Electric Cells and Combination of cells in Series and Parallel Questions in English

(N/A) cell consists of two electrodes,positive $(P)$ and negative $(N)$,partially dipped in an electrolyte. Chemical reactions produce positive and negative ions,creating a potential difference between the electrodes and the electrolyte.
The potential difference between the positive electrode and the electrolyte is $V_{+} (V_{+} > 0)$,and between the electrolyte and the negative electrode is $V_{-} (V_{-} < 0)$.
When no current flows in the circuit,the potential difference between the two terminals $P$ and $N$ is $\varepsilon = V_{+} - (-V_{-}) = V_{+} + V_{-}$. This is defined as the electromotive force (emf) of the cell.
Definition of emf: The energy gained by a unit positive charge when it moves from the negative to the positive terminal of the cell due to non-electric forces is called the emf of the cell.
When a resistor $R$ is connected to the cell,a current $I$ flows through the circuit. The potential difference $V$ across the resistor $R$ is $V = IR$.
Due to the internal resistance $r$ of the cell,there is a potential drop across it equal to $Ir$. Thus,the total emf $\varepsilon$ is the sum of the potential difference across the external resistor and the potential drop across the internal resistance:
$\varepsilon = V + Ir$
Substituting $V = IR$,we get $\varepsilon = IR + Ir = I(R + r)$.
Therefore,the relation is $V = \varepsilon - Ir$.

(B) The term 'Electromotive force' $(EMF)$ is a misnomer because it is not a force in the mechanical sense.
$EMF$ represents the work done by a non-electrostatic source (like a chemical reaction in a battery) to move a unit positive charge from the lower potential terminal to the higher potential terminal inside the source.
Mathematically, $EMF$ $(\varepsilon)$ is defined as the work done $(W)$ per unit charge $(q)$: $\varepsilon = \frac{W}{q}$.
Therefore, it represents the energy supplied per unit charge by the source.

(B) In an open circuit,no current flows through the cell,i.e.,$I = 0$.
The terminal voltage $V$ of a cell is given by the formula $V = E - Ir$,where $E$ is the electromotive force $(EMF)$ and $r$ is the internal resistance of the cell.
Substituting $I = 0$ into the equation,we get $V = E - (0)r = E$.
Therefore,in an open circuit,the terminal voltage is equal to the $EMF$ of the cell.

(N/A) The terminal voltage $(V)$ of a cell is the potential difference across its terminals when it is delivering current $(I)$ to an external circuit.
Let $E$ be the electromotive force (emf) of the cell and $r$ be its internal resistance.
When the cell is discharging,the potential drop across the internal resistance is given by $Ir$.
The terminal voltage is the emf minus the potential drop across the internal resistance.
Thus,the relation is: $V = E - Ir$.

(A) The internal resistance of a cell is defined as the opposition offered by the electrolyte and the electrodes of the cell to the flow of current through it when the circuit is closed.
It is denoted by the symbol $r$.
Unlike external resistance,which is due to components like resistors or bulbs,internal resistance is an inherent property of the chemical composition and physical state of the cell itself.

(N/A) The internal resistance $r$ of a cell can be neglected when the external resistance $R$ connected in the circuit is very large compared to the internal resistance $(R \gg r)$.
Explanation:
The terminal voltage $V$ of a cell is given by the formula $V = E - Ir$,where $E$ is the electromotive force $(EMF)$,$I$ is the current,and $r$ is the internal resistance.
Since $I = E / (R + r)$,we can write $V = E - [E / (R + r)] \cdot r = E \cdot [R / (R + r)]$.
If $R \gg r$,then $(R + r) \approx R$.
Substituting this into the equation,we get $V \approx E \cdot (R / R) = E$.
In this condition,the potential drop across the internal resistance is negligible,and the terminal voltage is approximately equal to the $EMF$ of the cell.

(A) The current $I$ drawn from a cell of electromotive force $E$ and internal resistance $r$ connected to an external resistance $R$ is given by the formula: $I = \frac{E}{R + r}$.
To maximize the current $I$,the denominator $(R + r)$ must be minimized.
Since the internal resistance $r$ is a constant property of the cell,the current is maximum when the external resistance $R$ is as small as possible.
Ideally,when the external resistance $R = 0$ (short circuit condition),the current becomes maximum,given by $I_{max} = \frac{E}{r}$.

(N/A) single cell often cannot provide the necessary voltage or current required for a specific circuit. In such conditions,by connecting more than one cell in different combinations,the desired voltage or current can be obtained. This process is known as the combination of cells.
There are three primary methods of combining cells:
$(1)$ Series connection: Used to increase the total electromotive force $(EMF)$ of the circuit.
$(2)$ Parallel connection: Used to increase the total current capacity and reduce the internal resistance of the circuit.
$(3)$ Mixed connection: Used to achieve an optimal balance between voltage and current requirements.

(N/A) When one terminal of one cell is connected to the opposite terminal of another cell,and the remaining two terminals are left free,such a connection is called a series connection of cells.
In the figure,a battery having $emf$ $\varepsilon_1$ and internal resistance $r_1$ is connected between points $A$ and $B$,and a battery with $emf$ $\varepsilon_2$ and internal resistance $r_2$ is connected between points $B$ and $C$.
Let the equivalent $emf$ between $A$ and $C$ be $\varepsilon_{eq}$ and the equivalent internal resistance be $r_{eq}$.
Let the potentials at points $A, B,$ and $C$ be $V(A), V(B),$ and $V(C)$ respectively.
The potential difference $(p.d.)$ across the first cell is $V_{AB} = V(A) - V(B) = \varepsilon_1 - I r_1$ ... $(1)$
The potential difference across the second cell is $V_{BC} = V(B) - V(C) = \varepsilon_2 - I r_2$ ... $(2)$
The total potential difference between $A$ and $C$ is:
$V_{AC} = V_{AB} + V_{BC} = (V(A) - V(B)) + (V(B) - V(C))$
$V_{AC} = (\varepsilon_1 - I r_1) + (\varepsilon_2 - I r_2) = (\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2)$ ... $(3)$
For the equivalent combination,the potential difference is:
$V_{AC} = \varepsilon_{eq} - I r_{eq}$ ... $(4)$
Comparing equations $(3)$ and $(4)$,we get:
$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$
$r_{eq} = r_1 + r_2$
If the cells are connected in series in an opposing condition (i.e.,positive terminal to positive terminal),the equivalent $emf$ becomes $\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2$ (assuming $\varepsilon_1 > \varepsilon_2$).

(N/A) If the positive terminals of given cells are connected at one point and the negative terminals are connected at another point,then such a connection is called a parallel connection of cells.
As shown in the figure,two cells with emf $\varepsilon_{1}$ and $\varepsilon_{2}$ and internal resistances $r_{1}$ and $r_{2}$ respectively are connected in parallel between points $B_{1}$ and $B_{2}$.
The current in the cell with emf $\varepsilon_{1}$ is $I_{1}$ and in the cell with emf $\varepsilon_{2}$ is $I_{2}$.
The total current at junction $B_{1}$ is $I = I_{1} + I_{2}$.
Let the potentials at $B_{1}$ and $B_{2}$ be $V(B_{1})$ and $V(B_{2})$ respectively. The potential difference $V = V(B_{1}) - V(B_{2})$ across each cell is given by:
$V = \varepsilon_{1} - I_{1}r_{1} \implies I_{1} = \frac{\varepsilon_{1} - V}{r_{1}}$
$V = \varepsilon_{2} - I_{2}r_{2} \implies I_{2} = \frac{\varepsilon_{2} - V}{r_{2}}$
Substituting these into the total current equation:
$I = \frac{\varepsilon_{1} - V}{r_{1}} + \frac{\varepsilon_{2} - V}{r_{2}} = \left( \frac{\varepsilon_{1}}{r_{1}} + \frac{\varepsilon_{2}}{r_{2}} \right) - V \left( \frac{1}{r_{1}} + \frac{1}{r_{2}} \right)$
Rearranging for $V$:
$V \left( \frac{r_{1} + r_{2}}{r_{1}r_{2}} \right) = \left( \frac{\varepsilon_{1}r_{2} + \varepsilon_{2}r_{1}}{r_{1}r_{2}} \right) - I$
$V = \left( \frac{\varepsilon_{1}r_{2} + \varepsilon_{2}r_{1}}{r_{1} + r_{2}} \right) - I \left( \frac{r_{1}r_{2}}{r_{1} + r_{2}} \right)$
Comparing this with the equivalent circuit equation $V = \varepsilon_{eq} - Ir_{eq}$,we get:
$\varepsilon_{eq} = \frac{\varepsilon_{1}r_{2} + \varepsilon_{2}r_{1}}{r_{1} + r_{2}}$ and $r_{eq} = \frac{r_{1}r_{2}}{r_{1} + r_{2}}$.

(N/A) Consider $n$ identical cells,each of emf $\varepsilon$ and internal resistance $r$,connected in series to form a row. There are $m$ such rows connected in parallel,and this combination is connected to an external resistor $R$.
$1$. Total emf of one row: $\varepsilon_{eq} = n\varepsilon$.
$2$. Total internal resistance of one row: $r_{eq} = nr$.
$3$. Since there are $m$ such rows in parallel,the total equivalent emf of the combination is $\varepsilon_{total} = n\varepsilon$.
$4$. The total internal resistance $r_{total}$ of the $m$ parallel rows is given by $\frac{1}{r_{total}} = \frac{1}{nr} + \frac{1}{nr} + \dots (m \text{ times}) = \frac{m}{nr}$. Thus,$r_{total} = \frac{nr}{m}$.
$5$. The total resistance of the circuit is $R_{total} = R + \frac{nr}{m} = \frac{mR + nr}{m}$.
$6$. The current $I$ flowing through the external resistor $R$ is given by $I = \frac{\varepsilon_{total}}{R_{total}} = \frac{n\varepsilon}{\frac{mR + nr}{m}} = \frac{mn\varepsilon}{mR + nr}$.

(N/A) Cells can be combined in three primary ways to achieve different voltage or current requirements:
$1$. Series Combination: Cells are connected end-to-end. The positive terminal of one cell is connected to the negative terminal of the next. The total electromotive force $(EMF)$ is the sum of individual EMFs $(E_{eq} = E_1 + E_2 + ... + E_n)$,and the total internal resistance is the sum of individual internal resistances $(r_{eq} = r_1 + r_2 + ... + r_n)$. This is used to increase the total voltage.
$2$. Parallel Combination: All positive terminals are connected together,and all negative terminals are connected together. The equivalent $EMF$ is given by $E_{eq} = \frac{\sum (E_i / r_i)}{\sum (1 / r_i)}$,and the equivalent internal resistance is $\frac{1}{r_{eq}} = \sum \frac{1}{r_i}$. This is used to increase the current capacity and reduce internal resistance.
$3$. Mixed Combination: $A$ combination where rows of cells in series are connected in parallel. This is used to obtain maximum power output in an external circuit.

(N/A) $1$. Series Connection of Cells: When cells are connected in series,the positive terminal of one cell is connected to the negative terminal of the next cell. In this arrangement,the total electromotive force $(EMF)$ is the sum of the individual EMFs of the cells $(E_{eq} = E_1 + E_2 + ... + E_n)$,and the total internal resistance is the sum of the individual internal resistances $(r_{eq} = r_1 + r_2 + ... + r_n)$. This configuration is used to increase the total voltage output.
$2$. Parallel Connection of Cells: When cells are connected in parallel,all positive terminals are connected together and all negative terminals are connected together. In this arrangement,the equivalent $EMF$ remains equal to the $EMF$ of a single cell (if all cells are identical,$E_{eq} = E$),while the total internal resistance decreases $(1/r_{eq} = 1/r_1 + 1/r_2 + ... + 1/r_n)$. This configuration is used to increase the current capacity and reduce the total internal resistance.

(N/A) For $n$ cells connected in parallel,each having an electromotive force (emf) $\varepsilon_i$ and internal resistance $r_i$,the equivalent emf $\varepsilon_{eq}$ is given by the formula:
$\varepsilon_{eq} = \frac{\sum_{i=1}^{n} \frac{\varepsilon_i}{r_i}}{\sum_{i=1}^{n} \frac{1}{r_i}}$
If all cells are identical with emf $\varepsilon$ and internal resistance $r$,the equivalent emf simplifies to:
$\varepsilon_{eq} = \varepsilon$

(N/A) When $n$ cells,each with electromotive force $E_i$ and internal resistance $r_i$,are connected in series,the total electromotive force $E_{eq}$ is the algebraic sum of the individual electromotive forces of the cells.
The equation for the equivalent emf is given by:
$E_{eq} = E_1 + E_2 + E_3 + ... + E_n = \sum_{i=1}^{n} E_i$
If all cells are identical,each with emf $E$,then the equivalent emf is:
$E_{eq} = nE$

(B) Let the two cells have electromotive forces $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$.
For series connection,the equivalent emf is $E_s = E_1 + E_2$.
For parallel connection,the equivalent emf is $E_p = \frac{E_1r_2 + E_2r_1}{r_1 + r_2}$.
Assuming the two cells are identical with $E_1 = E_2 = E$ and $r_1 = r_2 = r$:
In series,$E_s = E + E = 2E$.
In parallel,$E_p = \frac{Er + Er}{r + r} = \frac{2Er}{2r} = E$.
The ratio of equivalent emf in series to parallel is $\frac{E_s}{E_p} = \frac{2E}{E} = 2:1$.

(N/A) The equivalent internal resistance of the cells in series is:
$r = r_1 + r_2 = 2 + 8 = 10 \ \Omega$
Since $E_1 > E_2$,the equivalent $emf$ of the circuit is:
$E = E_1 - E_2 = 6 - 4 = 2 \ V$
Using Ohm's law,the current $I$ in the circuit is:
$I = \frac{E}{r} = \frac{2}{10} = 0.2 \ A$
The current flows from the higher potential cell $(E_1)$ to the lower potential cell $(E_2)$.
To find the potential difference $V_A - V_B$,we apply Kirchhoff's voltage law from $A$ to $B$:
$V_A - E_2 - I r_2 = V_B$
$V_A - V_B = E_2 + I r_2$
$V_A - V_B = 4 + (0.2 \times 8)$
$V_A - V_B = 4 + 1.6 = 5.6 \ V$

(C) The cells are connected in series. The total $emf$ of the circuit is $E_{eq} = E_{1} - E_{2} = 6 V - 4 V = 2 V$.
The total resistance of the circuit is $R_{eq} = r_{1} + r_{2} = 2 \Omega + 8 \Omega = 10 \Omega$.
The current $I$ flowing in the circuit is $I = \frac{E_{eq}}{R_{eq}} = \frac{2 V}{10 \Omega} = 0.2 A$.
To find the potential difference across points $X$ and $Y$,we apply Kirchhoff's voltage law from $X$ to $Y$ through the cell $E_{2}$:
$V_{X} - E_{2} - I \cdot r_{2} = V_{Y}$
$V_{X} - V_{Y} = E_{2} + I \cdot r_{2}$
$V_{X} - V_{Y} = 4 V + (0.2 A \times 8 \Omega) = 4 V + 1.6 V = 5.6 V$.
Thus,the potential difference across points $X$ and $Y$ is $5.6 V$.

(A) Let $n = 5$ be the number of cells,$E = 5\, V$ be the $emf$ of each cell,and $r = 1\, \Omega$ be the internal resistance of each cell.
For series combination,the total $emf$ is $nE$ and total internal resistance is $nr$. The current $i_1$ is given by:
$i_1 = \frac{nE}{R + nr} = \frac{5 \times 5}{R + 5 \times 1} = \frac{25}{R + 5}$
For parallel combination,the total $emf$ is $E$ and total internal resistance is $r/n$. The current $i_2$ is given by:
$i_2 = \frac{E}{R + r/n} = \frac{5}{R + 1/5} = \frac{5 \times 5}{5R + 1} = \frac{25}{5R + 1}$
Given $i_1 = i_2$,we have:
$\frac{25}{R + 5} = \frac{25}{5R + 1}$
$R + 5 = 5R + 1$
$4 = 4R$
$R = 1\, \Omega$

(B) The terminal voltage $V$ across a load resistance $R$ is given by the formula $V = \frac{E \cdot R}{R + r}$,where $E$ is the $emf$ and $r$ is the internal resistance of the cell.
For the first case: $1.25 = \frac{E \cdot 5}{5 + r} \implies 1.25(5 + r) = 5E \implies 6.25 + 1.25r = 5E \implies 1.25 + 0.25r = E \dots (i)$
For the second case: $1 = \frac{E \cdot 2}{2 + r} \implies 2 + r = 2E \implies 1 + 0.5r = E \dots (ii)$
Equating $(i)$ and $(ii)$:
$1.25 + 0.25r = 1 + 0.5r$
$0.25 = 0.25r \implies r = 1\, \Omega$
Substituting $r = 1$ into equation $(ii)$:
$E = 1 + 0.5(1) = 1.5\, V$
Given $E = \frac{x}{10}\, V$,we have:
$1.5 = \frac{x}{10} \implies x = 15$.

(C) The two $20 \,\Omega$ resistors are connected in parallel,so the equivalent external resistance $R$ is given by $\frac{1}{R} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$,which means $R = 10 \,\Omega$.
The two identical cells of emf $E = 1.5 \,V$ and internal resistance $r$ are connected in parallel. The equivalent emf $E_{eq} = E = 1.5 \,V$ and equivalent internal resistance $r_{eq} = \frac{r}{2}$.
The terminal voltage $V$ across the external resistance is given by $V = E_{eq} - I r_{eq}$,where $I$ is the total current.
Given $V = 1.2 \,V$,we have $1.2 = 1.5 - I \left(\frac{r}{2}\right)$,which implies $I \left(\frac{r}{2}\right) = 0.3$.
Also,the total current $I$ is given by $I = \frac{E_{eq}}{R + r_{eq}} = \frac{1.5}{10 + \frac{r}{2}}$.
Substituting $I$ into the equation $I \left(\frac{r}{2}\right) = 0.3$:
$\left(\frac{1.5}{10 + \frac{r}{2}}\right) \left(\frac{r}{2}\right) = 0.3$
$1.5 \left(\frac{r}{2}\right) = 0.3 \left(10 + \frac{r}{2}\right)$
$0.75r = 3 + 0.15r$
$0.6r = 3$
$r = \frac{3}{0.6} = 5 \,\Omega$.

(A) The total emf of the circuit is $2E$ and the total resistance is $R + r_{1} + r_{2}$.
The current $I$ in the circuit is given by:
$I = \frac{2E}{R + r_{1} + r_{2}} \quad \dots (i)$
The potential difference across the second cell (with emf $E$ and internal resistance $r_{2}$) is given by $V = E - Ir_{2}$.
Given that the potential difference across the second cell is zero:
$E - Ir_{2} = 0 \Rightarrow I = \frac{E}{r_{2}} \quad \dots (ii)$
Comparing the values of $I$ from equations $(i)$ and $(ii)$:
$\frac{E}{r_{2}} = \frac{2E}{R + r_{1} + r_{2}}$
$R + r_{1} + r_{2} = 2r_{2}$
$R = 2r_{2} - r_{2} - r_{1}$
$R = r_{2} - r_{1}$

(A) Let the $EMF$ of each cell be $E$ and the internal resistance be $r$. The external resistance is $R = 2 \,\Omega$.
For series combination,the total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $I_1$ is given by:
$I_1 = \frac{2E}{2r + R} = \frac{2E}{2r + 2} = \frac{E}{r + 1}$
For parallel combination,the total $EMF$ is $E$ and the total internal resistance is $r/2$. The current $I_2$ is given by:
$I_2 = \frac{E}{r/2 + R} = \frac{E}{r/2 + 2} = \frac{2E}{r + 4}$
Given that $I_1 = I_2$,we have:
$\frac{E}{r + 1} = \frac{2E}{r + 4}$
Canceling $E$ from both sides:
$\frac{1}{r + 1} = \frac{2}{r + 4}$
Cross-multiplying:
$r + 4 = 2(r + 1)$
$r + 4 = 2r + 2$
$r = 2 \,\Omega$.

(A) The total emf of the series combination is $E_{eq} = E + E = 2E$.
The total internal resistance is $r_{eq} = r_{1} + r_{2}$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{r_{1} + r_{2} + R}$.
The potential difference $V$ across the source with internal resistance $r_{1}$ is given by $V = E - I r_{1}$.
Given that $V = 0$,we have $E - I r_{1} = 0$,which implies $E = I r_{1}$.
Substituting the expression for $I$:
$E = \left( \frac{2E}{r_{1} + r_{2} + R} \right) r_{1}$
$1 = \frac{2 r_{1}}{r_{1} + r_{2} + R}$
$r_{1} + r_{2} + R = 2 r_{1}$
$R = 2 r_{1} - r_{1} - r_{2}$
$R = r_{1} - r_{2}$

(D) Let $E$ be the electromotive force $(EMF)$ of the cell and $r_i$ be its internal resistance.
The heat produced in a resistance $R$ is given by $H = I^2 R t = \left(\frac{E}{R+r_i}\right)^2 R t$.
According to the problem,the heat produced in both cases is the same for the same time $t$.
Therefore,$\frac{E^2 R}{(R+r_i)^2} t = \frac{E^2 r}{(r+r_i)^2} t$.
Canceling $E^2 t$ from both sides,we get $\frac{R}{(R+r_i)^2} = \frac{r}{(r+r_i)^2}$.
Taking the square root on both sides: $\frac{\sqrt{R}}{R+r_i} = \frac{\sqrt{r}}{r+r_i}$.
Cross-multiplying: $\sqrt{R}(r+r_i) = \sqrt{r}(R+r_i)$.
$\sqrt{R}r + \sqrt{R}r_i = \sqrt{r}R + \sqrt{r}r_i$.
Rearranging terms to solve for $r_i$: $r_i(\sqrt{R} - \sqrt{r}) = R\sqrt{r} - r\sqrt{R}$.
$r_i(\sqrt{R} - \sqrt{r}) = \sqrt{Rr}(\sqrt{R} - \sqrt{r})$.
Thus,$r_i = \sqrt{Rr}$.

(A) The total emf of the series combination is $1000E$ and the total internal resistance is $1000r$.
Since there is no external resistance,the current $i$ in the circuit is given by $i = \frac{\text{Total emf}}{\text{Total resistance}} = \frac{1000E}{1000r} = \frac{E}{r}$.
The potential drop across a single cell is given by $V = E - ir$.
Substituting the value of $i$,we get $V = E - (\frac{E}{r})r = E - E = 0$.
Since the potential drop across each individual cell is $0$,the total potential drop across any number of cells (including $399$ cells) connected in this series configuration will be $399 \times 0 = 0$.

(B) Let the $e.m.f.$ values of the two batteries be $E_1$ and $E_2$,and the total resistance of the circuit be $R_{total} = R + r_1 + r_2$.
When the batteries are connected in series with the same polarity,the total $e.m.f.$ is $E_1 + E_2$. The current is given by $I_1 = \frac{E_1 + E_2}{R_{total}} = 3.0 \,A$.
When the polarity of one battery is reversed,the total $e.m.f.$ becomes $E_1 - E_2$ (assuming $E_1 > E_2$). The current is given by $I_2 = \frac{E_1 - E_2}{R_{total}} = 1.0 \,A$.
Dividing the two equations: $\frac{I_1}{I_2} = \frac{E_1 + E_2}{E_1 - E_2} = \frac{3}{1} = 3$.
Cross-multiplying: $E_1 + E_2 = 3(E_1 - E_2) \implies E_1 + E_2 = 3E_1 - 3E_2$.
Rearranging terms: $4E_2 = 2E_1 \implies \frac{E_1}{E_2} = \frac{4}{2} = 2$.

Let $P$ be the number of parallel rows and $S$ be the number of cells in each row.
Given: $E = 6 \,V$,$r = 0.5 \,\Omega$,$R = 0.75 \,\Omega$,and $I = 24 \,A$.
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(A) For cells connected in parallel,the equivalent emf $E_{\text{eq}}$ is given by the formula:
$E_{\text{eq}} = \frac{\sum \frac{E_i}{r_i}}{\sum \frac{1}{r_i}}$
Given that all three cells are identical with $E = 10 \ V$ and $r = 3 \ \Omega$,we have:
$E_{\text{eq}} = \frac{\frac{10}{3} + \frac{10}{3} + \frac{10}{3}}{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}}$
$E_{\text{eq}} = \frac{10}{1} = 10 \ V$
Alternatively,for identical cells in parallel,the equivalent emf is equal to the emf of a single cell. Thus,the net emf is $10 \ V$.

(B) The terminal potential difference $V$ of a cell is given by the equation $V = E - Ir$,where $E$ is the electromotive force $(EMF)$ and $r$ is the internal resistance.
For the first case: $0.9 = E - 0.3r$ --- (Equation $1$)
For the second case: $1.0 = E - 0.25r$ --- (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$(1.0 - 0.9) = (E - 0.25r) - (E - 0.3r)$
$0.1 = -0.25r + 0.3r$
$0.1 = 0.05r$
$r = \frac{0.1}{0.05} = 2 \, \Omega$.
Thus,the internal resistance of the cell is $2 \, \Omega$.

(C) When cells are connected in series,the total internal resistance is the sum of the individual internal resistances,regardless of their polarity.
Since there are $5$ cells,each with an internal resistance $r$,the total equivalent internal resistance $R_{eq}$ is given by:
$R_{eq} = r + r + r + r + r = 5r$.
Therefore,the equivalent internal resistance is $5r$.

(B) Given: Total number of cells $(n) = 10$.
Potential of each cell $= E$; Internal resistance of each cell $= r$.
Total $EMF$ of the series circuit $= 10E$.
Total internal resistance of the circuit $= 10r$.
According to Ohm's law,the current $(I)$ flowing through the circuit is given by $I = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{10E}{10r} = \frac{E}{r}$.
An ideal voltmeter connected across three cells measures the potential difference across those three cells.
The potential difference $(V)$ across $3$ cells is given by $V = I \times (3r)$.
Substituting the value of $I$,we get $V = \left(\frac{E}{r}\right) \times 3r = 3E$.
Therefore,the voltmeter will read $3E$.

(B) To find the current through the external resistor $R$,we first find the equivalent emf $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ of the two cells connected in parallel.
The formula for equivalent emf of cells in parallel is:
$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$
Given $E_1 = 12 \, V$,$r_1 = 3 \, \Omega$ and $E_2 = 6 \, V$,$r_2 = 6 \, \Omega$. Note that the cells are connected such that their polarities oppose each other in the loop,so we use:
$E_{eq} = \frac{\frac{12}{3} - \frac{6}{6}}{\frac{1}{3} + \frac{1}{6}} = \frac{4 - 1}{\frac{2+1}{6}} = \frac{3}{\frac{3}{6}} = \frac{3}{0.5} = 6 \, V$
The equivalent internal resistance is:
$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \implies r_{eq} = 2 \, \Omega$
Now,the circuit simplifies to a single cell of $6 \, V$ and $2 \, \Omega$ in series with an external resistor $R = 4 \, \Omega$.
The current $i$ is given by:
$i = \frac{E_{eq}}{r_{eq} + R} = \frac{6}{2 + 4} = \frac{6}{6} = 1 \, A$

(A) Let the $EMF$ of each cell be $\varepsilon$ and the internal resistance be $r$.
For series combination,the total $EMF$ is $2\varepsilon$ and the total internal resistance is $2r$. The current $i$ in the external resistance $R = 5\,\Omega$ is:
$i = \frac{2\varepsilon}{R + 2r} = \frac{2\varepsilon}{5 + 2r} \dots (1)$
For parallel combination,the total $EMF$ is $\varepsilon$ and the total internal resistance is $\frac{r}{2}$. The current $i$ in the external resistance $R = 5\,\Omega$ is:
$i = \frac{\varepsilon}{R + \frac{r}{2}} = \frac{\varepsilon}{5 + \frac{r}{2}} \dots (2)$
Since the current is the same in both cases,equate $(1)$ and $(2)$:
$\frac{2\varepsilon}{5 + 2r} = \frac{\varepsilon}{5 + \frac{r}{2}}$
Dividing both sides by $\varepsilon$ and cross-multiplying:
$2(5 + \frac{r}{2}) = 5 + 2r$
$10 + r = 5 + 2r$
$r = 5\,\Omega$
Thus,the internal resistance of each cell is $5\,\Omega$.

(A) Given:
$emf (E) = 4\,V$
Internal resistance $(r) = 0.5\,\Omega$
External resistance $(R) = 7.5\,\Omega$
First,calculate the current $(I)$ flowing through the circuit using Ohm's law:
$I = \frac{E}{R + r} = \frac{4}{7.5 + 0.5} = \frac{4}{8} = 0.5\,A$
The terminal potential difference $(V)$ is given by the formula:
$V = E - Ir$
$V = 4 - (0.5 \times 0.5)$
$V = 4 - 0.25$
$V = 3.75\,V$
Alternatively,$V = IR = 0.5 \times 7.5 = 3.75\,V$.

(A) The total emf of two cells connected in series is $E_{eq} = 1.5\,V + 1.5\,V = 3.0\,V$.
Let $r$ be the internal resistance of each cell. The total internal resistance in the series circuit is $R_{int} = r + r = 2r$.
The total resistance of the circuit is $R_{total} = R + R_{int} = 10 + 2r$.
The current $I$ flowing through the circuit is given by $I = \frac{E_{eq}}{R_{total}} = \frac{3}{10 + 2r}$.
The voltage across the $10\,\Omega$ resistor is given as $V = 1.5\,V$.
Using Ohm's law,$V = I \times R$,we have $1.5 = \left( \frac{3}{10 + 2r} \right) \times 10$.
$1.5(10 + 2r) = 30$.
$15 + 3r = 30$.
$3r = 15$.
$r = 5\,\Omega$.

(B) The total electromotive force (emf) of the circuit is $E_{eq} = E_1 - E_2 = 8 \ V - 2 \ V = 6 \ V$.
The total resistance of the circuit is $R_{eq} = r_1 + r_2 = 2 \ \Omega + 4 \ \Omega = 6 \ \Omega$.
The current flowing in the circuit is $I = \frac{E_{eq}}{R_{eq}} = \frac{6 \ V}{6 \ \Omega} = 1 \ A$.
Since the cell $E_2$ is being charged (current enters its positive terminal),its terminal potential difference $V$ is given by $V = E_2 + Ir_2$.
Substituting the values,we get $V = 2 \ V + (1 \ A \times 4 \ \Omega) = 2 \ V + 4 \ V = 6 \ V$.

(D) Let the $EMF$ of each battery be $E$ and internal resistance be $r = 1 \Omega$.
For series connection,the total $EMF$ is $2E$ and total internal resistance is $2r = 2 \Omega$. The power dissipated in $R$ is $J_1 = (\frac{2E}{R+2})^2 R$.
For parallel connection,the total $EMF$ is $E$ and total internal resistance is $r/2 = 0.5 \Omega$. The power dissipated in $R$ is $J_2 = (\frac{E}{R+0.5})^2 R$.
Given $J_1 = 2.25 J_2$,we have $(\frac{2E}{R+2})^2 R = 2.25 (\frac{E}{R+0.5})^2 R$.
Dividing both sides by $E^2 R$,we get $\frac{4}{(R+2)^2} = 2.25 \frac{1}{(R+0.5)^2}$.
Taking the square root of both sides,$\frac{2}{R+2} = \frac{1.5}{R+0.5}$.
Cross-multiplying gives $2(R+0.5) = 1.5(R+2)$,which simplifies to $2R + 1 = 1.5R + 3$.
Solving for $R$,we get $0.5R = 2$,so $R = 4 \Omega$.

(B) The circuit consists of two batteries in a loop. The equivalent electromotive force $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ can be calculated using the formula for parallel combination of cells:
$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$
Given $E_1 = 6 \text{ V}$, $r_1 = 1 \text{ }\Omega$, $E_2 = 3 \text{ V}$, $r_2 = 2 \text{ }\Omega$ (note the polarity, the $3 \text{ V}$ battery is connected in opposition to the $6 \text{ V}$ battery).
$E_{eq} = \frac{\frac{6}{1} - \frac{3}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{6 - 1.5}{1.5} = \frac{4.5}{1.5} = 3 \text{ V}$.
Alternatively, using Kirchhoff's loop rule for the loop:
$I = \frac{6 - 3}{1 + 2} = \frac{3}{3} = 1 \text{ A}$.
The potential difference across $AB$ is $V_{AB} = E_1 - I r_1 = 6 - (1 \times 1) = 5 \text{ V}$.
Checking with the second branch: $V_{AB} = E_2 + I r_2 = 3 + (1 \times 2) = 5 \text{ V}$.
Thus, the voltage across $AB$ is $5 \text{ V}$.

(D) For two batteries with emfs $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ connected in parallel,the equivalent emf $\varepsilon_{eq}$ is given by $\varepsilon_{eq} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$.
This value of $\varepsilon_{eq}$ always lies between $\varepsilon_1$ and $\varepsilon_2$. Thus,Statement-$I$ is false because it is not necessarily smaller than both.
The equivalent internal resistance $r_{eq}$ is given by $\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$,which implies $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$.
Since $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$,it is clear that $r_{eq} < r_1$ and $r_{eq} < r_2$. Thus,Statement-$II$ is true.

(C) Case $1$: Series connection.
Equivalent emf,$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2 = 1 \ V + 2 \ V = 3 \ V$.
Equivalent internal resistance,$r_{eq} = r_1 + r_2 = 2 \ \Omega + 1 \ \Omega = 3 \ \Omega$.
Total resistance,$R_{total} = R + r_{eq} = 6 \ \Omega + 3 \ \Omega = 9 \ \Omega$.
Current $I_1 = \frac{\varepsilon_{eq}}{R_{total}} = \frac{3 \ V}{9 \ \Omega} = \frac{1}{3} \ A$.
Case $2$: Parallel connection.
Equivalent emf,$\varepsilon_{eq} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{2.5}{1.5} = \frac{5}{3} \ V$.
Equivalent internal resistance,$r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3} \ \Omega$.
Total resistance,$R_{total} = R + r_{eq} = 6 \ \Omega + \frac{2}{3} \ \Omega = \frac{20}{3} \ \Omega$.
Current $I_2 = \frac{\varepsilon_{eq}}{R_{total}} = \frac{5/3}{20/3} = \frac{5}{20} = \frac{1}{4} \ A$.
Ratio $\frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac{4}{3}$.
Given $\frac{I_1}{I_2} = \frac{x}{3}$,therefore $x = 4$.

(B) The circuit consists of two identical cells, each with an electromotive force $(EMF)$ $E = 3 \text{ V}$ and internal resistance $r = 3 \Omega$, connected in parallel. These are connected to an external resistor $R = 6 \Omega$.
For cells in parallel, the equivalent $EMF$ $(E_{eq})$ is given by:
$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{3/3 + 3/3}{1/3 + 1/3} = \frac{1 + 1}{2/3} = \frac{2}{2/3} = 3 \text{ V}$.
The equivalent internal resistance $(r_{eq})$ is given by:
$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \implies r_{eq} = 1.5 \Omega$.
The total resistance of the circuit is $R_{total} = R + r_{eq} = 6 + 1.5 = 7.5 \Omega$.
The current through the $6 \Omega$ resistor is given by Ohm's Law:
$I = \frac{E_{eq}}{R_{total}} = \frac{3}{7.5} = \frac{30}{75} = \frac{2}{5} \text{ A}$.

(B) The formula for the equivalent e.m.f. of the parallel combination of batteries is given by $\varepsilon_{eq} = r_{eq} \left( \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} \right)$.
First,we calculate the equivalent internal resistance $r_{eq}$:
$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{2} + \frac{1}{1} = \frac{3}{2} \Omega^{-1}$.
Therefore,$r_{eq} = \frac{2}{3} \Omega$.
Now,substitute the values into the formula for equivalent e.m.f.:
$\varepsilon_{eq} = \frac{2}{3} \left( \frac{12}{2} + \frac{6}{1} \right) = \frac{2}{3} (6 + 6) = \frac{2}{3} \times 12 = 8 \text{ V}$.
The voltmeter measures the potential difference across the parallel combination,which is equal to the equivalent e.m.f. of the circuit.
Thus,the reading of the voltmeter is $8 \text{ V}$.

(A) The total $EMF$ of the series combination is $E_{eq} = E + E = 2E$.
The total resistance of the circuit is $R_{total} = r_1 + r_2 + R$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{r_1 + r_2 + R}$.
The potential difference across cell $E_1$ is given by $V_1 = E - Ir_1$.
Given that $V_1 = 0$,we have $E - Ir_1 = 0$,which implies $E = Ir_1$.
Substituting the value of $I$,we get $E = \left( \frac{2E}{r_1 + r_2 + R} \right) r_1$.
Dividing both sides by $E$,we get $1 = \frac{2r_1}{r_1 + r_2 + R}$.
Rearranging the terms,$r_1 + r_2 + R = 2r_1$.
Therefore,$R = 2r_1 - r_1 - r_2 = r_1 - r_2$.

(D) For two cells with emf $\varepsilon_1$ and $\varepsilon_2$ and internal resistances $r_1$ and $r_2$ connected in parallel,the equivalent emf $\varepsilon_{eq}$ is given by the formula:
$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$
Given values:
$\varepsilon_1 = 2 \ V, r_1 = 0.1 \ \Omega$
$\varepsilon_2 = 4 \ V, r_2 = 0.2 \ \Omega$
Substituting these values into the formula:
$\varepsilon_{eq} = \frac{(2 \times 0.2) + (4 \times 0.1)}{0.1 + 0.2}$
$\varepsilon_{eq} = \frac{0.4 + 0.4}{0.3}$
$\varepsilon_{eq} = \frac{0.8}{0.3} \approx 2.67 \ V$
Thus,the correct option is $D$.

(A) The total number of batteries is $n = 4$. Each battery has an $EMF$ $\varepsilon = 1.5 \ V$ and internal resistance $r = 0.1 \ \Omega$.
When batteries are connected in series,if one battery is connected in reverse,its $EMF$ opposes the others.
Resultant $EMF$ $\varepsilon' = (\varepsilon + \varepsilon + \varepsilon) - \varepsilon = 2\varepsilon$.
$\varepsilon' = 2 \times 1.5 \ V = 3 \ V$.
Since internal resistances are always in series regardless of the polarity of the $EMF$,the resultant internal resistance $r'$ is the sum of all individual internal resistances.
$r' = r + r + r + r = 4r$.
$r' = 4 \times 0.1 \ \Omega = 0.4 \ \Omega$.
Therefore,the resultant $EMF$ is $3 \ V$ and the resultant internal resistance is $0.4 \ \Omega$.

(D) For series connection:
The total electromotive force is $2\varepsilon$ and the total internal resistance is $2r$.
The current flowing through the circuit is given by:
$I_1 = \frac{2\varepsilon}{R + 2r} = \frac{2\varepsilon}{2 + 2r}$
For parallel connection:
The total electromotive force is $\varepsilon$ and the total internal resistance is $\frac{r}{2}$.
The current flowing through the circuit is given by:
$I_2 = \frac{\varepsilon}{R + \frac{r}{2}} = \frac{\varepsilon}{2 + \frac{r}{2}} = \frac{2\varepsilon}{4 + r}$
Since the current is the same in both connections,$I_1 = I_2$:
$\frac{2\varepsilon}{2 + 2r} = \frac{2\varepsilon}{4 + r}$
$4 + r = 2 + 2r$
$r = 2 \ \Omega$
Thus,the internal resistance of each cell is $2 \ \Omega$.

(D) Given: Electromotive force $\varepsilon = 4 \text{ V}$, internal resistance $r = 0.1 \ \Omega$, and external resistance $R = 3.9 \ \Omega$.
First, calculate the current $I$ flowing through the circuit using Ohm's law:
$I = \frac{\varepsilon}{R + r} = \frac{4}{3.9 + 0.1} = \frac{4}{4} = 1 \text{ A}$.
The terminal voltage $V$ across the cell is given by the potential difference across the external resistor $R$:
$V = I \times R$
$V = 1 \text{ A} \times 3.9 \ \Omega = 3.9 \text{ V}$.
Alternatively, using the formula $V = \varepsilon - Ir$:
$V = 4 - (1 \times 0.1) = 4 - 0.1 = 3.9 \text{ V}$.
Therefore, the correct option is $D$.