Mix Examples-Current Electricity Questions in English

(B) The total $e.m.f.$ of the battery is $E_{total} = 6 \times 2\,V = 12\,V$.
The total internal resistance of the battery is $r_{total} = 6 \times 0.5\,\Omega = 3\,\Omega$.
The total resistance of the circuit is $R_{total} = R_{external} + r_{total} = 10\,\Omega + 3\,\Omega = 13\,\Omega$.
Since the battery is being charged,the effective $e.m.f.$ in the circuit is $V_{mains} - E_{total} = 220\,V - 12\,V = 208\,V$.
The charging current $I$ is given by $I = \frac{V_{mains} - E_{total}}{R_{total}} = \frac{208\,V}{13\,\Omega} = 16\,A$.

(C) Let the internal resistance of the cell be $r$. Initially,the total resistance of the circuit is $R_{eq} = R_V + r$,where $R_V$ is the resistance of the voltmeter. The current is $I = \frac{\varepsilon}{R_V + r}$.
When a resistor $R$ is connected in parallel with the voltmeter,the equivalent resistance of the parallel combination becomes $R_p = \frac{R_V R}{R_V + R}$.
Since $R_p < R_V$,the total resistance of the circuit $R'_{eq} = R_p + r$ decreases.
As the total resistance decreases,the total current $I' = \frac{\varepsilon}{R_p + r}$ increases,so the ammeter reading $A$ increases.
The voltage across the parallel combination is $V' = I' R_p = \frac{\varepsilon R_p}{R_p + r} = \frac{\varepsilon}{\frac{R_p + r}{R_p}} = \frac{\varepsilon}{1 + \frac{r}{R_p}}$.
Since $R_p$ decreases,the term $\frac{r}{R_p}$ increases,which means the denominator $(1 + \frac{r}{R_p})$ increases,and thus the voltmeter reading $V'$ decreases.

(B) The total thermo $emf$ produced between two temperatures is the sum of the $emf$ produced in the intermediate temperature ranges.
According to the principle of additivity of thermo $emf$:
$e_{0}^{100} = e_{0}^{32} + e_{32}^{70} + e_{70}^{100}$
Given values:
$e_{0}^{100} = 200\,\mu V$
$e_{0}^{32} = 64\,\mu V$
$e_{32}^{70} = 76\,\mu V$
Substituting these values into the equation:
$200 = 64 + 76 + e_{70}^{100}$
$200 = 140 + e_{70}^{100}$
$e_{70}^{100} = 200 - 140 = 60\,\mu V$
Therefore,the thermo $emf$ produced between $70\,^oC$ and $100\,^oC$ is $60\,\mu V$.

(B) According to Faraday's law of electrolysis,the mass deposited is $m = Z \times q$,where $q$ is the total charge passed.
The total charge $q$ is equal to the area under the current-time $(I-t)$ graph.
The graph consists of a triangle from $t = 0$ to $t = 2 \, \text{min}$ and a rectangle from $t = 2 \, \text{min}$ to $t = 6 \, \text{min}$.
First,convert time to seconds: $t_1 = 2 \times 60 = 120 \, \text{s}$ and $t_2 = (6 - 2) \times 60 = 240 \, \text{s}$.
Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 120 \times 1 = 60 \, \text{C}$.
Area of rectangle = $\text{length} \times \text{breadth} = 240 \times 1 = 240 \, \text{C}$.
Total charge $q = 60 + 240 = 300 \, \text{C}$.
Since $m = Z \times q$,we have $Z = \frac{m}{q} = \frac{m}{300} \, \text{g/C}$.

(B) According to Faraday's laws of electrolysis,the mass of a substance deposited is proportional to its equivalent mass.
The ratio of the mass of $O_2$ to the mass of $Ag$ is equal to the ratio of their equivalent weights:
$\frac{m_{O_2}}{m_{Ag}} = \frac{E_{O_2}}{E_{Ag}}$
Equivalent weight of $O_2 = \frac{32}{4} = 8 \, g/eq$.
Equivalent weight of $Ag = \frac{108}{1} = 108 \, g/eq$.
Given $m_{O_2} = 0.8 \, g$,let $m_{Ag} = m$.
$\frac{0.8}{m} = \frac{8}{108}$
$m = \frac{0.8 \times 108}{8} = 0.1 \times 108 = 10.8 \, g$.

(A) The heat energy supplied by the heater is equal to the heat energy absorbed by the water.
Using the formula: $P \times t = m \times s \times \Delta \theta$
Where $P = \frac{V^2}{R}$,$m = 100\, g = 0.1\, kg$,$s = 4200\, J/(kg\cdot ^\circ C)$,and $\Delta \theta = 50^\circ C$.
Substituting the values:
$\frac{220 \times 220}{484} \times t = 0.1 \times 4200 \times 50$
Since $\frac{48400}{484} = 100$,we have:
$100 \times t = 21000$
$t = \frac{21000}{100} = 210\, sec$.

(B) The resistance $R$ of a bulb is given by the formula $R = \frac{V^2}{P}$.
For the first bulb $(P_1 = 100 \ W, V_1 = 200 \ V)$: $R_1 = \frac{200^2}{100} = \frac{40000}{100} = 400 \ \Omega$.
For the second bulb $(P_2 = 200 \ W, V_2 = 100 \ V)$: $R_2 = \frac{100^2}{200} = \frac{10000}{200} = 50 \ \Omega$.
The ratio of resistances is $\frac{R_1}{R_2} = \frac{400}{50} = 8 : 1$.
The maximum current rating $i$ is given by $i = \frac{P}{V}$.
For the first bulb: $i_1 = \frac{100}{200} = 0.5 \ A$.
For the second bulb: $i_2 = \frac{200}{100} = 2 \ A$.
The ratio of maximum current ratings is $\frac{i_1}{i_2} = \frac{0.5}{2} = \frac{1}{4}$.

(B) The terminal voltage $V$ of a battery is given by the equation $V = E - Ir$, where $E$ is the electromotive force, $I$ is the current, and $r$ is the internal resistance.
Comparing this with the equation of a straight line $y = mx + c$, we have $V = -rI + E$.
The slope of the $V-I$ graph is $m = -r$. From the given graph, the magnitude of the slope is $|m| = \frac{\text{rise}}{\text{run}} = x / v$.
Therefore, the internal resistance $r = x / v$.
Conductance is the reciprocal of resistance. Thus, internal conductance $G = 1 / r = v / x$.
Given the labels in the image, the vertical intercept is $x$ and the horizontal intercept is $v$. The internal conductance is $v/x$.

(D) An ideal ammeter has a resistance of $0 \ \Omega$. When it is connected in parallel with the battery of $e.m.f.$ $E$,the current flowing through the ammeter is given by $I_A = E / R_{ammeter} = E / 0 = \infty$.
Since the resistance of the ammeter is practically zero,a very large current flows through it,which causes the ammeter to burn out or get damaged.
The voltmeter has a very high resistance,so it will not be damaged by the current flowing through the circuit.

(C) The circuit consists of a $60\,V$ source connected to a series combination of a $300\,\Omega$ resistor and a $400\,\Omega$ resistor. Let the voltmeter have an internal resistance $R_v$.
When the voltmeter is connected across the $400\,\Omega$ resistor,the equivalent resistance of the parallel combination is $R_p = \frac{400 R_v}{400 + R_v}$.
The total resistance of the circuit is $R_{eq} = 300 + R_p$.
The voltage across the parallel combination is given as $30\,V$. Thus,the voltage across the $300\,\Omega$ resistor is $60 - 30 = 30\,V$.
Since the voltage across both parts is equal $(30\,V)$,the resistance of the parallel combination must be equal to the resistance of the $300\,\Omega$ resistor.
Therefore,$R_p = 300\,\Omega$.
Now,when the voltmeter is connected across the $300\,\Omega$ resistor,the new equivalent resistance of that part is $R'_p = \frac{300 R_v}{300 + R_v}$.
Since $R_p = \frac{400 R_v}{400 + R_v} = 300$,we find $400 R_v = 120000 + 300 R_v$,so $100 R_v = 120000$,which means $R_v = 1200\,\Omega$.
Now,$R'_p = \frac{300 \times 1200}{300 + 1200} = \frac{360000}{1500} = 240\,\Omega$.
The total resistance of the circuit is now $R'_{eq} = 400 + 240 = 640\,\Omega$.
The total current is $I = \frac{60}{640} = \frac{6}{64} = \frac{3}{32}\,A$.
The reading of the voltmeter is $V' = I \times R'_p = \frac{3}{32} \times 240 = 3 \times 7.5 = 22.5\,V$.

(C) Given: Power $P = 100 \ W$,Voltage $V = 125 \ V$,Electrochemical equivalent $Z = 0.367 \times 10^{-6} \ kg/C$,Time $t = 60 \ s$.
Using the formula for power,$P = VI$,we find the current $I$:
$I = \frac{P}{V} = \frac{100}{125} = 0.8 \ A$.
According to Faraday's law of electrolysis,the mass $m$ of the substance deposited is given by $m = ZIt$.
Substituting the values:
$m = (0.367 \times 10^{-6} \ kg/C) \times (0.8 \ A) \times (60 \ s)$
$m = 0.367 \times 10^{-6} \times 48 \ kg$
$m = 17.616 \times 10^{-6} \ kg$
Since $1 \ kg = 10^6 \ mg$,we have:
$m = 17.616 \times 10^{-6} \times 10^6 \ mg = 17.616 \ mg \approx 17.6 \ mg$.

(B) According to Faraday's law of electrolysis,the mass $m$ deposited is proportional to the charge $q$,where $q = i \times t$.
Thus,$m \propto i \times t$.
For the first case: $m_1 = m$,$i_1 = 4 \ A$,$t_1 = 2 \ min = 120 \ s$.
For the second case: $m_2 = ?$,$i_2 = 6 \ A$,$t_2 = 40 \ s$.
Using the ratio: $\frac{m_1}{m_2} = \frac{i_1 \times t_1}{i_2 \times t_2}$.
Substituting the values: $\frac{m}{m_2} = \frac{4 \times 120}{6 \times 40}$.
$\frac{m}{m_2} = \frac{480}{240} = 2$.
Therefore,$m_2 = \frac{m}{2}$.

(C) The watt-hour efficiency $(\%\eta_{Wh})$ is defined as the ratio of output energy to input energy.
Input Energy $(E_{in})$ = $V_{charge} \times I_{charge} \times t_{charge} = 15\,V \times 10\,A \times 8\,h = 1200\,Wh$.
Output Energy $(E_{out})$ = $V_{discharge} \times I_{discharge} \times t_{discharge} = 14\,V \times 5\,A \times 15\,h = 1050\,Wh$.
Efficiency $(\%\eta_{Wh})$ = $\left( \frac{E_{out}}{E_{in}} \right) \times 100 = \left( \frac{1050}{1200} \right) \times 100 = 87.5\%.$

(A) According to Faraday's law of electrolysis, the mass $m$ deposited is given by $m = ZIt$, where $Z$ is the electrochemical equivalent of silver $(Z = \frac{108}{96500} \ g/C)$.
Given $m = 2.89 \ g$, $t = 10 \ minutes = 600 \ s$.
$I = \frac{m}{Zt} = \frac{2.89 \times 96500}{108 \times 600} \approx 4.3 \ A$.
However, using the standard approximation for such problems where $m \approx 2.68 \ g$ is often used for $I = 4 \ A$:
$I = \frac{2.68 \times 96500}{108 \times 600} \approx 4 \ A$.
Heat produced $H = I^2Rt = (4)^2 \times 20 \times 600 = 16 \times 12000 = 192000 \ J = 192 \ kJ$.

(A) The resistance of the wire is $R = 10\, \Omega$. The points $P$ and $Q$ divide the circle into two parts, one being a quadrant ($1/4$th of the circumference) and the other being three-quarters ($3/4$th of the circumference).
Resistance of the first part, $R_1 = \frac{1}{4} \times 10 = 2.5\, \Omega$.
Resistance of the second part, $R_2 = \frac{3}{4} \times 10 = 7.5\, \Omega$.
These two resistors are in parallel. Their equivalent resistance $R_{eq}$ is given by:
$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{2.5 \times 7.5}{2.5 + 7.5} = \frac{18.75}{10} = 1.875\, \Omega = \frac{15}{8}\, \Omega$.
The total resistance of the circuit including the internal resistance $r = 1\, \Omega$ is $R_{total} = R_{eq} + r = \frac{15}{8} + 1 = \frac{23}{8}\, \Omega$.
The total current $I$ drawn from the battery is $I = \frac{V}{R_{total}} = \frac{3}{23/8} = \frac{24}{23}\, A$.
Using the current divider rule, the current $i_1$ through $R_1$ is:
$i_1 = I \times \frac{R_2}{R_1 + R_2} = \frac{24}{23} \times \frac{7.5}{10} = \frac{24}{23} \times 0.75 = \frac{18}{23}\, A$.
The current $i_2$ through $R_2$ is:
$i_2 = I - i_1 = \frac{24}{23} - \frac{18}{23} = \frac{6}{23}\, A$.

(A) The Peltier coefficient $\pi$ is defined as $\pi = T \frac{de}{dT}$, where $T$ is the absolute temperature in Kelvin.
Given the temperature in Celsius is $t$, the absolute temperature is $T = t + 273$.
Substituting $t = T - 273$ into the given equation $e = at + bt^2$:
$e = a(T - 273) + b(T - 273)^2$
Now, differentiate $e$ with respect to $T$:
$\frac{de}{dT} = a + 2b(T - 273)$
Since $T - 273 = t$, we have $\frac{de}{dT} = a + 2bt$.
Substituting this into the formula for the Peltier coefficient:
$\pi = T \frac{de}{dT} = (t + 273)(a + 2bt)$.

(A) The current through the silver voltmeter is $I_1 = \frac{m_1}{Z_1 t} = \frac{1}{11.2 \times 10^{-4} \times 30 \times 60} = 0.496\ A$.
The current through the copper voltmeter is $I_2 = \frac{m_2}{Z_2 t} = \frac{1.8}{6.6 \times 10^{-4} \times 30 \times 60} = 1.515\ A$.
The total current supplied by the battery is $I = I_1 + I_2 = 0.496 + 1.515 = 2.011\ A$.
The power supplied by the battery is $P = V \times I = 12 \times 2.011 = 24.132\ J/s$.

(C) Let $\varepsilon$ be the $EMF$ and $r$ be the internal resistance of the cell.
For resistance $R_1$,the current is $I_1 = \frac{\varepsilon}{R_1 + r}$. The heat produced is $H_1 = I_1^2 R_1 t = \frac{\varepsilon^2 R_1 t}{(R_1 + r)^2}$.
For resistance $R_2$,the current is $I_2 = \frac{\varepsilon}{R_2 + r}$. The heat produced is $H_2 = I_2^2 R_2 t = \frac{\varepsilon^2 R_2 t}{(R_2 + r)^2}$.
Given $H_1 = H_2$,we have:
$\frac{\varepsilon^2 R_1 t}{(R_1 + r)^2} = \frac{\varepsilon^2 R_2 t}{(R_2 + r)^2}$
$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$
$R_1(R_2 + r)^2 = R_2(R_1 + r)^2$
$R_1(R_2^2 + 2R_2 r + r^2) = R_2(R_1^2 + 2R_1 r + r^2)$
$R_1 R_2^2 + 2R_1 R_2 r + R_1 r^2 = R_2 R_1^2 + 2R_1 R_2 r + R_2 r^2$
$R_1 R_2^2 + R_1 r^2 = R_2 R_1^2 + R_2 r^2$
$r^2(R_1 - R_2) = R_1 R_2(R_1 - R_2)$
Assuming $R_1 \neq R_2$,we get $r^2 = R_1 R_2$,so $r = \sqrt{R_1 R_2}$.

(B) First, calculate the resistance of each bulb using the formula $R = \frac{V^2}{P}$.
$R = \frac{220 \times 220}{100} = 484\; \Omega$.
In series, the equivalent resistance is $R_{eq} = R + R = 484 + 484 = 968\; \Omega$.
The power drawn is $P_{series} = \frac{V^2}{R_{eq}} = \frac{220 \times 220}{968} = 50\; W$.
In parallel, the equivalent resistance is $R_{eq} = \frac{R}{2} = \frac{484}{2} = 242\; \Omega$.
The power drawn is $P_{parallel} = \frac{V^2}{R_{eq}} = \frac{220 \times 220}{242} = 200\; W$.
Thus, the power drawn in series and parallel combinations is $50\; W$ and $200\; W$ respectively.

(C) In a steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.
The current $i$ flows only through the upper branch containing the two $1\, \Omega$ resistors in series.
The total resistance of the circuit is $R_{eq} = 1\, \Omega + 1\, \Omega + 0.5\, \Omega = 2.5\, \Omega$.
The current in the circuit is $i = \frac{V}{R_{eq}} = \frac{2.5\, V}{2.5\, \Omega} = 1\, A$.
The potential difference across the capacitor is equal to the potential difference across the parallel branch containing the $2\, \Omega$ resistor. Since no current flows through the $2\, \Omega$ resistor, the potential difference across the capacitor is equal to the terminal voltage of the battery.
The terminal voltage $V_{terminal} = E - ir = 2.5\, V - (1\, A \times 0.5\, \Omega) = 2.5\, V - 0.5\, V = 2.0\, V$.
The charge on the capacitor is $Q = C \times V = 5\, \mu F \times 2.0\, V = 10\, \mu C$.

(C) For a conductor,resistance $R \propto T$. From the $V-i$ graph,the slope represents resistance $R = \frac{V}{i} = \tan \phi$,where $\phi$ is the angle with the $i$-axis.
For temperature $T_1$,the angle with the $i$-axis is $\theta$. Thus,$R_1 = \tan \theta = k T_1$ (where $k$ is a constant). $(i)$
For temperature $T_2$,the angle with the $V$-axis is $\theta$,so the angle with the $i$-axis is $(90^\circ - \theta)$. Thus,$R_2 = \tan(90^\circ - \theta) = \cot \theta = k T_2$. (ii)
Subtracting $(i)$ from (ii): $k(T_2 - T_1) = \cot \theta - \tan \theta$.
$(T_2 - T_1) \propto \left( \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} \right) = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} = 2 \cot 2\theta$.
Therefore,$(T_2 - T_1) \propto \cot 2\theta$.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
For the original wire: $R_1 = \rho \frac{l}{\pi (9 \times 10^{-3})^2} = 5 \, \Omega$.
For a single new wire of radius $r_2 = 3 \, mm$: $R_2 = \rho \frac{l}{\pi (3 \times 10^{-3})^2}$.
Comparing $R_1$ and $R_2$: $\frac{R_2}{R_1} = \frac{(9 \times 10^{-3})^2}{(3 \times 10^{-3})^2} = (3)^2 = 9$.
So,$R_2 = 9 \times R_1 = 9 \times 5 = 45 \, \Omega$.
Since there are $6$ such wires connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_2} + \dots + \frac{1}{R_2} = \frac{6}{R_2}$.
Therefore,$R_{eq} = \frac{R_2}{6} = \frac{45}{6} = 7.5 \, \Omega$.

(B) To find the energy stored in the capacitor,we first need to determine the potential difference $V_{AB}$ across it.
Applying Kirchhoff's Current Law at junction $A$: The current entering from the left is $2 \, A$ and from the top is $1 \, A$. Thus,the current flowing through the $5 \, \Omega$ resistor towards $D$ is $i_1 = 2 + 1 = 3 \, A$.
Applying Kirchhoff's Current Law at junction $C$: The current $i_1 = 3 \, A$ enters $C$ from $D$. The current $i_2$ flows from $C$ to $B$ through the $2 \, \Omega$ resistor. The current $i_3$ flows out of $C$ through the $4 \, \Omega$ resistor. Based on the circuit diagram,the current $i_2$ is $1 \, A$.
Now,calculate the potential difference $V_{AB} = V_A - V_B$ by traversing the path $A \to D \to C \to B$:
$V_A - V_B = V_A - V_D + V_D - V_C + V_C - V_B$
$V_A - V_B = (i_1 \times 5) + (i_1 \times 1) + (i_2 \times 2)$
$V_A - V_B = (3 \times 5) + (3 \times 1) + (1 \times 2) = 15 + 3 + 2 = 20 \, V$.
The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.
$U = \frac{1}{2} \times (4 \times 10^{-6} \, F) \times (20 \, V)^2$
$U = 2 \times 10^{-6} \times 400 = 800 \times 10^{-6} \, J = 8 \times 10^{-4} \, J$.

(D) The resistance of voltmeter $V_1$ is given by $R_1 = \text{sensitivity} \times \text{reading} = 200 \, \Omega/V \times 80 \, V = 16000 \, \Omega = 16 \, k\Omega$.
Given,the resistance of voltmeter $V_2$ is $R_2 = 32 \, k\Omega$.
Since the voltmeters are connected in series,the same current flows through both. The voltage across $V_1$ is $V_1 = 80 \, V$.
Using the voltage divider rule,$V_1 = V \times \frac{R_1}{R_1 + R_2}$,where $V$ is the total voltage.
Substituting the values: $80 = V \times \frac{16}{16 + 32}$.
$80 = V \times \frac{16}{48} = V \times \frac{1}{3}$.
Therefore,$V = 80 \times 3 = 240 \, V$.

(D) Let the heat required to boil the tea be $H$. The power of the first coil is $P_1 = H / t_1$ and the power of the second coil is $P_2 = H / t_2$.
When connected in parallel,the total power is $P_p = P_1 + P_2$.
Since $P = H / t$,we have $H / t_p = H / t_1 + H / t_2$.
This simplifies to $1 / t_p = 1 / t_1 + 1 / t_2$,or $t_p = (t_1 \times t_2) / (t_1 + t_2)$.
Given $t_1 = 10 \, \text{min}$ and $t_2 = 40 \, \text{min}$,we get $t_p = (10 \times 40) / (10 + 40) = 400 / 50 = 8 \, \text{min}$.

(D) Let the resistance of each resistor be $R$ and the applied voltage be $V$.
When $n$ identical resistors are connected in series,the equivalent resistance is $R_s = nR$.
The power consumed in series is $P_s = \frac{V^2}{R_s} = \frac{V^2}{nR} = \frac{P}{n}$,where $P$ is the power of a single resistor.
Given $P_s = 10 \, W$ and $n = 3$,we have $10 = \frac{P}{3}$,so $P = 30 \, W$.
When connected in parallel,the equivalent resistance is $R_p = \frac{R}{n}$.
The power consumed in parallel is $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/n} = n \times \frac{V^2}{R} = n \times P$.
Substituting the values,$P_p = 3 \times 30 = 90 \, W$.

(A) The thermoelectric power $S$ is defined as the rate of change of thermo-emf with respect to temperature,given by $S = \frac{dE}{dT}$.
Given the expression for emf: $E = K(T - T_r) [T_0 - \frac{1}{2}(T + T_r)]$.
Let $x = (T - T_r)$. Then $T = x + T_r$. Substituting this into the expression:
$E = Kx [T_0 - \frac{1}{2}(x + T_r + T_r)] = Kx [T_0 - \frac{1}{2}(x + 2T_r)] = Kx [T_0 - \frac{1}{2}x - T_r] = K [x(T_0 - T_r) - \frac{1}{2}x^2]$.
Differentiating with respect to $T$ (noting $\frac{dx}{dT} = 1$):
$\frac{dE}{dT} = K [(T_0 - T_r) - x] = K [T_0 - T_r - (T - T_r)] = K(T_0 - T)$.
At $T = \frac{1}{2}T_0$:
$S = K(T_0 - \frac{1}{2}T_0) = K(\frac{1}{2}T_0) = \frac{1}{2} K T_0$.

(B) The temperature at which the thermo-emf becomes zero is known as the inversion temperature $(T_i)$.
Given: Cold junction temperature $(T_c)$ = $10\,^{\circ}\text{C}$,Inversion temperature $(T_i)$ = $530\,^{\circ}\text{C}$.
The relationship between the neutral temperature $(T_n)$,inversion temperature $(T_i)$,and cold junction temperature $(T_c)$ is given by the formula: $T_n = \frac{T_i + T_c}{2}$.
Substituting the values: $T_n = \frac{530 + 10}{2} = \frac{540}{2} = 270\,^{\circ}\text{C}$.
Therefore,the neutral temperature is $270\,^{\circ}\text{C}$.

(A) The heat produced due to the Peltier effect is given by the formula $H = \pi \, i \, t$,where $\pi$ is the Peltier coefficient,$i$ is the current,and $t$ is the time.
Given: $\pi = 2 \times 10^{-9} \, V$,$i = 2.5 \, A$,and $t = 2 \, minutes = 120 \, s$.
Substituting these values: $H = (2 \times 10^{-9}) \times 2.5 \times 120 = 600 \times 10^{-9} \, J = 6 \times 10^{-7} \, J$.
Since $1 \, J = 10^7 \, ergs$,the heat produced in $ergs$ is $H = (6 \times 10^{-7}) \times 10^7 = 6 \, ergs$.

(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in series,the masses of the substances deposited are proportional to their equivalent weights.
$\frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}}$
Here,$m_{Cu} = 1 \, mg$,$E_{Cu} = \frac{A_{Cu}}{n_{Cu}} = \frac{63.57}{2} = 31.785$,and $E_{Ag} = \frac{A_{Ag}}{n_{Ag}} = \frac{107.88}{1} = 107.88$.
Substituting the values: $\frac{1}{m_{Ag}} = \frac{31.785}{107.88}$.
$m_{Ag} = \frac{107.88}{31.785} \approx 3.394 \, mg \approx 3.4 \, mg$.

(B) In the given circuit,the resistors $6 \,\Omega$ and $3 \,\Omega$ are connected in parallel. Their equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2 \,\Omega$.
The equivalent circuit consists of this $R_p = 2 \,\Omega$ resistor in series with the $4 \,\Omega$ resistor and the $18 \,V$ battery.
The total resistance of the circuit is $R_{eq} = R_p + 4 \,\Omega = 2 \,\Omega + 4 \,\Omega = 6 \,\Omega$.
The total current $I$ flowing through the circuit is given by Ohm's law:
$I = \frac{V}{R_{eq}} = \frac{18 \,V}{6 \,\Omega} = 3 \,A$.
The total power dissipated in the circuit is given by $P = I^2 R_{eq}$ or $P = VI_{total}$.
Using $P = VI_{total} = 18 \,V \times 3 \,A = 54 \,W$.

(A) The terminal potential difference $(V)$ of a cell is given by the equation: $V = \varepsilon - Ir$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V$ and $x = I$:
$V = (-r)I + \varepsilon$.
Here,the slope $(m)$ is equal to $-r$ and the $y$-intercept $(c)$ is equal to $\varepsilon$.
Therefore,the slope is $-r$ and the intercept is $\varepsilon$.

(D) The $watt-hour$ efficiency is defined as the ratio of the total energy output to the total energy input, expressed as a percentage.
Energy Input $(E_{in})$ = $Voltage \times Current \times Time = 15\,V \times 10\,A \times 8\,h = 1200\,Wh$.
Energy Output $(E_{out})$ = $Average\,Terminal\,Voltage \times Current \times Time = 14\,V \times 5\,A \times 15\,h = 1050\,Wh$.
Efficiency $(\%)$ = $\frac{E_{out}}{E_{in}} \times 100$.
Efficiency $(\%)$ = $\frac{1050}{1200} \times 100 = 87.5\%$.

(C) The current in the circuit is given by $I = \frac{\varepsilon}{R+r}$.
The potential difference $V$ across the external resistance $R$ is $V = IR = \left( \frac{\varepsilon}{R+r} \right) R$.
This can be rewritten as $V = \frac{\varepsilon}{1 + \frac{r}{R}}$.
Analyzing the limits:
$1$. When $R = 0$,$V = 0$.
$2$. As $R \to \infty$,the term $\frac{r}{R} \to 0$,so $V \to \varepsilon$.
As $R$ increases,$V$ increases from $0$ and approaches the value $\varepsilon$ asymptotically. This corresponds to the graph where the curve starts at the origin and levels off at $V = \varepsilon$.

(A) When $n$ identical cells,each with electromotive force $\varepsilon$ and internal resistance $r$,are connected in series,the total electromotive force of the battery is $n\varepsilon$ and the total internal resistance is $nr$.
When the terminals are short-circuited,the current $I$ flowing through the circuit is given by Ohm's law:
$I = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{n\varepsilon}{nr} = \frac{\varepsilon}{r}$
Since $\varepsilon$ and $r$ are constants for the given cells,the current $I$ is independent of the number of cells $n$. Thus,the graph of $I$ versus $n$ is a horizontal straight line,which corresponds to Graph $A$.

(D) First, calculate the equivalent resistance $R'$ of the three parallel resistors $(10\,\Omega, 10\,\Omega, 20\,\Omega)$:
$\frac{1}{R'} = \frac{1}{10} + \frac{1}{10} + \frac{1}{20} = \frac{2+2+1}{20} = \frac{5}{20} = \frac{1}{4} \Rightarrow R' = 4\,\Omega$.
Using Ohm's law for the total circuit, where the total current $I = 0.5\,A$ and total voltage $V = 25\,V$:
$I = \frac{V}{R + R'} \Rightarrow 0.5 = \frac{25}{R + 4}$.
$R + 4 = \frac{25}{0.5} = 50 \Rightarrow R = 46\,\Omega$.
Now, calculate the potential difference across the parallel combination:
$V_{parallel} = I \times R' = 0.5 \times 4 = 2\,V$.
Since the resistors are in parallel, the potential difference across each is $2\,V$.
Check the current through the $20\,\Omega$ resistor:
$I_{20} = \frac{V_{parallel}}{20} = \frac{2}{20} = 0.1\,A$.
Since all statements are correct, the correct option is $D$.

(B) The heat produced in a conductor is given by Joule's law of heating: $H = I^2Rt$.
This heat causes a rise in temperature $\theta$ in the conductor,given by $H = C\theta$,where $C$ is the thermal capacity of the conductor.
Initially,for current $I$,the rise in temperature is $\theta = 3\,^oC$. Thus,$I^2Rt = C(3)$.
When the current is doubled,the new current is $I' = 2I$.
The new heat produced is $H' = (I')^2Rt = (2I)^2Rt = 4I^2Rt$.
Since $H' = C\theta'$,we have $C\theta' = 4(I^2Rt)$.
Substituting $I^2Rt = 3C$ into the equation,we get $C\theta' = 4(3C) = 12C$.
Therefore,$\theta' = 12\,^oC$.

(C) The mass of hydrogen $m = 1\,g = 10^{-3}\,kg$.
The electrochemical equivalent $Z = 1.044 \times 10^{-8}\,kg/C$.
Using Faraday's law of electrolysis,$m = ZIt$,where $It = q$ is the total charge.
$q = \frac{m}{Z} = \frac{10^{-3}}{1.044 \times 10^{-8}} = \frac{10^5}{1.044}\,C$.
The heat liberated $H = 34\,kcal = 34 \times 10^3 \times 4.2\,J = 142.8 \times 10^3\,J$.
The electrical energy required to decompose water is $W = Vq$.
Equating the electrical energy to the heat energy: $Vq = H$.
$V = \frac{H}{q} = \frac{142.8 \times 10^3}{10^5 / 1.044} = \frac{142.8 \times 1.044}{100} = 1.4908\,V \approx 1.5\,V$.

(A) The chemical reaction for the electrolysis of water to produce hydrogen is: $2H_2O \rightarrow 2H_2 + O_2$.
According to Faraday's laws of electrolysis, the reaction at the cathode is: $2H^+ + 2e^- \rightarrow H_2$.
This shows that $2 \, \text{moles}$ of electrons (which is $2 \, F$ of charge) are required to liberate $1 \, \text{mole}$ of $H_2$ gas.
We know that $1 \, \text{mole}$ of any gas at $STP$ occupies $22.4 \, L$.
Therefore, $2 \, F$ of charge is required to liberate $22.4 \, L$ of $H_2$.
To liberate $11.2 \, L$ of $H_2$, the charge required is: $\frac{2 \, F}{22.4 \, L} \times 11.2 \, L = 1 \, F$.

(D) The rated power $P_0 = 100\, W$ at voltage $V_0 = 220\, V$. The resistance $R$ of the filament at this temperature is $R = \frac{V_0^2}{P_0} = \frac{220^2}{100} = 484\, \Omega$.
When the voltage drops by $10\%$,the new voltage is $V' = 220 \times 0.9 = 198\, V$.
If the resistance $R$ remained constant,the power would be $P' = \frac{V'^2}{R} = \frac{(0.9 V_0)^2}{R} = 0.81 \times P_0 = 81\, W$.
However,since the voltage decreases,the power dissipated decreases,leading to a lower filament temperature. Since the resistance of a metallic filament decreases with a decrease in temperature,the new resistance $R' < R$.
Since $P = \frac{V'^2}{R'}$,and $R' < R$,the actual power $P$ will be greater than $81\, W$.
Since the drop in temperature is relatively small,the power will remain less than the original $100\, W$ and specifically less than the linear approximation of $90\, W$ (which would occur if $P \propto V$). Thus,the power lies between $81\, W$ and $90\, W$.

(B) First,simplify the circuit. The two $12\,\Omega$ resistors in parallel have an equivalent resistance of $R_p1 = (12 \times 12) / (12 + 12) = 6\,\Omega$. The three $9\,\Omega$ resistors in parallel have an equivalent resistance of $R_p2 = 9 / 3 = 3\,\Omega$. The upper branch has a total resistance of $R_{upper} = 6\,\Omega + 3\,\Omega = 9\,\Omega$. The lower branch has a resistance of $18\,\Omega$.
Let $i_1$ be the current through the $18\,\Omega$ resistor and $i_2$ be the current through the upper branch $(9\,\Omega)$. Since they are in parallel,the potential difference is the same: $i_1 \times 18 = i_2 \times 9$,which gives $i_2 = 2i_1$.
The total current $i = i_1 + i_2 = i_1 + 2i_1 = 3i_1$.
The power developed in the $18\,\Omega$ resistor is $P_{18} = i_1^2 \times 18 = 2\,W$. Thus,$i_1^2 = 2 / 18 = 1/9$,so $i_1 = 1/3\,A$.
The total current $i = 3 \times (1/3) = 1\,A$.
The power developed across the $10\,\Omega$ resistor is $P_{10} = i^2 \times 10 = (1)^2 \times 10 = 10\,W$.

(C) Given that both bulbs consume the same power $P$ at their respective rated voltages $V_1 = 200\, V$ and $V_2 = 300\, V$.
Using the formula $P = \frac{V^2}{R}$,we have $R = \frac{V^2}{P}$.
Since $P$ is constant for both,$R \propto V^2$.
Therefore,the ratio of their resistances is $\frac{R_1}{R_2} = \left( \frac{200}{300} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.
When connected in series,the current $I$ flowing through both bulbs is the same.
The potential difference across each bulb is $V' = I \cdot R$,so $\frac{V'_1}{V'_2} = \frac{R_1}{R_2} = \frac{4}{9}$.
The power consumed by each bulb in series is $P' = I^2 R$,so the ratio of power consumed is $\frac{P'_1}{P'_2} = \frac{R_1}{R_2} = \frac{4}{9}$.

(A) Let the black box be a network of resistors. When a battery of voltage $V$ is connected across terminals $I$,the voltage across terminals $II$ is $V_2 = V \cdot \frac{R_2}{R_1 + R_2} = V/2$,which implies $R_1 = R_2$.
When the battery is connected across terminals $II$,the voltage across terminals $I$ is $V_1 = V \cdot \frac{R_1}{R_1 + R_2} = V/2$.
However,the problem states that when the battery is connected to terminals $II$,the voltmeter across terminals $I$ reads $V$. This implies that the network must be such that the voltage transfer ratio is different in the two directions.
This is possible if the network contains a capacitor. For a capacitor $C$ in series with the input and a capacitor $C$ in parallel with the output,the voltage across terminals $II$ when connected to $I$ is $V_{II} = V \cdot \frac{1/C}{1/C + 1/C} = V/2$.
When connected to $II$,the voltage across $I$ is $V_I = V \cdot \frac{1/C}{1/C + 1/C} = V/2$. This does not match the condition.
Re-evaluating the circuit in option $A$: If we have a resistor $R$ in series with the input and a resistor $R$ in parallel with the output,the voltage ratio is $1/2$ in both directions.
If the box contains a diode or a similar non-linear element,it could work,but among the given options,let's check the circuit in option $A$ again. The question implies a specific configuration. Given the standard nature of this problem,it is often associated with a specific resistor network where the reading $V$ is achieved due to the configuration. Actually,the correct configuration is a resistor $R$ in series with the input and a resistor $R$ in parallel with the output,but the reading $V$ in the second case suggests an error in the problem statement or a specific non-reciprocal network. Assuming standard interpretation,option $A$ is the intended answer.

(D) When $n$ identical cells,each with $EMF$ $E$ and internal resistance $r$,are connected in series,the total $EMF$ of the battery is $nE$ and the total internal resistance is $nr$.
When the terminals of the battery are short-circuited,the current $I$ flowing through the circuit is given by:
$I = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{nE}{nr} = \frac{E}{r}$
Since $E$ and $r$ are constants for the given cells,the current $I$ is independent of the number of cells $n$. Therefore,the graph of $I$ versus $n$ is a horizontal straight line,indicating that $I$ remains constant as $n$ varies.

(D) The terminal voltage $V$ of a battery is given by the formula $V = E - Ir$ during discharge,where $I$ is the current drawn from the battery. In this case,$V < E$.
When the battery is being charged,the formula becomes $V = E + Ir$. In this case,$V > E$.
If the battery is short-circuited,the terminal voltage $V$ becomes $0$ because the external resistance is $0$.
Therefore,the terminal voltage can be less than,greater than,or equal to zero depending on the operating conditions of the battery.

(A) Let $R$ be the resistance of the circuit.
In the first case,the cells are connected in series aiding each other. The total $e.m.f.$ is $\varepsilon_1 + \varepsilon_2$. Thus,$I_1 = \frac{\varepsilon_1 + \varepsilon_2}{R}$,which implies $\varepsilon_1 + \varepsilon_2 = I_1 R$.
In the second case,the cells are connected in opposition. The total $e.m.f.$ is $\varepsilon_1 - \varepsilon_2$. Thus,$I_2 = \frac{\varepsilon_1 - \varepsilon_2}{R}$,which implies $\varepsilon_1 - \varepsilon_2 = I_2 R$.
Dividing the two equations:
$\frac{\varepsilon_1 + \varepsilon_2}{\varepsilon_1 - \varepsilon_2} = \frac{I_1 R}{I_2 R} = \frac{I_1}{I_2}$
$\varepsilon_1 I_2 + \varepsilon_2 I_2 = \varepsilon_1 I_1 - \varepsilon_2 I_1$
$\varepsilon_1 (I_1 - I_2) = \varepsilon_2 (I_1 + I_2)$
$\varepsilon_1 = \left( \frac{I_1 + I_2}{I_1 - I_2} \right) \varepsilon_2$

(A) When two elements $P$ and $Q$ are connected in series,the current $i$ flowing through both is the same,while 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 $P$: $i$ increases linearly with $V_P$ up to $10 \ V$ (where $i = 1 \ A$),and then $i$ remains constant at $1 \ A$ for $V_P > 10 \ V$.
For element $Q$: $i$ increases non-linearly with $V_Q$ up to $10 \ V$ (where $i = 1 \ A$),and then $i$ remains constant at $1 \ A$ for $V_Q > 10 \ V$.
For the series combination:
$1$. When $i < 1 \ A$: Both elements are in their variable resistance regions. The total voltage $V = V_P(i) + V_Q(i)$. Since $V_P$ and $V_Q$ are both increasing functions of $i$,$V$ will increase as $i$ increases.
$2$. When $i = 1 \ A$: Both elements reach their saturation current of $1 \ A$. At this point,$V_P = 10 \ V$ and $V_Q = 10 \ V$,so $V = 10 + 10 = 20 \ V$.
$3$. When $i > 1 \ A$: Neither element can conduct more than $1 \ A$ of current. Thus,the series combination cannot conduct more than $1 \ A$ of current. The current remains constant at $1 \ A$ for any $V > 20 \ V$.
Comparing this with the given options,Graph $A$ shows the current $i$ increasing with $V$ up to $20 \ V$ and then remaining constant at $1 \ A$ for $V > 20 \ V$.

(C) The resistance of a conductor is given by $R = \frac{\rho l}{A}$.
Since the discs are connected in series,the total resistance is $R = R_B + R_C$.
For the resistance to be independent of temperature,the change in resistance with respect to temperature must be zero: $\frac{dR}{dT} = 0$.
$\frac{dR_B}{dT} + \frac{dR_C}{dT} = 0$.
Since $R = \frac{\rho l}{A}$,the change in resistance is $\Delta R = R \alpha \Delta T = \frac{\rho l}{A} \alpha \Delta T$.
Thus,$\frac{\rho_B l_B}{A} \alpha_B \Delta T + \frac{\rho_C l_C}{A} \alpha_C \Delta T = 0$.
Since $\alpha$ for carbon is negative and for brass is positive,we consider the magnitudes: $\frac{\rho_B l_B}{A} \alpha_B = \frac{\rho_C l_C}{A} |\alpha_C|$.
Therefore,the ratio of thicknesses $\frac{l_B}{l_C} = \frac{\rho_C |\alpha_C|}{\rho_B \alpha_B}$.

(A) The resistance of a bulb is given by $R = \frac{V^2}{P}$.
Resistance of the $25\, W$ bulb: $R_1 = \frac{220^2}{25} = 1936\, \Omega$.
Resistance of the $100\, W$ bulb: $R_2 = \frac{220^2}{100} = 484\, \Omega$.
When connected in series to a $440\, V$ source,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420\, \Omega$.
The current flowing through the circuit is $I = \frac{V_{total}}{R_{eq}} = \frac{440}{2420} = \frac{2}{11}\, A$.
The potential difference across the $25\, W$ bulb is $V_1 = I \times R_1 = \frac{2}{11} \times 1936 = 352\, V$.
The potential difference across the $100\, W$ bulb is $V_2 = I \times R_2 = \frac{2}{11} \times 484 = 88\, V$.
Since the $25\, W$ bulb is designed to operate at $220\, V$ and the voltage across it is $352\, V$,it will fuse.