]_0^{a/2b}$.
Substituting the upper limit $t = \frac{a}{2b}$:
$H = R \left[ a^2(\frac{a}{2b}) - 2ab(\frac{a^2}{4b^2}) + \frac{4b^2}{3}(\frac{a^3}{8b^3}) \right]$
$H = R \left[ \frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b} \right] = \frac{a^3R}{6b}$.

(A) Case $1$: When the resistance of the voltmeter is considered infinite (ideal voltmeter).
In this case,the voltmeter does not draw any current. The circuit consists of a $600 \, \Omega$ resistor and a $400 \, \Omega$ resistor in series connected to a $12 \, V$ source.
The voltage across the $400 \, \Omega$ resistor is given by the voltage divider rule:
$V_{1} = 12 \times \frac{400}{600 + 400} = 12 \times \frac{400}{1000} = 4.8 \, V$.
Case $2$: When the resistance of the voltmeter $(1200 \, \Omega)$ is taken into account.
The voltmeter is in parallel with the $400 \, \Omega$ resistor. The equivalent resistance of this parallel combination is:
$R_{p} = \frac{400 \times 1200}{400 + 1200} = \frac{480000}{1600} = 300 \, \Omega$.
Now,the circuit consists of a $600 \, \Omega$ resistor in series with the $300 \, \Omega$ equivalent resistance.
The voltage across the parallel combination (which is the voltmeter reading) is:
$V_{2} = 12 \times \frac{300}{600 + 300} = 12 \times \frac{300}{900} = 4 \, V$.
The percentage error is calculated as:
$\text{Error} = \frac{V_{1} - V_{2}}{V_{1}} \times 100 \% = \frac{4.8 - 4}{4.8} \times 100 \% = \frac{0.8}{4.8} \times 100 \% = \frac{1}{6} \times 100 \% = 16.67 \%$.

(D) The terminal potential difference $V$ of a cell is given by the formula $V = E - Ir$,where $E$ is the electromotive force,$I$ is the current drawn,and $r$ is the internal resistance.
Given that the internal resistance $r$ is proportional to the current $I$,we can write $r = kI$,where $k$ is a positive constant.
Substituting this into the potential difference equation,we get $V = E - I(kI) = E - kI^2$.
This equation $V = E - kI^2$ represents a downward-opening parabola starting from $V = E$ at $I = 0$.
Comparing this with the given options,the graph in option $D$ represents a downward-opening curve starting from the $V$-axis,which matches the derived relationship.

(C) The resistance of a bulb is given by $R = V^2 / P$.
For the first bulb, $R_1 = 220^2 / 40$.
For the second bulb, $R_2 = 220^2 / 60$.
When connected in series, the effective resistance is $R_s = R_1 + R_2$. The power consumed is $P_1 = V^2 / (R_1 + R_2)$.
When connected in parallel, the effective resistance is $R_p = (R_1 R_2) / (R_1 + R_2)$. The power consumed is $P_2 = V^2 / R_p = V^2 (R_1 + R_2) / (R_1 R_2)$.
The ratio is $P_1 / P_2 = [V^2 / (R_1 + R_2)] / [V^2 (R_1 + R_2) / (R_1 R_2)] = (R_1 R_2) / (R_1 + R_2)^2$.
Substituting the values: $R_1 = 1210\, \Omega$ and $R_2 = 806.67\, \Omega$.
$P_1 / P_2 = (40 \times 60) / (40 + 60)^2 = 2400 / 10000 = 0.24$.

(B) The thermo-e.m.f. $(E)$ in a thermocouple is given by the relation $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the junctions.
At the neutral temperature $(\theta_n)$,the thermo-e.m.f. is maximum.
Mathematically,for the e.m.f. to be maximum,the first derivative of $E$ with respect to temperature must be zero.
Therefore,$\frac{dE}{d\theta} = 0$ at $\theta = \theta_n$.
This derivative $\frac{dE}{d\theta}$ represents the rate of increase of thermo-e.m.f. with temperature.
Thus,at the neutral temperature,the rate of increase of thermo-e.m.f. is zero.

(D) According to Faraday's law of electrolysis,the mass $m$ of a substance deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.
Given that the electrochemical equivalent $Z$ is numerically equal to the mass $m$ deposited,we have $m = Z$.
Substituting this into the equation: $Z = ZIt$.
Dividing both sides by $Z$ (assuming $Z \neq 0$),we get $1 = It$.
Given $t = 0.25\,s$,we have $1 = I \times 0.25$.
Therefore,$I = \frac{1}{0.25} = 4\,A$.

(B) Let $I_1$ be the current through the $5\,\Omega$ resistance and $I_2$ be the current through the $(6+9)\,\Omega$ branch.
Given that the heat developed in the $5\,\Omega$ resistance is $20.00\,cal/s$,we have:
$P_1 = I_1^2 R_1 = 20.00\,cal/s$
$I_1^2 \times 5 = 20 \implies I_1^2 = 4 \implies I_1 = 2\,A$.
The potential difference across points $C$ and $D$ is $V_{CD} = I_1 \times 5 = 2 \times 5 = 10\,V$.
The current through the upper branch $(6\,\Omega + 9\,\Omega)$ is $I_2 = \frac{V_{CD}}{6+9} = \frac{10}{15} = \frac{2}{3}\,A$.
The total current $I$ flowing through the $2\,\Omega$ resistance is $I = I_1 + I_2 = 2 + \frac{2}{3} = \frac{8}{3}\,A$.
The heat developed per second in the $2\,\Omega$ resistance is $P = I^2 R = \left(\frac{8}{3}\right)^2 \times 2 = \frac{64}{9} \times 2 = \frac{128}{9} \approx 14.22\,cal/s$.

(D) In circuit $I$,the switch (key) is open,which means the circuit is incomplete. Therefore,no current flows through the resistor $(I = 0 \ A)$. According to Ohm's law,the voltage drop across the resistor is $V = I \times R = 0 \times 2 = 0 \ V$. Thus,the voltmeter reads $0 \ V$.
In circuit $II$,the switch is closed,completing the circuit. The current flows through the resistor. Since the internal resistance of the battery is not mentioned,we assume it to be ideal. The entire electromotive force $(2 \ V)$ appears across the $2 \ \Omega$ resistor. Therefore,the voltmeter reads $2 \ V$.

(D) The positive terminal of a battery is the point of the highest potential,not the lowest. Therefore,the Assertion is incorrect.
In an external circuit,current flows from the positive terminal (higher potential) to the negative terminal (lower potential). The Reason states that current flows towards the point of higher potential,which is also incorrect. Thus,both the Assertion and the Reason are incorrect.

(A) The electric power of a heater is significantly higher than that of a bulb. Since $P = V^2/R$,we have $R \propto 1/P$,meaning the resistance of the heater is much lower than that of the bulb.
When the heater is switched on in parallel,the total current drawn from the source increases. Due to the internal resistance of the source (or line resistance),the terminal voltage across the bulb drops,causing it to become dim.
As the heater coil heats up,its resistance increases due to the positive temperature coefficient of the material. Consequently,the current drawn by the heater decreases,the terminal voltage recovers,and the bulb's brightness increases (dimness decreases).

(D) In the given circuit, the battery $E$, ammeter $A$, resistor $R$, and voltmeter $V$ are connected in series.
An ideal voltmeter has infinite resistance, which prevents any current from flowing through the circuit.
Since the circuit is effectively open due to the infinite resistance of the ideal voltmeter, the current $I$ flowing through the ammeter is $I = \frac{E}{R + R_V} = \frac{E}{R + \infty} = 0$.
Therefore, the ammeter reading is $0$.
The voltmeter is connected in series with the rest of the circuit. The potential difference across the ideal voltmeter will be the entire $EMF$ of the battery, $V = E$. Thus, the voltmeter reading is not zero.
Therefore, the Assertion is incorrect, and the Reason is also incorrect.

(B) Let the resistance of each bulb be $R$.
Case $(i)$: All six bulbs are glowing. Section $A$ has $3$ bulbs in parallel,and section $B$ has $3$ bulbs in parallel. These two sections are in series. The equivalent resistance is $R_{eq_1} = (R/3) + (R/3) = 2R/3$. The power consumed is $P_1 = E^2 / R_{eq_1} = 3E^2 / 2R$.
Case $(ii)$: Two bulbs from section $A$ are in parallel (resistance $R/2$) and one bulb from section $B$ is in series with this combination. The equivalent resistance is $R_{eq_2} = R/2 + R = 3R/2$. The power consumed is $P_2 = E^2 / R_{eq_2} = 2E^2 / 3R$.
The ratio of power consumption is $P_1 : P_2 = (3E^2 / 2R) : (2E^2 / 3R) = 9 : 4$.

$(A)$ When a steady current flows in a metallic conductor of non-uniform cross-section, the current $(I)$ flowing through the conductor remains constant because charge cannot accumulate at any point. Since $I = J A = n e v_d A$, where $J$ is current density, $A$ is cross-sectional area, and $v_d$ is drift speed, $J$, $E$ (electric field), and $v_d$ are inversely proportional to $A$. Thus, they vary with the cross-section.
$(b)$ No, Ohm's law is not universally applicable. Elements that do not obey Ohm's law are called non-ohmic conductors. Examples include vacuum diodes, semiconductors, and junction diodes.
$(c)$ According to the relation $I = \frac{\varepsilon}{R + r}$, where $\varepsilon$ is the electromotive force and $r$ is the internal resistance, to obtain a high current $(I)$ from a low voltage source $(\varepsilon)$, the internal resistance $(r)$ must be very low.
$(d)$ $A$ high tension $(HT)$ supply must have a very large internal resistance to limit the current to a safe value. If the internal resistance were low, a short circuit would cause an extremely high current to flow, which could be dangerous and damage the equipment.

(A) Power consumption in $1$ day ($5$ hours) is $10$ units.
Power consumed in $1$ hour is $P = 10/5 = 2$ units = $2000$ $W$.
Since $P = VI$,the current $I = P/V = 2000/220 \approx 9.09$ $A$.
Resistance of the copper wire $R = \rho l / A = \rho l / (\pi r^2) = (1.7 \times 10^{-8} \times 10) / (3.14 \times (10^{-3})^2) = 1.7 \times 10^{-7} / 3.14 \times 10^{-6} \approx 0.054$ $\Omega$.
Power dissipated in the wire $P_{loss} = I^2 R = (9.09)^2 \times 0.054 \approx 4.46$ $W$.
Fraction of power lost = $P_{loss} / P_{total} = 4.46 / 2000 \approx 0.00223$ or $0.223\%$.
If aluminium is used,$\rho_{Al} = 2.7 \times 10^{-8} \, \Omega m$.
Since $R \propto \rho$,the resistance increases by a factor of $2.7/1.7 \approx 1.59$.
Thus,the power loss in the aluminium wire will be approximately $1.59$ times higher than in the copper wire.

(A) In the series connection,the equivalent resistance of the resistors is $R_{eq,s} = nR = 10n \; \Omega$.
The total resistance of the circuit is $R_{total,s} = 10 + 10n \; \Omega$.
The current in the series circuit is $I_s = \frac{20}{10 + 10n} = \frac{2}{1 + n} \; A$.
In the parallel connection,the equivalent resistance of the resistors is $R_{eq,p} = \frac{R}{n} = \frac{10}{n} \; \Omega$.
The total resistance of the circuit is $R_{total,p} = 10 + \frac{10}{n} = \frac{10n + 10}{n} \; \Omega$.
The current in the parallel circuit is $I_p = \frac{20}{\frac{10n + 10}{n}} = \frac{20n}{10(n + 1)} = \frac{2n}{n + 1} \; A$.
Given that the current increases $20$ times,we have $I_p = 20 I_s$.
Substituting the expressions: $\frac{2n}{n + 1} = 20 \times \frac{2}{n + 1}$.
Canceling the common terms,we get $2n = 40$,which implies $n = 20$.

(C) First, simplify the circuit. The parallel combination of two $4\, \Omega$ resistors is $2\, \Omega$. The parallel combination of two $8\, \Omega$ resistors is $4\, \Omega$. The parallel combination of two $12\, \Omega$ resistors is $6\, \Omega$.
The upper branch has resistances in series: $2\, \Omega$ (from $4\, \Omega || 4\, \Omega$) + $2\, \Omega$ + $R_{p1}$ (where $R_{p1} = 15\, \Omega || 10\, \Omega = 6\, \Omega$). Total upper branch resistance $R_1 = 2 + 2 + 6 = 10\, \Omega$.
The lower branch has resistances in series: $4\, \Omega$ (from $8\, \Omega || 8\, \Omega$) + $6\, \Omega$ (from $12\, \Omega || 12\, \Omega$) = $10\, \Omega$.
The two branches are in parallel, so their equivalent resistance is $R_{eq_branches} = (10\, \Omega || 10\, \Omega) = 5\, \Omega$.
Total circuit resistance $R_{total} = 5\, \Omega + 1\, \Omega$ (internal) $= 6\, \Omega$.
Total current $I = V / R_{total} = 12\, V / 6\, \Omega = 2\, A$.
This current splits equally into the two $10\, \Omega$ branches, so $1\, A$ flows through the upper branch.
The voltage across the $15\, \Omega || 10\, \Omega$ parallel combination is $V_{AB} = I_{upper} \times R_{p1} = 1\, A \times 6\, \Omega = 6\, V$.

(B) Step $1$: Calculate the mass of the ice.
$m = \rho A \ell = 10^3\, kg/m^3 \times 10^{-4}\, m^2 \times 1\, m = 0.1\, kg$.
Step $2$: Calculate the total energy $Q$ required to raise the temperature of ice from $-10^{\circ}C$ to $0^{\circ}C$ and then melt it.
$Q = mc_{ice}\Delta T + mL_f$
$Q = 0.1 \times (2 \times 10^3 \times 10) + 0.1 \times (3.33 \times 10^5)$
$Q = 2 \times 10^3 + 3.33 \times 10^4 = 3.53 \times 10^4\, J$.
Step $3$: Calculate the time $t$ using the heat produced by the resistor.
$Q = I^2Rt$
$3.53 \times 10^4 = (0.5)^2 \times (4 \times 10^3) \times t$
$3.53 \times 10^4 = 0.25 \times 4000 \times t$
$3.53 \times 10^4 = 1000 \times t$
$t = 35.3\, s$.

(A) The drift velocity is given by $v_{d} = \frac{e E}{m} \tau$. Thus, $(A) \rightarrow (R)$.
The electrical resistivity $\rho$ is defined as the ratio of electric field to current density, $\rho = \frac{E}{J}$. Thus, $(B) \rightarrow (S)$.
The relaxation period $\tau$ is related to resistivity by $\rho = \frac{m}{n e^{2} \tau}$, which implies $\tau = \frac{m}{n e^{2} \rho}$. Thus, $(C) \rightarrow (P)$.
The current density $J$ is given by $J = n e v_{d}$. Thus, $(D) \rightarrow (Q)$.
Therefore, the correct matching is $(A)-(R), (B)-(S), (C)-(P), (D)-(Q)$.

(C) Analysis of Statement $I$:
When a wire of resistance $R = 80\,\Omega$ is cut into $4$ equal parts,the resistance of each part becomes $r = R/4 = 80/4 = 20\,\Omega$.
When these $4$ resistors are connected in parallel,the equivalent resistance $R_{eq}$ is given by:
$1/R_{eq} = 1/r + 1/r + 1/r + 1/r = 4/r = 4/20 = 1/5$.
Thus,$R_{eq} = 5\,\Omega$. Statement $I$ is correct.
Analysis of Statement $II$:
When resistors are connected in parallel,the potential difference $V$ across each is the same.
The thermal energy (heat) developed in time $t$ is given by $H = (V^2/R)t$.
Therefore,$H \propto 1/R$.
The ratio of thermal energy developed in $3\,R$ and $2\,R$ is:
$H_{3R} / H_{2R} = (V^2 / 3R) / (V^2 / 2R) = 2R / 3R = 2/3$.
The ratio is $2:3$,not $3:2$. Statement $II$ is incorrect.
Conclusion: Statement $I$ is correct but statement $II$ is incorrect.

(A) When a steady current $I$ flows through a junction between two different metals,charge accumulation occurs at the interface due to the difference in their conductivities.
The current density $J$ is given by $J = \sigma E$,where $\sigma$ is the conductivity and $E$ is the electric field. Since the current $I$ and the cross-sectional area $A$ are the same for both rods,the current density $J = I/A$ is constant.
At the junction,the continuity of current requires that the normal component of the current density be continuous. However,the electric field $E = J/\sigma$ must change because the conductivity $\sigma$ of copper is higher than that of iron $(\sigma_{Cu} > \sigma_{Fe})$.
According to Gauss's law,the surface charge density $\rho_s$ at the interface is given by $\rho_s = \epsilon_0 (E_{Fe} - E_{Cu})$. Since $\sigma_{Cu} > \sigma_{Fe}$,it follows that $E_{Fe} > E_{Cu}$. Thus,$\rho_s$ is positive,meaning positive charge accumulates on the copper side and negative charge accumulates on the iron side of the junction to maintain the steady current flow.
Therefore,the correct representation is shown in option $(A)$ and $(B)$ (as they are identical in the provided images). Based on the standard interpretation,the positive charge accumulates on the side of higher conductivity $(Cu)$ and negative charge on the side of lower conductivity $(Fe)$.