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

(D) For cells connected in parallel,the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formula:
$\frac{\varepsilon_{eq}}{r_{eq}} = \sum \frac{\varepsilon_i}{r_i} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} + \frac{\varepsilon_3}{r_3}$
Substituting the given values:
$\frac{\varepsilon_{eq}}{r_{eq}} = \frac{1.2}{0.1} + \frac{1.4}{0.2} + \frac{1.5}{0.3}$
$\frac{\varepsilon_{eq}}{r_{eq}} = 12 + 7 + 5$
$\frac{\varepsilon_{eq}}{r_{eq}} = 24 \ V \Omega^{-1}$
Thus,the correct option is $D$.

(D) For two batteries connected in parallel,the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formulas:
$\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2}$
and
$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \implies r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$
Substituting $r_{eq}$ into the first equation:
$\varepsilon_{eq} = \left( \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2} \right) \times \left( \frac{r_1 r_2}{r_1 + r_2} \right)$
$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$
Since $\varepsilon_2 > \varepsilon_1$,the equivalent emf $\varepsilon_{eq}$ is a weighted average of $\varepsilon_1$ and $\varepsilon_2$,which implies that $\varepsilon_1 < \varepsilon_{eq} < \varepsilon_2$.

(D) The equivalent emf of two cells connected in parallel is given by the formula:
$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$
This can be rewritten as:
$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{E_1 r_2 + E_1 r_1 - E_1 r_1 + E_2 r_1}{r_1 + r_2} = \frac{E_1(r_1 + r_2) + r_1(E_2 - E_1)}{r_1 + r_2} = E_1 + \frac{r_1}{r_1 + r_2}(E_2 - E_1)$
Since $E_2 > E_1$ and $r_1, r_2 > 0$,it is clear that $E_{eq} > E_1$.
Similarly,$E_{eq} = E_2 - \frac{r_2}{r_1 + r_2}(E_2 - E_1)$. Since $E_2 > E_1$ and $r_1, r_2 > 0$,it is clear that $E_{eq} < E_2$.
Thus,$E_1 < E_{eq} < E_2$.
Given $r_2 > r_1$,the term $\frac{r_1}{r_1 + r_2} < \frac{1}{2}$.
Therefore,$E_{eq}$ is closer to $E_1$ than to $E_2$.

(C) Total number of cells $n = 10$. Each cell has emf $E = 2 \, V$ and internal resistance $r = 1 \, \Omega$.
When two cells are connected in reverse, they cancel the emf of two other cells. Thus, the effective number of cells contributing to the net emf is $n - 2m$, where $m$ is the number of wrongly connected cells.
Here, $m = 2$, so effective cells = $10 - 2(2) = 6$ cells.
Net emf $E_{\text{net}} = (10 - 2 \times 2) \times E = 6 \times 2 = 12 \, V$.
The total internal resistance remains the sum of all cells: $R_{\text{int}} = 10 \times r = 10 \times 1 = 10 \, \Omega$.
The external resistance $R = 10 \, \Omega$.
Total resistance $R_{\text{total}} = R_{\text{int}} + R = 10 + 10 = 20 \, \Omega$.
Using Ohm's law, the current $I = \frac{E_{\text{net}}}{R_{\text{total}}} = \frac{12}{20} = 0.6 \, A$.

(A) Given that there are $10$ identical cells connected in series,each with electromotive force $E$ and internal resistance $r$.
Total electromotive force of the circuit $= 10 E$.
Total internal resistance of the circuit $= 10 r$.
Using Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{10 E}{10 r} = \frac{E}{r}$.
The potential difference $V$ across $3$ cells is the sum of the potential drops across them.
For $3$ cells in series,the total internal resistance is $3 r$.
The potential difference across these $3$ cells is $V = I \times (3 r)$.
Substituting the value of $I$,we get $V = \frac{E}{r} \times 3 r = 3 E$.

(C) Since the two cells are connected in series,the total emf is $E_{total} = E + E = 2E$,where $E$ is the emf of each cell.
The total resistance of the circuit is $R_{total} = R + r_{1} + r_{2}$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_{1} + r_{2}}$.
The terminal potential difference $V_{1}$ across the cell with internal resistance $r_{1}$ 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 $E = \left( \frac{2E}{R + r_{1} + r_{2}} \right) r_{1}$.
Dividing both sides by $E$,we get $1 = \frac{2r_{1}}{R + r_{1} + r_{2}}$.
Rearranging the terms,$R + r_{1} + r_{2} = 2r_{1}$.
Therefore,$R = 2r_{1} - r_{1} - r_{2} = r_{1} - r_{2}$.

(D) The terminal potential difference $V$ of a cell is given by the equation: $V = E - Ir$,where $E$ is the $EMF$ and $r$ is the internal resistance.
This equation is in the form of a straight line $y = mx + c$,where $V$ is on the y-axis and $I$ is on the x-axis.
$1$. When the current $I = 0$,the terminal potential difference $V = E$. From the graph,at $I = 0$,$V = 3 \text{ V}$. Therefore,$E = 3 \text{ V}$.
$2$. When the terminal potential difference $V = 0$,the current is $I = 6 \text{ A}$. Substituting these values into the equation: $0 = E - Ir \implies 0 = 3 - (6)r$.
$3$. Solving for $r$: $6r = 3 \implies r = \frac{3}{6} = 0.5 \ \Omega$.
Thus,the $EMF$ is $3 \text{ V}$ and the internal resistance is $0.5 \ \Omega$.

(C) When $n$ identical cells,each of emf $E$ and internal resistance $r$,are connected in series,the total emf is $nE$ and the total internal resistance is $nr$.
If $m$ cells are connected with reversed polarity,the effective emf $E'$ is given by the formula:
$E' = (n - 2m)E$
Here,the total number of cells $n = 4$ and the number of wrongly connected cells $m = 1$.
Substituting these values:
$E' = (4 - 2 \times 1)E = (4 - 2)E = 2E$
The internal resistance of cells in series is additive regardless of their polarity. Therefore,the effective internal resistance $r_{eq}$ remains:
$r_{eq} = n \times r = 4r$
Thus,the equivalent emf is $2E$ and the effective internal resistance is $4r$.

(D) The two cells are connected in opposition,so the net electromotive force (emf) of the circuit is $E_{net} = E_{1} - E_{2}$.
The total resistance of the circuit is the sum of the internal resistances and the external resistance,which is $R_{total} = r_{1} + r_{2} + R$.
According to Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{E_{net}}{R_{total}} = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$.
The terminal potential difference $V$ across the external resistance $R$ is given by $V = I \times R$.
Substituting the value of $I$,we get $V = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$.

(D) Let $E$ be the $EMF$ and $r$ be the internal resistance of each cell. The external resistance is $R = 2 \Omega$.
Case $1$: Cells in series.
The total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $I$ is given by:
$I = \frac{2E}{R + 2r}$ ... $(i)$
Case $2$: Cells in parallel.
The total $EMF$ is $E$ and the total internal resistance is $\frac{r}{2}$. The current $I$ is given by:
$I = \frac{E}{R + \frac{r}{2}} = \frac{2E}{2R + r}$ ... (ii)
Since the current $I$ is the same in both cases,we equate $(i)$ and (ii):
$\frac{2E}{R + 2r} = \frac{2E}{2R + r}$
$R + 2r = 2R + r$
$r = R$
Given $R = 2 \Omega$,therefore $r = 2 \Omega$.

(B) Let the $EMF$ of each battery be $E$ and internal resistance be $r = 1 \Omega$.
In series connection,total $EMF$ $= 2E$ and total internal resistance $= 2r = 2 \Omega$.
The current $I_s = \frac{2E}{R + 2}$.
The power $P_1 = I_s^2 R = \left(\frac{2E}{R + 2}\right)^2 R$.
In parallel connection,total $EMF$ $= E$ and total internal resistance $= \frac{r}{2} = 0.5 \Omega$.
The current $I_p = \frac{E}{R + 0.5}$.
The power $P_2 = I_p^2 R = \left(\frac{E}{R + 0.5}\right)^2 R$.
Given $P_1 = 2.25 P_2$,so $\left(\frac{2E}{R + 2}\right)^2 R = 2.25 \left(\frac{E}{R + 0.5}\right)^2 R$.
Dividing both sides by $E^2 R$,we get $\frac{4}{(R + 2)^2} = 2.25 \frac{1}{(R + 0.5)^2}$.
Taking square root on both sides: $\frac{2}{R + 2} = \frac{1.5}{R + 0.5}$.
$2(R + 0.5) = 1.5(R + 2) \implies 2R + 1 = 1.5R + 3$.
$0.5R = 2 \implies R = 4 \Omega$.

(D) Given: $E_1 = 1.2 \ V$,$r_1 = 2 \ \Omega$,$E_2 = 1.5 \ V$,$r_2 = 1 \ \Omega$.
When two cells are connected in parallel with like poles together,the equivalent emf $E_{\text{eq}}$ is given by the formula:
$E_{\text{eq}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$
Substituting the values:
$E_{\text{eq}} = \frac{\frac{1.2}{2} + \frac{1.5}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{0.6 + 1.5}{0.5 + 1} = \frac{2.1}{1.5} = 1.4 \ V$.

(D) Let the electromotive force $(EMF)$ of the cell be $E$ and its internal resistance be $r$.
According to Ohm's law,the current $I$ in a circuit with external resistance $R$ is given by $I = \frac{E}{R + r}$.
For the first case: $1 = \frac{E}{2.5 + r} \implies E = 2.5 + r$ (Equation $1$).
For the second case: $0.5 = \frac{E}{10 + r} \implies E = 0.5(10 + r) = 5 + 0.5r$ (Equation $2$).
Equating the two expressions for $E$:
$2.5 + r = 5 + 0.5r$
$r - 0.5r = 5 - 2.5$
$0.5r = 2.5$
$r = \frac{2.5}{0.5} = 5 \ \Omega$.
Therefore,the internal resistance of the cell is $5 \ \Omega$.

(B) Let $E = 2.16 \text{ V}$ be the emf of each cell and $r_1, r_2$ be their internal resistances. The external resistance is $R = 19.6 \text{ } \Omega$.
The current $I$ in the series circuit is given by $I = \frac{E + E}{r_1 + r_2 + R} = \frac{4.32}{r_1 + r_2 + 19.6}$.
For cell $P$,the terminal voltage is $V_P = E - I r_1$. Given $V_P = 2 \text{ V}$,we have $2 = 2.16 - I r_1$,so $I r_1 = 0.16$.
For cell $Q$,the terminal voltage is $V_Q = E - I r_2$. Given $V_Q = 1.92 \text{ V}$,we have $1.92 = 2.16 - I r_2$,so $I r_2 = 0.24$.
Taking the ratio of these two equations:
$\frac{I r_1}{I r_2} = \frac{0.16}{0.24} = \frac{16}{24} = \frac{2}{3}$.
Thus,the ratio of internal resistances is $r_1 : r_2 = 2 : 3$.

(C) Let the total internal resistance be $r = r_1 + r_2$.
When the cells are in series with the same polarity,the total emf is $E_1 + E_2$. The current $I_1$ is given by $I_1 = \frac{E_1 + E_2}{R + r} = 5 \ A$.
So,$E_1 + E_2 = 5(R + r) \quad (1)$.
When the polarity of $E_2$ is reversed,the total emf becomes $E_1 - E_2$ (assuming $E_1 > E_2$). The current $I_2$ is given by $I_2 = \frac{E_1 - E_2}{R + r} = 2 \ A$.
So,$E_1 - E_2 = 2(R + r) \quad (2)$.
Dividing equation $(1)$ by equation $(2)$,we get:
$\frac{E_1 + E_2}{E_1 - E_2} = \frac{5}{2}$.
Applying componendo and dividendo:
$\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{5 + 2}{5 - 2}$.
$\frac{2E_1}{2E_2} = \frac{7}{3}$.
Therefore,$\frac{E_1}{E_2} = \frac{7}{3}$.

(C) Let $E$ be the $EMF$ of each cell and $r$ be the internal resistance of each cell.
For $n$ cells in series,the total $EMF$ is $nE$ and the total internal resistance is $nr$. The current $I_s$ is given by $I_s = \frac{nE}{R + nr}$.
For $n$ cells in parallel,the total $EMF$ is $E$ and the total internal resistance is $r/n$. The current $I_p$ is given by $I_p = \frac{E}{R + r/n} = \frac{nE}{nR + r}$.
Given that $I_s = I_p$,we have $\frac{nE}{R + nr} = \frac{nE}{nR + r}$.
This implies $R + nr = nR + r$.
Rearranging the terms,we get $nr - r = nR - R$,which simplifies to $r(n - 1) = R(n - 1)$.
Assuming $n \neq 1$,we divide both sides by $(n - 1)$ to get $r = R$.

(B) Let $n$ be the number of cells connected in series and $m$ be the number of parallel rows,each containing $n$ cells. The total number of cells is $N = nm$.
The total emf of the combination is $E_{eq} = nE = n(1.5)$.
The total internal resistance is $r_{eq} = \frac{nr}{m} = \frac{n(1)}{m} = \frac{n}{m}$.
The current $I$ in the external resistance $R = 30 \ \Omega$ is given by $I = \frac{nE}{R + \frac{nr}{m}}$.
Given $I = 1.5 \ A$,$E = 1.5 \ V$,$R = 30 \ \Omega$,and $r = 1 \ \Omega$:
$1.5 = \frac{n(1.5)}{30 + \frac{n}{m}} \implies 30 + \frac{n}{m} = n \implies 30 = n(1 - \frac{1}{m}) = n(\frac{m-1}{m})$.
Since $N = nm$,we have $n = \frac{N}{m}$. Substituting this: $30 = \frac{N}{m} \cdot \frac{m-1}{m} = N \frac{m-1}{m^2}$.
To minimize $N$,we need to maximize the function $f(m) = \frac{m-1}{m^2}$. Setting $f'(m) = 0$ gives $m=2$.
For $m=2$,$30 = N \frac{2-1}{2^2} = N \frac{1}{4} \implies N = 120$.

(A) $1$. In one row,there are $n$ cells in series. The total emf of one row is $nE$ and the total internal resistance of one row is $nr$.
$2$. There are $m$ such rows connected in parallel. The total emf of the parallel combination remains $nE$ because all rows are identical and connected in parallel.
$3$. The equivalent internal resistance $r_{eq}$ of $m$ rows in parallel,where each row has resistance $nr$,is given by $\frac{1}{r_{eq}} = \frac{1}{nr} + \frac{1}{nr} + ... (m \text{ times}) = \frac{m}{nr}$. Thus,$r_{eq} = \frac{nr}{m}$.
$4$. The total current $I$ flowing through the external load resistance $R$ is $I = \frac{nE}{R + r_{eq}} = \frac{nE}{R + \frac{nr}{m}} = \frac{mnE}{mR + nr}$.
$5$. Since there are $m$ identical rows in parallel,the total current $I$ is divided equally among the $m$ rows. Therefore,the current in each row (and thus in each cell of that row) is $I_{cell} = \frac{I}{m} = \frac{1}{m} \times \frac{mnE}{mR + nr} = \frac{nE}{nr + mR}$.

(C) Equivalent emf,$E_{eq} = E + E = 2E$
Equivalent resistance,$R_{eq} = r_1 + r_2 + R$
Current flowing through the circuit,$i = \frac{2E}{r_1 + r_2 + R}$
Potential difference across the first cell is given by $V_1 = E - ir_1$.
Given that $V_1 = 0$,we have:
$0 = E - ir_1$
$E = ir_1$
Substituting the value of $i$:
$E = \left( \frac{2E}{r_1 + r_2 + R} \right) r_1$
$1 = \frac{2r_1}{r_1 + r_2 + R}$
$r_1 + r_2 + R = 2r_1$
$R = 2r_1 - r_1 - r_2$
$R = r_1 - r_2$

(B) When a cell of $\operatorname{emf} E$ and internal resistance $r$ is connected to an external resistor of resistance $R$,the circuit forms a closed loop.
According to Ohm's law for the entire circuit,the current $I$ is given by $I = \frac{E}{R + r}$.
The terminal potential difference $V$ across the cell is the potential difference across the external resistor $R$,which is $V = I R$.
Alternatively,considering the internal drop,the terminal potential difference is the $\operatorname{emf}$ minus the potential drop across the internal resistance: $V = E - I r$.
Both expressions represent the terminal potential difference of the cell.

(B) Let $E$ be the electromotive force $(EMF)$ of the battery and $r$ be its internal resistance.
The terminal voltage $V$ across a resistor $R$ is given by $V = I R$,where $I = \frac{E}{R+r}$.
Thus,$V = E \left( \frac{R}{R+r} \right)$,which can be rewritten as $E = V + I r = V + \left( \frac{V}{R} \right) r$.
For the first case: $R_1 = 16 \ \Omega$,$V_1 = 12 \ V$.
$E = 12 + \left( \frac{12}{16} \right) r = 12 + 0.75 r$ --- $(i)$
For the second case: $R_2 = 10 \ \Omega$,$V_2 = 11 \ V$.
$E = 11 + \left( \frac{11}{10} \right) r = 11 + 1.1 r$ --- (ii)
Equating $(i)$ and (ii):
$12 + 0.75 r = 11 + 1.1 r$
$12 - 11 = 1.1 r - 0.75 r$
$1 = 0.35 r$
$r = \frac{1}{0.35} = \frac{100}{35} = \frac{20}{7} \ \Omega$.

(D) Let the electromotive force of the cell be $E$ and its internal resistance be $r$. The current flowing through the circuit is $I = 2 \ A$.
When the current flows in the normal direction,the terminal potential difference is given by $V_1 = E - Ir$.
Substituting the given values: $20 = E - 2r$ --- (Equation $1$).
When the direction of the current is reversed,the cell is being charged,so the terminal potential difference is given by $V_2 = E + Ir$.
Substituting the given values: $30 = E + 2r$ --- (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(E + 2r) - (E - 2r) = 30 - 20$.
$4r = 10$.
$r = 2.5 \ \Omega$.
Thus,the internal resistance of the cell is $2.5 \ \Omega$.

(B) The circuit consists of three cells connected in parallel. Let the emfs be $\varepsilon_1 = 10 \text{ V}$,$\varepsilon_2 = 7 \text{ V}$,and $\varepsilon_3 = 20 \text{ V}$,with internal resistances $r_1 = 3 \Omega$,$r_2 = 0.6 \Omega$,and $r_3 = 2 \Omega$.
Since the third cell is connected with reversed polarity relative to the first two,the equivalent emf $E_{\text{eq}}$ and equivalent internal resistance $r_{\text{eq}}$ are given by:
$\frac{E_{\text{eq}}}{r_{\text{eq}}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} - \frac{\varepsilon_3}{r_3}$
$\frac{E_{\text{eq}}}{r_{\text{eq}}} = \frac{10}{3} + \frac{7}{0.6} - \frac{20}{2} = \frac{10}{3} + \frac{35}{3} - 10 = \frac{45}{3} - 10 = 15 - 10 = 5 \text{ A}$
For parallel combination,the equivalent internal resistance is:
$\frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{3} + \frac{1}{0.6} + \frac{1}{2} = \frac{1}{3} + \frac{5}{3} + \frac{1}{2} = 2 + 0.5 = 2.5 \text{ S}$
$r_{\text{eq}} = \frac{1}{2.5} = 0.4 \Omega$
Now,$E_{\text{eq}} = \left(\frac{E_{\text{eq}}}{r_{\text{eq}}}\right) \times r_{\text{eq}} = 5 \times 0.4 = 2 \text{ V}$
Thus,$E = 2 \text{ V}$ and $r = 0.4 \Omega$.

(C) For cells connected in parallel,the equivalent emf $E_{\text{eq}}$ and equivalent internal resistance $r_{\text{eq}}$ are given by the formula: $\frac{E_{\text{eq}}}{r_{\text{eq}}} = \sum \frac{E_i}{r_i}$ and $\frac{1}{r_{\text{eq}}} = \sum \frac{1}{r_i}$.
First,calculate the initial value of $\frac{E_{\text{eq}}}{r_{\text{eq}}}$:
$\frac{E_{\text{eq}}}{r_{\text{eq}}} = \frac{5}{1} + \frac{8}{2} + \frac{10}{3} = 5 + 4 + 3.33 = \frac{37}{3} \ \text{A}$.
When $E_3$ is changed to $E_{3N}$,the equivalent emf becomes $2E_{\text{eq}}$. The new equivalent internal resistance $r_{\text{eq}}$ remains the same as it depends only on the internal resistances.
Thus,the new equation is: $\frac{2E_{\text{eq}}}{r_{\text{eq}}} = \frac{5}{1} + \frac{8}{2} + \frac{E_{3N}}{3} = 9 + \frac{E_{3N}}{3} = \frac{27 + E_{3N}}{3}$.
Since $\frac{E_{\text{eq}}}{r_{\text{eq}}} = \frac{37}{3}$,we have $\frac{2E_{\text{eq}}}{r_{\text{eq}}} = \frac{74}{3}$.
Equating the two expressions: $\frac{27 + E_{3N}}{3} = \frac{74}{3}$.
$27 + E_{3N} = 74$.
$E_{3N} = 74 - 27 = 47 \ V$.

(D) The circuit consists of two parallel branches, each containing three ideal batteries in series.
Each branch has an equivalent emf of $E_{eq} = 3 \times 20 \, V = 60 \, V$.
Since the two branches are connected in parallel, the total equivalent emf of the battery combination remains $E_{total} = 60 \, V$.
As the batteries are ideal, they have no internal resistance.
Therefore, the current $i$ flowing through the external resistance $R = 4 \, \Omega$ is given by Ohm's law:
$i = \frac{E_{total}}{R} = \frac{60 \, V}{4 \, \Omega} = 15 \, A$.

(B) The equivalent emf $(E_{eq})$ of two batteries in parallel is given by $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$.
Here, $E_1 = 18 \, V$, $r_1 = 3 \, \Omega$, $E_2 = 10 \, V$, and $r_2 = 1 \, \Omega$.
Since the batteries are connected with opposite polarities (as seen in the figure), we take $E_2 = -10 \, V$.
$E_{eq} = \frac{18/3 + (-10)/1}{1/3 + 1/1} = \frac{6 - 10}{4/3} = \frac{-4}{4/3} = -3 \, V$.
The equivalent internal resistance is $r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{3 \times 1}{3 + 1} = 0.75 \, \Omega$.
The potential difference across the parallel combination is the terminal voltage $V = |E_{eq}| = 3 \, V$. However, looking at the circuit as a loop, the current $i$ flows from the $18 \, V$ battery to the $10 \, V$ battery: $i = \frac{18 - 10}{3 + 1} = \frac{8}{4} = 2 \, A$.
The potential difference across the $18 \, V$ battery is $V = E_1 - i r_1 = 18 - (2 \times 3) = 18 - 6 = 12 \, V$.
The potential difference across the $10 \, V$ battery is $V = E_2 + i r_2 = 10 + (2 \times 1) = 12 \, V$.
Thus, the voltmeter reading is $12 \, V$.

(D) Let the number of wrongly connected cells in the $12$-cell battery be $m$.
Each wrongly connected cell cancels the emf of one correctly connected cell.
Thus,the effective emf of the $12$-cell battery is $E_{12} = (12 - m)E - mE = (12 - 2m)E$.
When the $12$-cell battery and the $2$-cell battery (emf $2E$) aid each other,the total emf is $E_{total} = (12 - 2m)E + 2E = (14 - 2m)E$.
The current is $i_1 = \frac{(14 - 2m)E}{R} = 3 \text{ A}$ ...$(i)$
When they oppose each other,the total emf is $E_{total} = (12 - 2m)E - 2E = (10 - 2m)E$.
The current is $i_2 = \frac{(10 - 2m)E}{R} = 2 \text{ A}$ ...(ii)
Dividing equation $(i)$ by (ii):
$\frac{3}{2} = \frac{14 - 2m}{10 - 2m}$
$3(10 - 2m) = 2(14 - 2m)$
$30 - 6m = 28 - 4m$
$2 = 2m$
$m = 1$.
Therefore,the number of wrongly connected cells is $1$.

(A) The terminal voltage $V_R$ across the external resistance $R$ is given by the formula:
$V_R = \frac{E}{(R + r)} \times R$
Dividing the numerator and denominator by $R$,we get:
$V_R = \frac{E}{(1 + r/R)}$
For the battery to act as a constant voltage source,the terminal voltage $V_R$ should be approximately equal to the emf $E$,regardless of the variations in the external load $R$.
This condition is satisfied when $r/R$ is very small,which implies $r << R$.

(A) When four cells of emf $E$ are connected in series,the total emf is $4E$.
If one cell is connected in reverse,its emf opposes the others.
Therefore,the effective emf of the circuit is $E_{eff} = E + E + E - E = 2E$.
The total internal resistance of the four cells in series is $r_{total} = r + r + r + r = 4r$.
The total resistance of the circuit is $R_{total} = 4r + R$.
Using Ohm's law,the current $I$ in the external circuit is given by $I = \frac{E_{eff}}{R_{total}} = \frac{2E}{4r + R}$.

(B) The batteries are connected in parallel. The equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ of the combination are given by:
$E_{eq} = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}}$
$\frac{1}{r_{eq}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}$
Given $E_{1} = 1 \text{ V}, r_{1} = 1 \Omega$; $E_{2} = 2 \text{ V}, r_{2} = 2 \Omega$; $E_{3} = 3 \text{ V}, r_{3} = 1 \Omega$.
First, calculate the equivalent internal resistance:
$\frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{1} = 1 + 0.5 + 1 = 2.5 \Omega^{-1} = \frac{5}{2} \Omega^{-1}$
$r_{eq} = \frac{2}{5} \Omega = 0.4 \Omega$
Now, calculate the equivalent emf:
$E_{eq} = r_{eq} \times \left( \frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}} \right)$
$E_{eq} = 0.4 \times \left( \frac{1}{1} + \frac{2}{2} + \frac{3}{1} \right) = 0.4 \times (1 + 1 + 3) = 0.4 \times 5 = 2.0 \text{ V}$
Thus, the potential difference between points $P$ and $Q$ is $2.0 \text{ V}$.

(A) For two identical cells of emf $E$ and internal resistance $r$ connected in parallel,the equivalent emf $E_{eq} = E$ and the equivalent internal resistance $r_{eq} = r / 2$.
The current $I$ flowing through the external resistance $R$ is given by $I = \frac{E}{R + r_{eq}} = \frac{E}{R + r/2} = \frac{2E}{2R + r}$.
The power $P$ developed across the external resistance $R$ is $P = I^2 R = \left( \frac{2E}{2R + r} \right)^2 R$.
To maximize power,we set $\frac{dP}{dR} = 0$,which leads to the condition $R = r_{eq}$.
Substituting $r_{eq} = r / 2$,we get $R = r / 2$.

(C) The total e.m.f. of the two cells connected in series is $E + E = 2E$.
The total resistance of the circuit is $R + r_1 + r_2$.
The current $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_1 + r_2}$.
The potential difference $V$ across the first cell (with internal resistance $r_1$) is given by $V = E - Ir_1$.
We are given that the potential difference across the first cell is zero,so $E - Ir_1 = 0$,which implies $E = Ir_1$.
Substituting the value of $I$ into this equation:
$E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$
$1 = \frac{2r_1}{R + r_1 + r_2}$
$R + r_1 + r_2 = 2r_1$
$R = 2r_1 - r_1 - r_2$
$R = r_1 - r_2$

(D) When the two cells are connected in series,the total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $i_{1}$ through the external resistor $R = 6 \ \Omega$ is given by: $i_{1} = \frac{2E}{R + 2r} = \frac{2E}{6 + 2r}$.
When the two cells are connected in parallel,the total $EMF$ remains $E$ and the total internal resistance is $\frac{r}{2}$. The current $i_{2}$ through the external resistor $R = 6 \ \Omega$ is given by: $i_{2} = \frac{E}{R + \frac{r}{2}} = \frac{E}{6 + \frac{r}{2}}$.
Given that $i_{1} = i_{2}$,we equate the expressions: $\frac{2E}{6 + 2r} = \frac{E}{6 + \frac{r}{2}}$.
Dividing both sides by $E$ and simplifying: $\frac{2}{6 + 2r} = \frac{1}{6 + \frac{r}{2}}$.
Cross-multiplying: $2(6 + \frac{r}{2}) = 6 + 2r$.
$12 + r = 6 + 2r$.
Solving for $r$: $r = 12 - 6 = 6 \ \Omega$.

(A) Let $E$ be the electromotive force (emf) and $r$ be the internal resistance of the cell.
According to Ohm's law for a complete circuit,$E = I(R + r)$.
For the first case: $E = 0.25(5 + r)$.
For the second case: $E = 0.5(2 + r)$.
Since the emf $E$ is constant,we equate the two expressions:
$0.25(5 + r) = 0.5(2 + r)$
Dividing both sides by $0.25$:
$5 + r = 2(2 + r)$
$5 + r = 4 + 2r$
$r = 5 - 4 = 1\text{ }\Omega$.
Therefore,the internal resistance of the cell is $1\text{ }\Omega$.

(C) For cells in parallel,the equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by:
$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$ and $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$.
Given $E_1 = 1$ $V$,$r_1 = 2 \Omega$,$E_2 = 2$ $V$,$r_2 = 1 \Omega$.
$E_{eq} = \frac{1/2 + 2/1}{1/2 + 1/1} = \frac{2.5}{1.5} = \frac{5}{3}$ $V$.
$r_{eq} = \frac{2 \times 1}{2 + 1} = \frac{2}{3} \Omega$.
Given current $I = 1$ $A$ through external resistance $R$,$I = \frac{E_{eq}}{R + r_{eq}} \Rightarrow 1 = \frac{5/3}{R + 2/3}$.
$R + 2/3 = 5/3 \Rightarrow R = 1 \Omega$.
If the polarity of one cell is reversed,the new equivalent emf $E'_{eq} = \frac{E_1/r_1 - E_2/r_2}{1/r_1 + 1/r_2} = \frac{0.5 - 2}{1.5} = \frac{-1.5}{1.5} = -1$ $V$.
The magnitude of the new current $I' = \frac{|E'_{eq}|}{R + r_{eq}} = \frac{1}{1 + 2/3} = \frac{1}{5/3} = \frac{3}{5}$ $A$.
Comparing $I' = \frac{3}{5}$ $A$ with $\frac{\alpha}{5}$ $A$,we get $\alpha = 3$.