Circuit Solving for current and Voltage Questions in English

(C) The heating effect in a fuse wire is governed by Joule's law of heating,given by $H = I^2 R t$,where $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
For a fuse to melt at a specific current $I$,the heat generated must be sufficient to raise the temperature to the melting point.
Since the heat generated depends on the resistance $R$,which is proportional to the resistivity $\rho$ and inversely proportional to the square of the radius $r$,these factors are critical.
The current $I$ is the threshold value the fuse is designed for.
The length $l$ of the wire does not affect the current-carrying capacity or the melting point of the fuse wire,as the heat generated per unit length is what matters for the fuse's operation. Thus,the length is immaterial.

(C) In a steady state,the branches containing capacitors act as open circuits and can be neglected.
Thus,the current flows only through the resistors in the bottom and middle branches.
The equivalent resistance of the circuit is $R_{eq} = 2\,\Omega + 2\,\Omega = 4\,\Omega$.
The power consumed by the circuit is given by $P = \frac{V^2}{R_{eq}}$.
Substituting the values,$P = \frac{(2)^2}{4} = \frac{4}{4} = 1\,W$.

(C) When resistors are connected in parallel,the potential difference $(V)$ across each resistor is the same.
The power dissipated $(P)$ in a resistor is given by the formula $P = \frac{V^2}{R}$.
Since $V$ is constant,the power dissipated is inversely proportional to the resistance: $P \propto \frac{1}{R}$.
Given the ratio of resistances is $\frac{R_1}{R_2} = \frac{2}{1}$.
Therefore,the ratio of power dissipated is $\frac{P_1}{P_2} = \frac{R_2}{R_1} = \frac{1}{2}$.

(C) The resistance of the lamp when it is hot (at rated power) is given by the formula $R_{Hot} = \frac{V^2}{P}$.
Substituting the given values: $R_{Hot} = \frac{200 \times 200}{100} = 400\,\Omega$.
It is given that the hot resistance is $10$ times the cold resistance: $R_{Hot} = 10 \times R_{Cold}$.
Therefore,the cold resistance (resistance when not in use) is $R_{Cold} = \frac{R_{Hot}}{10}$.
$R_{Cold} = \frac{400}{10} = 40\,\Omega$.

(B) Given:
$e.m.f.$ of the battery $(E)$ = $3\, V$
Internal resistance $(r_{int})$ = $1.0\, \Omega$
Current in the circuit $(I)$ = $1.5\, A$
Let the resistance of the voltameter be $R$.
According to Ohm's law for a complete circuit:
$E = I(R + r_{int})$
Substituting the given values:
$3 = 1.5(R + 1.0)$
$R + 1.0 = \frac{3}{1.5}$
$R + 1.0 = 2$
$R = 2 - 1.0 = 1\, \Omega$
Therefore,the resistance of the voltameter is $1\, \Omega$.

(B) The current flowing through a voltameter is determined by the external circuit connected to it. According to the principle of continuity in a series circuit,the current entering the voltameter must be equal to the current leaving it. Therefore,the current inside the copper voltameter is the same as the current drawn from the battery in the external circuit.

(A) The electrochemical equivalent $(Z)$ of a substance is defined by the formula $Z = \frac{M}{nF}$,where $M$ is the molar mass of the substance,$n$ is the valency (number of electrons transferred),and $F$ is Faraday's constant.
Since $M$,$n$,and $F$ are properties intrinsic to the material and the chemical reaction,the electrochemical equivalent depends solely on the nature of the material.
Therefore,the correct option is $A$.

(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the charge.
Since the masses deposited are equal,we have $m_1 = m_2$,which implies $z_1 q_1 = z_2 q_2$.
From this,the ratio of charges is $\frac{q_1}{q_2} = \frac{z_2}{z_1}$.
We know that the total charge $q = q_1 + q_2$.
We want to find the charge $q_2$ flowing through the silver voltameter.
From the ratio,$q_1 = q_2 \frac{z_2}{z_1}$.
Substituting this into the total charge equation: $q = q_2 \frac{z_2}{z_1} + q_2 = q_2 (1 + \frac{z_2}{z_1})$.
Solving for $q_2$,we get $q_2 = \frac{q}{1 + \frac{z_2}{z_1}}$.

(B) The thermo $e.m.f.$ $(E)$ in a thermocouple as a function of temperature $(T)$ is given by the relation: $E = aT + \frac{1}{2}bT^2$.
To find the temperature at which the $e.m.f.$ is maximum,we differentiate $E$ with respect to $T$ and set it to zero:
$\frac{dE}{dT} = a + bT = 0$.
Solving for $T$,we get $T = -\frac{a}{b}$. This temperature is defined as the neutral temperature $(T_n)$.
At the neutral temperature,the thermo $e.m.f.$ reaches its maximum value.

(C) The Thomson effect describes the phenomenon where an electromotive force is generated in a single homogeneous conductor when a temperature gradient exists between its two ends.
This potential difference arises due to the redistribution of charge carriers caused by the temperature difference.
Therefore,the electromotive force generated between the two ends of a single conductor at different temperatures is known as the Thomson electromotive force.

(A) The thermo $e.m.f.$ produced by a thermocouple is given by $e = S \times \theta$,where $S$ is the sensitivity $(30 \,\mu V/^\circ C)$ and $\theta$ is the temperature difference.
Given,the minimum current $i = 3 \times 10^{-7} \, A$ and resistance $R = 50 \,\Omega$.
Using Ohm's law,$i = \frac{e}{R}$,we have $i = \frac{S \times \theta}{R}$.
Substituting the values: $3 \times 10^{-7} = \frac{(30 \times 10^{-6}) \times \theta}{50}$.
Solving for $\theta$: $\theta = \frac{3 \times 10^{-7} \times 50}{30 \times 10^{-6}}$.
$\theta = \frac{150 \times 10^{-7}}{30 \times 10^{-6}} = 5 \times 10^{-1} = 0.5^\circ C$.

(D) The rated power is $P_1 = 100\,W$ at $V_1 = 220\,V$. The resistance at this temperature is $R_1 = \frac{V_1^2}{P_1}$.
When connected to $V_2 = 220 \times 0.8\,V$,the new power is $P_2 = \frac{V_2^2}{R_2}$.
Taking the ratio: $\frac{P_2}{P_1} = \frac{V_2^2}{V_1^2} \times \frac{R_1}{R_2} = (0.8)^2 \times \frac{R_1}{R_2}$.
Since the voltage decreases,the temperature of the filament decreases,which causes the resistance to decrease $(R_2 < R_1)$. Therefore,$\frac{R_1}{R_2} > 1$.
This implies $P_2 > (0.8)^2 \times P_1 = 100 \times (0.8)^2\,W$.
Also,since $P = VI$,and both $V$ and $I$ decrease compared to the rated values,the power $P_2$ must be less than the power calculated if resistance remained constant at $R_1$ (which would be $100 \times 0.8\,W$).
Thus,the actual power $P_2$ lies between $100 \times (0.8)^2\,W$ and $100 \times 0.8\,W$.

(B) The circuit consists of two parallel branches. The upper branch has a total resistance of $R_2 = 4\, \Omega + 6\, \Omega = 10\, \Omega$. The lower branch has a resistance of $R_1 = 5\, \Omega$.
Since the branches are in parallel, the potential difference $V$ across both branches is the same.
The current in the branches is inversely proportional to their resistance: $\frac{i_1}{i_2} = \frac{R_2}{R_1} = \frac{10}{5} = \frac{2}{1}$.
The heat produced per second (power) is given by $P = i^2 R$.
For the $5\, \Omega$ resistor, $P_5 = i_1^2 \times 5 = 10\, \text{cal/s}$.
For the $4\, \Omega$ resistor, the current is $i_2$. The power dissipated in the upper branch is $P_{\text{upper}} = i_2^2 \times R_2 = i_2^2 \times 10$.
Using the ratio $\frac{i_1}{i_2} = 2$, we have $i_1 = 2i_2$. Substituting this into the power equation for the $5\, \Omega$ resistor: $(2i_2)^2 \times 5 = 10 \Rightarrow 20i_2^2 = 10 \Rightarrow i_2^2 = 0.5$.
The power dissipated in the $4\, \Omega$ resistor is $P_4 = i_2^2 \times 4 = 0.5 \times 4 = 2\, \text{cal/s}$.

(C) The heat required to boil the water is given by $H = \frac{V^2}{R} t$,where $V$ is the supply voltage,$R$ is the resistance of the heating wire,and $t$ is the time taken.
Since the supply voltage $V$ and the amount of heat $H$ required are the same in both cases,we have $H = \frac{V^2}{R_1} t_1 = \frac{V^2}{R_2} t_2$.
This implies $\frac{t_1}{R_1} = \frac{t_2}{R_2}$,or $\frac{t_2}{t_1} = \frac{R_2}{R_1}$ ..... $(i)$
We know that the resistance of a wire is proportional to its length,$R \propto l$,so $\frac{R_2}{R_1} = \frac{l_2}{l_1}$ ..... $(ii)$
Substituting $(ii)$ into $(i)$,we get $\frac{t_2}{t_1} = \frac{l_2}{l_1}$.
Given that the new length $l_2 = \frac{2}{3} l_1$,we have $\frac{l_2}{l_1} = \frac{2}{3}$.
Substituting the values,$\frac{t_2}{15} = \frac{2}{3}$,which gives $t_2 = 15 \times \frac{2}{3} = 10\,\text{min}$.

(D) The circuit consists of two parallel branches connected across the same potential difference $V$.
Resistance of the upper branch, $R_1 = 2\,\Omega + 3\,\Omega = 5\,\Omega$.
Resistance of the lower branch, $R_2 = 6\,\Omega + 4\,\Omega = 10\,\Omega$.
Since the branches are in parallel, the potential difference $V$ across both is the same.
Let $i_1$ be the current in the upper branch and $i_2$ be the current in the lower branch.
$V = i_1 R_1 = i_2 R_2 \implies i_1(5) = i_2(10) \implies i_1 = 2i_2$.
The heat generated per second (power) in a resistor is given by $P = i^2 R$.
Given, heat generated in $6\,\Omega$ resistor is $P_2 = i_2^2 \times 6 = 60\, \text{cal/s}$.
Thus, $i_2^2 = 10$.
We need to find the heat generated in the $3\,\Omega$ resistor, $P_1 = i_1^2 \times 3$.
Substituting $i_1 = 2i_2$, we get $P_1 = (2i_2)^2 \times 3 = 4i_2^2 \times 3 = 12i_2^2$.
Substituting $i_2^2 = 10$, we get $P_1 = 12 \times 10 = 120\, \text{cal/s}$.

(A) $1$. The power consumed by the heater is $P = 110\, W$ and its resistance is $R_h = 110\, \Omega$. Using the formula $P = \frac{V_h^2}{R_h}$,we find the voltage across the heater:
$110 = \frac{V_h^2}{110} \Rightarrow V_h^2 = 12100 \Rightarrow V_h = 110\, V$.
$2$. The current through the heater is $i_1 = \frac{V_h}{R_h} = \frac{110}{110} = 1\, A$.
$3$. The total voltage is $220\, V$. Since the heater combination is in series with an $11\, \Omega$ resistor,the voltage across the $11\, \Omega$ resistor is $V_s = 220 - 110 = 110\, V$.
$4$. The total current $i$ flowing through the $11\, \Omega$ resistor is $i = \frac{V_s}{11} = \frac{110}{11} = 10\, A$.
$5$. By Kirchhoff's Current Law,the current through $R$ is $i_2 = i - i_1 = 10 - 1 = 9\, A$.
$6$. Since $R$ is in parallel with the heater,the voltage across $R$ is also $110\, V$. Thus,$R = \frac{V_h}{i_2} = \frac{110}{9} \approx 12.22\, \Omega$.

(B) For normal brightness of each bulb,the voltage across each bulb must be $1.5\,V$ and the current through each bulb must be $0.5\,A$.
Since the four bulbs are connected in parallel,the total current $i$ flowing through the circuit is the sum of the currents through each bulb: $i = 0.5\,A + 0.5\,A + 0.5\,A + 0.5\,A = 2\,A$.
The voltage across the parallel combination of bulbs is $1.5\,V$.
Since the battery voltage is $12\,V$,the voltage across the resistor $R$ is $V_R = 12\,V - 1.5\,V = 10.5\,V$.
Using Ohm's law for the resistor $R$,$V_R = i \times R$,we get $10.5\,V = 2\,A \times R$.
Therefore,$R = 10.5 / 2 = 21/4\,\Omega$.

(D) The resistance of a bulb is given by $R = \frac{V^2}{P}$.
For bulb $B_1$ $(100\, W)$,$R_1 = \frac{V^2}{100}$.
For bulbs $B_2$ and $B_3$ $(60\, W)$,$R_2 = R_3 = \frac{V^2}{60}$.
Bulbs $B_1$ and $B_2$ are in series,and this combination is in parallel with bulb $B_3$ across the $250\, V$ source.
The power dissipated in $B_3$ is $W_3 = \frac{(250)^2}{R_3} = \frac{(250)^2}{V^2/60} = 60 \times \frac{(250)^2}{V^2}$.
The current through the series branch ($B_1$ and $B_2$) is $I = \frac{250}{R_1 + R_2} = \frac{250}{\frac{V^2}{100} + \frac{V^2}{60}} = \frac{250}{V^2(\frac{3+5}{300})} = \frac{250 \times 300}{8V^2} = \frac{75000}{8V^2} = \frac{9375}{V^2}$.
The power in $B_1$ is $W_1 = I^2 R_1 = (\frac{9375}{V^2})^2 \times \frac{V^2}{100} = \frac{9375^2}{100 V^2}$.
The power in $B_2$ is $W_2 = I^2 R_2 = (\frac{9375}{V^2})^2 \times \frac{V^2}{60} = \frac{9375^2}{60 V^2}$.
Comparing $W_1$ and $W_2$,since $R_1 < R_2$,$W_1 < W_2$. Since $B_3$ is connected directly across the source,it operates at its rated voltage,while $B_1$ and $B_2$ operate at lower voltages,thus $W_2 < W_3$. Therefore,$W_1 < W_2 < W_3$.

(D) Initially, the total resistance of the circuit is $R_1 = 2\, \Omega + 3\, \Omega = 5\, \Omega$. The current through the voltameter is $i_1 = \frac{V}{5}$.
When a $2\, \Omega$ resistor is connected in parallel with the $2\, \Omega$ voltameter, the equivalent resistance of this parallel combination is $R_p = \frac{2 \times 2}{2 + 2} = 1\, \Omega$.
The new total resistance of the circuit is $R_2 = 1\, \Omega + 3\, \Omega = 4\, \Omega$. The main current in the circuit is $i = \frac{V}{4}$.
Since the parallel resistors are equal ($2\, \Omega$ each), the current $i$ splits equally. Thus, the current through the voltameter is $i_2 = \frac{i}{2} = \frac{V}{8}$.
The rate of deposition of silver is proportional to the current, $R \propto i$. The percentage change in the rate is $\frac{i_2 - i_1}{i_1} \times 100$.
Percentage change $= \frac{\frac{V}{8} - \frac{V}{5}}{\frac{V}{5}} \times 100 = \frac{\frac{5V - 8V}{40}}{\frac{V}{5}} \times 100 = \frac{-3V}{40} \times \frac{5}{V} \times 100 = -37.5\, \%$.
Therefore, the rate of deposition decreases by $37.5\, \%$.

(B) The thermo $e.m.f.$ $(e)$ produced by the thermocouple is given by the product of the thermo $e.m.f.$ coefficient $(\alpha)$ and the temperature difference $(\Delta \theta)$: $e = \alpha \Delta \theta$.
According to Ohm's law, the $e.m.f.$ is also equal to the product of the current $(i)$ and the resistance $(R)$: $e = iR$.
Equating the two expressions: $\alpha \Delta \theta = iR$.
Given: $\alpha = 25\,\mu V/^{\circ}C = 25 \times 10^{-6}\,V/^{\circ}C$, $i = 10^{-5}\,A$, and $R = 40\,\Omega$.
Substituting the values: $(25 \times 10^{-6}) \times \Delta \theta = 10^{-5} \times 40$.
$\Delta \theta = \frac{40 \times 10^{-5}}{25 \times 10^{-6}} = \frac{400}{25} = 16\,^{\circ}C$.
Therefore, the smallest temperature difference that can be detected is $16\,^{\circ}C$.

(B) The rated power of the bulb is $P = 500\, W$ and the rated voltage is $V_{rated} = 100\, V$.
The current $i$ flowing through the bulb when it operates at its rated power is given by $i = \frac{P}{V_{rated}} = \frac{500}{100} = 5\, A$.
Since the bulb is connected in series with a resistor $R$ to a $200\, V$ supply,the total voltage is divided between the bulb and the resistor.
The voltage across the bulb is $100\, V$,so the voltage across the resistor $R$ must be $V_R = 200\, V - 100\, V = 100\, V$.
Using Ohm's law for the resistor $R$,we have $V_R = i \times R$.
Substituting the values,$100 = 5 \times R$.
Therefore,$R = \frac{100}{5} = 20\, \Omega$.

(D) In a thermocouple,the thermoelectric $EMF$ $(E)$ is given by $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the junctions.
Under normal conditions (temperature below the inversion temperature),the current flows from metal $X$ to metal $Y$ through the cold junction.
When the temperature of the hot junction exceeds the temperature of inversion,the direction of the thermoelectric $EMF$ reverses.
Consequently,the current direction reverses,flowing from $X$ to $Y$ through the hot junction,which is equivalent to flowing from $Y$ to $X$ through the cold junction.
Therefore,both options $(b)$ and $(c)$ describe the same physical state.

(B) According to Faraday's laws of electrolysis,the mass $m$ deposited on the cathode is given by $m = Zit$,where $Z$ is the electrochemical equivalent,$i$ is the current through the voltameter,and $t$ is the time.
Case $1$: Series connection.
The total resistance is $R_s = 2\,\Omega + 3\,\Omega = 5\,\Omega$.
The current through the voltameter is $i_1 = \frac{V}{R_s} = \frac{10\,V}{5\,\Omega} = 2\,A$.
The mass deposited is $m_1 = Z i_1 t = 1\,g$.
Case $2$: Parallel connection.
The voltameter is connected directly across the $10\,V$ battery.
The current through the voltameter is $i_2 = \frac{V}{R_v} = \frac{10\,V}{2\,\Omega} = 5\,A$.
The mass deposited is $m_2 = Z i_2 t$.
Taking the ratio: $\frac{m_2}{m_1} = \frac{i_2}{i_1} = \frac{5\,A}{2\,A} = 2.5$.
Since $m_1 = 1\,g$,we have $m_2 = 2.5\,g$.
The increase in the deposited mass is $\Delta m = m_2 - m_1 = 2.5\,g - 1\,g = 1.5\,g$.

(C) The mass of copper deposited is given by Faraday's law of electrolysis: $m = Z i t$.
Also,mass can be expressed as $m = \text{Density} (\rho) \times \text{Volume} (V) = \rho \times A \times x$,where $A$ is the area and $x$ is the thickness.
Equating the two expressions: $\rho A x = Z i t$.
Therefore,the thickness $x = \frac{Z i t}{A \rho}$.
Given values:
$Z = 0.00033\, g/C = 0.00033 \times 10^{-3}\, kg/C = 3.3 \times 10^{-7}\, kg/C$
$i = 1.5\, A$
$t = 20\, min = 20 \times 60 = 1200\, s$
$A = 50\, cm^2 = 50 \times 10^{-4}\, m^2$
$\rho = 9000\, kg/m^3$
Substituting the values:
$x = \frac{0.00033 \times 10^{-3} \times 1.5 \times 1200}{50 \times 10^{-4} \times 9000}$
$x = \frac{0.00033 \times 1.5 \times 1.2}{50 \times 9} = \frac{0.000594}{450} = 1.32 \times 10^{-6}\, m$ (Wait,re-calculating: $x = \frac{3.3 \times 10^{-7} \times 1.5 \times 1200}{50 \times 10^{-4} \times 9000} = \frac{5.94 \times 10^{-4}}{45} = 1.32 \times 10^{-5}\, m$).
Thus,the thickness is approximately $1.3 \times 10^{-5}\, m$.

(A) Let the temperature of the molten metal be $t\,^{\circ}C$.
The thermo-emf generated is $e = 10 \times 10^{-6} \times t\,\text{V} = 10^{-5}t\,\text{V}$.
The total resistance of the circuit is $R_{total} = R_{thermocouple} + R_{galvanometer} = 1.6\,\Omega + 8\,\Omega = 9.6\,\Omega$.
The current in the circuit is $i = \frac{e}{R_{total}} = \frac{10^{-5}t}{9.6}\,\text{A}$.
The galvanometer reads the potential difference across its own resistance,which is $V_G = 8\,\text{mV} = 8 \times 10^{-3}\,\text{V}$.
The current through the galvanometer is also $i = \frac{V_G}{R_G} = \frac{8 \times 10^{-3}}{8} = 10^{-3}\,\text{A}$.
Equating the two expressions for current: $\frac{10^{-5}t}{9.6} = 10^{-3}$.
Solving for $t$: $t = \frac{10^{-3} \times 9.6}{10^{-5}} = 9.6 \times 10^2 = 960\,^{\circ}C$.

(A) The Peltier coefficient $\pi$ is given by the relation $\pi = T \frac{de}{dT}$,where $T$ is the absolute temperature in Kelvin.
Given that one junction is at $0\,^{\circ}C$,the absolute temperature $T$ is related to the Celsius temperature $t$ by $T = t + 273$.
Substituting $t = T - 273$ into the given emf equation $e = at + bt^2$:
$e = a(T - 273) + b(T - 273)^2$.
Differentiating with respect to $T$:
$\frac{de}{dT} = a + 2b(T - 273)$.
Since $T - 273 = t$,we have $\frac{de}{dT} = a + 2bt$.
Now,calculating the Peltier coefficient:
$\pi = T \frac{de}{dT} = (t + 273)(a + 2bt)$.

(B) The rated voltage for both appliances is assumed to be $220\,V$.
Resistance of the bulb,$R_{Bulb} = \frac{V^2}{P} = \frac{220^2}{100} = 484\,\Omega$.
Resistance of the geyser,$R_{Geyser} = \frac{V^2}{P} = \frac{220^2}{1000} = 48.4\,\Omega$.
$(i)$ When only the bulb is $ON$,the circuit consists of the wiring resistance $(6\,\Omega)$ in series with the bulb $(484\,\Omega)$.
The potential drop across the bulb is $V_{Bulb1} = 220 \times \frac{484}{484 + 6} = 220 \times \frac{484}{490} \approx 217.3\,V$.
$(ii)$ When the geyser is also switched $ON$,the bulb and geyser are in parallel. Their equivalent resistance is $R_p = \frac{484 \times 48.4}{484 + 48.4} = \frac{23425.6}{532.4} = 44\,\Omega$.
The circuit now consists of the wiring resistance $(6\,\Omega)$ in series with the parallel combination $(44\,\Omega)$.
The potential drop across the parallel combination (and thus the bulb) is $V_{Bulb2} = 220 \times \frac{44}{44 + 6} = 220 \times \frac{44}{50} = 193.6\,V$.
The change in potential drop is $\Delta V = V_{Bulb1} - V_{Bulb2} = 217.3 - 193.6 = 23.7\,V$.
This is nearly $23\,V$.

(B) The charging current $i$ is given by the formula $i = \frac{V_{supply} - V_{battery}}{R_{ext} + r_{int}}$.
Substituting the given values: $i = \frac{24 - 12}{2 + 1} = \frac{12}{3} = 4\, A$.
The energy stored in the battery is given by $E = V_{battery} \times i \times t$.
Given $E = 360\, Wh$, $V_{battery} = 12\, V$, and $i = 4\, A$.
$360 = 12 \times 4 \times t$.
$360 = 48 \times t$.
$t = \frac{360}{48} = 7.5\, hours$.

(B) The circuit consists of a $12 \, \Omega$ resistor in series with a parallel combination of two branches. One branch contains a $5 \, \Omega$ resistor,and the other branch contains resistors of $9 \, \Omega$ and $6 \, \Omega$ in series (total $15 \, \Omega$).
Let $i_1$ be the current through the $5 \, \Omega$ resistor and $i_2$ be the current through the $9 \, \Omega$ and $6 \, \Omega$ branch.
Since the branches are in parallel,the potential difference across them is equal: $i_1 \times 5 = i_2 \times (9 + 6) = i_2 \times 15$.
Therefore,$\frac{i_1}{i_2} = \frac{15}{5} = 3$,which means $i_1 = 3i_2$.
The power developed in the $5 \, \Omega$ resistor is $P_5 = i_1^2 \times 5 = 45 \, W$.
$i_1^2 = \frac{45}{5} = 9 \Rightarrow i_1 = 3 \, A$.
Since $i_1 = 3i_2$,we have $3 = 3i_2$,so $i_2 = 1 \, A$.
The total current $i$ flowing through the $12 \, \Omega$ resistor is $i = i_1 + i_2 = 3 + 1 = 4 \, A$.
The power developed in the $12 \, \Omega$ resistor is $P_{12} = i^2 \times 12 = (4)^2 \times 12 = 16 \times 12 = 192 \, W$.

(A) The total charge $q$ passed through the voltameter is equal to the area under the current-time graph.
Area $= \text{Area of trapezium} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Area $= \frac{1}{2} \times (10 + 30) \times 100 \times 10^{-3} \, \text{A} = \frac{1}{2} \times 40 \times 0.1 = 2 \, \text{C}$.
According to Faraday's law of electrolysis,$m = z q$,where $z$ is the $ECE$ of copper.
Given that the mass deposited is $m$ for a charge $q = 2 \, \text{C}$,we have:
$m = z \times 2$
$z = \frac{m}{2}$.

(D) If at any instant,the current through the circuit is $i$,then applying Kirchhoff's voltage law to the circuit (assuming a source with $EMF$ $E$ and internal resistance $R$):
$iR + e = E$
Rearranging the equation,we get:
$e = E - iR$
This equation is of the form $y = mx + c$,where $y = e$,$x = i$,$m = -R$ (slope),and $c = E$ (intercept).
Since the slope is negative $(-R)$ and the intercept is positive $(E)$,the graph between $e$ and $i$ will be a straight line with a negative slope and a positive intercept on the $e$-axis.
Therefore,the correct graph is represented by option $D$.

(B) The heat produced $H$ in a wire with resistance $R$ under constant voltage $V$ is given by $H = \frac{V^2}{R} t$. For a fixed time $t$,$H \propto \frac{1}{R}$.
To double the heat $(H' = 2H)$,the resistance must be halved $(R' = R/2)$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.
If the length $l$ is halved $(l' = l/2)$ and the radius $r$ is kept constant,$R' = \rho \frac{l/2}{\pi r^2} = R/2$. This satisfies the condition.
However,checking the options provided: If the length $l$ is halved $(l' = l/2)$ and radius $r$ is halved $(r' = r/2)$,then $R' = \rho \frac{l/2}{\pi (r/2)^2} = \rho \frac{l/2}{\pi r^2 / 4} = 2R$. This does not work.
If the length $l$ is doubled $(l' = 2l)$ and radius $r$ is doubled $(r' = 2r)$,then $R' = \rho \frac{2l}{\pi (2r)^2} = \rho \frac{2l}{4\pi r^2} = R/2$. This satisfies the condition.
Therefore,the correct option is $B$.

(B) Let the left junction be $A$ and the right junction be $B$. The circuit is a bridge. The left part consists of two parallel branches: one with $2\Omega$ and $1\Omega$ in series (total $3\Omega$) and another with $4\Omega$ in series with the $PQ$ branch. However,looking at the bridge structure,the left side has two parallel branches connected to $P$ and $Q$: one branch has $2\Omega$ and $4\Omega$ resistors. The potential at $P$ and $Q$ can be calculated. The equivalent resistance of the left part is $R_L = (2 || 4) = 8/6 = 4/3 \Omega$. The equivalent resistance of the right part is $R_R = (2 || 4) = 4/3 \Omega$. The central part has $1\Omega$ (between $P$ and $S$) and $1\Omega$ (between $Q$ and $T$) and $4\Omega$ (between $Q$ and $T$). Actually,the bridge is balanced if the ratio of resistances is equal. The left side has $2\Omega$ and $4\Omega$ connected to the source. The potential at $P$ is $V_P = 12 \times (4/(2+4)) = 8V$ and $V_Q = 12 \times (2/(2+4)) = 4V$. The right side is symmetric,so $V_S = 4V$ and $V_T = 8V$. Since $V_P = 8V$ and $V_S = 4V$,current flows from $P$ to $S$. The branch $PQ$ has a $1\Omega$ resistor. Since $V_P=8V$ and $V_Q=4V$,current flows through $PQ$. Thus,option $A$ is incorrect. Calculating $I_1$: The total resistance is $R_{eq} = 4\Omega$. $I_1 = V/R_{eq} = 12/4 = 3A$. Option $B$ is correct. Since $V_S = 4V$ and $V_Q = 4V$,the potential at $S$ is equal to the potential at $Q$. Thus,$C$ is incorrect. Therefore,the correct option is $B$.

(B) First,calculate the resistance of each bulb using $R = V^2 / P$.
For the first bulb $(P_1 = 25\, W, V = 220\, V)$: $R_1 = (220)^2 / 25 = 48400 / 25 = 1936\, \Omega$.
For the second bulb $(P_2 = 100\, W, V = 220\, V)$: $R_2 = (220)^2 / 100 = 48400 / 100 = 484\, \Omega$.
When connected in series to a $440\, V$ supply,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420\, \Omega$.
The current in the series circuit is $I = V / R_{eq} = 440 / 2420 = 44 / 242 = 2 / 11\, A$.
The voltage drop across the first bulb is $V_1 = I \times R_1 = (2 / 11) \times 1936 = 2 \times 176 = 352\, V$.
The voltage drop across the second bulb is $V_2 = I \times R_2 = (2 / 11) \times 484 = 2 \times 44 = 88\, V$.
Since the voltage drop across the $25\, W$ bulb $(352\, V)$ exceeds its rated voltage $(220\, V)$,it will fuse.

(B) In a parallel circuit,the potential difference $V$ across each resistor is the same.
The power dissipated in a resistor is given by $P = \frac{V^2}{R}$.
Since $V$ is constant,$P \propto \frac{1}{R}$.
Therefore,$\frac{P_1}{P_2} = \frac{R_2}{R_1}$.
Given $P_1 = 6\,W$ for $R_1 = 6\,\Omega$,and we need to find $P_2$ for $R_2 = 4\,\Omega$.
Substituting the values: $\frac{6}{P_2} = \frac{4}{6}$.
$\frac{6}{P_2} = \frac{2}{3}$.
$P_2 = \frac{6 \times 3}{2} = 9\,W$.

(B) The mass deposited is given by Faraday's law: $m = Z \cdot i \cdot t$,where $Z$ is the electrochemical equivalent.
Given that $96500 \ C$ deposits $1 \ g$,the electrochemical equivalent $Z$ for the substance is $Z = \frac{1 \ g}{96500 \ C} = \frac{1}{96500} \ g/C$.
For copper $(Cu^{2+})$,the molar mass is approximately $63.5 \ g/mol$ and valency $n=2$,so the equivalent mass is $31.75 \ g/mol$. However,using the provided context where $1 \ g$ is deposited by $96500 \ C$,we use $Z = \frac{32}{96500} \ g/C$ (based on the provided solution logic).
Given $m = 20 \ mg = 20 \times 10^{-3} \ g$,$i = 0.15 \ A$.
Substituting the values: $20 \times 10^{-3} = \left( \frac{32}{96500} \right) \times 0.15 \times t$.
$t = \frac{20 \times 10^{-3} \times 96500}{32 \times 0.15} = \frac{1930}{4.8} \approx 402.08 \ s$.
$402.08 \ s = 6 \ min \ 42 \ s$.

(C) According to Faraday's laws 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.
For two different substances deposited by the same current for the same time,the ratio of their masses is equal to the ratio of their electrochemical equivalents:
$\frac{m_{Cu}}{m_{Ag}} = \frac{Z_{Cu}}{Z_{Ag}}$
Given: $m_{Cu} = 14 \ g$,$Z_{Cu} = 7 \times 10^{-6}$,$Z_{Ag} = 1.2 \times 10^{-6}$.
Substituting the values:
$\frac{14}{m_{Ag}} = \frac{7 \times 10^{-6}}{1.2 \times 10^{-6}}$
$m_{Ag} = \frac{14 \times 1.2}{7} = 2 \times 1.2 = 2.4 \ g$.

(B) The heat produced in a conductor is given by Joule's law of heating: $H = i^2 Rt$.
Given values are: Current $i = 2 \ A$,Heat $H = 80 \ J$,and Time $t = 10 \ s$.
Rearranging the formula to solve for resistance $R$: $R = \frac{H}{i^2 t}$.
Substituting the values: $R = \frac{80}{(2)^2 \times 10} = \frac{80}{4 \times 10} = \frac{80}{40} = 2 \ \Omega$.
Therefore,the resistance of the conductor is $2 \ \Omega$.

(B) The reading of the voltmeter when no current is drawn gives the electromotive force $(EMF)$ of the battery,so $E = 5\,V$.
When the ammeter is connected,it measures the short-circuit current,which is given by $I_{sc} = E/r$,where $r$ is the internal resistance.
Thus,$10 = 5/r$,which gives $r = 5/10 = 0.5\,\Omega$.
When a resistor $R = 2\,\Omega$ is connected across the battery,the total resistance in the circuit is $R_{total} = R + r = 2 + 0.5 = 2.5\,\Omega$.
The current $I$ flowing through the resistor is given by Ohm's law: $I = E / R_{total} = 5 / 2.5 = 2\,A$.

(A) Given that the galvanometer reading is zero,$I_g = 0$.
This implies that the current $I$ flowing through the $400 \ \Omega$ resistor and the resistor $X$ is the same.
The total resistance of the circuit loop is $R_{total} = 400 + X$.
Using Ohm's law,the current $I$ is given by $I = \frac{10}{400 + X}$.
The potential difference across the resistor $X$ is equal to the $EMF$ of the second cell,which is $2 \ V$.
Therefore,$I \cdot X = 2$.
Substituting the expression for $I$: $\frac{10 \cdot X}{400 + X} = 2$.
$10X = 2(400 + X)$.
$10X = 800 + 2X$.
$8X = 800$.
$X = 100 \ \Omega$.

(A) In the given circuit,the battery of $2 \ V$ is connected to a triangular network of three resistors,each of $3 \ \Omega$.
Looking at the circuit,the current from the battery splits at the top node. One path goes through one $3 \ \Omega$ resistor,and the other path goes through two $3 \ \Omega$ resistors connected in series.
Thus,the resistance of the first branch is $R_1 = 3 \ \Omega$.
The resistance of the second branch is $R_2 = 3 \ \Omega + 3 \ \Omega = 6 \ \Omega$.
These two branches are in parallel with each other across the $2 \ V$ battery.
The equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \ \Omega^{-1}$.
Therefore,$R_{eq} = 2 \ \Omega$.
The total current $I$ flowing from the battery is:
$I = \frac{V}{R_{eq}} = \frac{2 \ V}{2 \ \Omega} = 1 \ A$.

(C) cube has $12$ edges. Let the resistance of each edge be $R$. When we calculate the equivalent resistance between two diagonally opposite corners,the current splits into three paths at the first corner.
These three resistors are in parallel,each having resistance $R$,so their equivalent resistance is $R/3$.
Then,the current flows through six resistors in parallel,each having resistance $R$,so their equivalent resistance is $R/6$.
Finally,the current flows through three resistors in parallel,each having resistance $R$,so their equivalent resistance is $R/3$.
The total equivalent resistance $R_{eq}$ is the sum of these series combinations:
$R_{eq} = R/3 + R/6 + R/3 = (2R + R + 2R) / 6 = 5R/6$.

(D) The total resistance of the circuit is $R_{total} = 100 \, \Omega + 100 \, \Omega + 80 \, \Omega + 20 \, \Omega = 300 \, \Omega$.
The current flowing through the circuit is $I = \frac{V}{R_{total}} = \frac{48 \, V}{300 \, \Omega} = 0.16 \, A$.
The potential difference across $PQ$ (which corresponds to the $20 \, \Omega$ resistor) is $V_{PQ} = I \times R = 0.16 \, A \times 20 \, \Omega = 3.2 \, V$.

(C) The heat produced in a resistor is given by $Q = I^2 R t$. Since the battery has an electromotive force $E$ and internal resistance $r$,the current $I$ is given by $I = \frac{E}{R+r}$.
Given that the heat $Q$ produced is the same for both resistors in the same time $t$,we have:
$Q = \left( \frac{E}{R_1 + r} \right)^2 R_1 t = \left( \frac{E}{R_2 + r} \right)^2 R_2 t$
Canceling $E^2$ and $t$ from both sides:
$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$
Taking the square root of both sides:
$\frac{\sqrt{R_1}}{R_1 + r} = \frac{\sqrt{R_2}}{R_2 + r}$
$\sqrt{R_1}(R_2 + r) = \sqrt{R_2}(R_1 + r)$
$\sqrt{R_1} R_2 + r\sqrt{R_1} = \sqrt{R_2} R_1 + r\sqrt{R_2}$
$r(\sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_2} R_1 - \sqrt{R_1} R_2$
$r(\sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_1 R_2}(\sqrt{R_1} - \sqrt{R_2})$
$r = \sqrt{R_1 R_2}$

(D) Let the potential at point $A$ be $V_A = 0 \, V$. Then the potential at point $C$ is $V_C = 0 \, V$ because the circuit is symmetric and the two batteries of $5 \, V$ are connected in opposition relative to the central node.
Since the potential at point $A$ is $0 \, V$ and the potential at point $C$ is $0 \, V$,the potential difference across the $12 \, \Omega$ resistor connected between $E$ and $F$ (which are connected to $A$ and $C$ respectively) is $V_E - V_F = 0 \, V - 0 \, V = 0 \, V$.
According to Ohm's law,$I = V/R = 0 \, V / 12 \, \Omega = 0 \, A$.
Thus,the current passing through the $12 \, \Omega$ resistor is $0 \, A$.

(B) The formula for current in a circuit is $I = \frac{E}{R + r}$.
Given $E = 4 \ V$,$R = 2 \ \Omega$,and $I = 1 \ A$.
Substituting the values: $1 = \frac{4}{2 + r}$.
Solving for $r$: $2 + r = 4$,which gives $r = 2 \ \Omega$.
When the terminals are short-circuited,the external resistance $R = 0$.
The short-circuit current is $I_{sc} = \frac{E}{r} = \frac{4 \ V}{2 \ \Omega} = 2 \ A$.

(B) The circuit consists of three parallel branches connected between points $A$ and $B$.
However,the provided solution image shows a simplified loop where the potential difference between $A$ and $B$ is determined by the branches.
Based on the provided solution image,the effective electromotive force $(emf)$ is $2 \, V + 2 \, V = 4 \, V$ and the total resistance is $2 \, \Omega$.
Using Ohm's law,$I = \frac{V}{R} = \frac{4 \, V}{2 \, \Omega} = 2 \, A$.
Since the current flows from the positive terminal of the batteries towards $B$,the current through the branch containing $E_2$ (or the equivalent path) is $2 \, A$ from $A$ to $B$.

(B) The mass $m$ deposited at the cathode is given by Faraday's law: $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current through the voltameter,and $t$ is time.
Case $1$ (Series): The total resistance is $R_s = 2\,\Omega + 3\,\Omega = 5\,\Omega$. The current through the voltameter is $I_s = V / R_s = 10\,V / 5\,\Omega = 2\,A$. Given $m_s = 1\,g$.
Case $2$ (Parallel): The voltameter $(2\,\Omega)$ and the resistor $(3\,\Omega)$ are connected in parallel across the $10\,V$ battery. The current through the voltameter is $I_p = V / R_v = 10\,V / 2\,\Omega = 5\,A$.
Since $m \propto I$ for the same time $t$,we have $m_p / m_s = I_p / I_s$.
$m_p = (5\,A / 2\,A) \times 1\,g = 2.5\,g$.
The increase in mass is $\Delta m = m_p - m_s = 2.5\,g - 1\,g = 1.5\,g$.

(C) The circuit consists of two resistors of $64 \, \Omega$ and $32 \, \Omega$ connected in series across a potential difference of $60 \, V$.
Using the voltage divider rule,the potential difference across the $32 \, \Omega$ resistor is given by:
$V_J - V_G = \left( \frac{R_2}{R_1 + R_2} \right) \times V_{total}$
$V_J - V_G = \left( \frac{32}{64 + 32} \right) \times 60 \, V$
$V_J - V_G = \left( \frac{32}{96} \right) \times 60 \, V$
$V_J - V_G = \left( \frac{1}{3} \right) \times 60 \, V = 20 \, V$
Since the potential at $G$ is $V_G = 0 \, V$,the potential at $J$ is $V_J = 20 \, V$.

(B) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.
The circuit simplifies to a $6\,V$ battery connected in series with a $2.8\,\Omega$ resistor and a parallel combination of $2\,\Omega$ and $3\,\Omega$ resistors.
The equivalent resistance of the parallel part is $R_p = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2\,\Omega$.
The total resistance of the circuit is $R_{eq} = 2.8\,\Omega + 1.2\,\Omega = 4.0\,\Omega$.
The total current supplied by the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{4\,\Omega} = 1.5\,A$.
Using the current divider rule,the current through the $2\,\Omega$ resistor is $I_2 = I \times \frac{3}{2 + 3} = 1.5 \times \frac{3}{5} = 1.5 \times 0.6 = 0.9\,A$.