Physics M.C.Q (1 Marks)

$0.5 = \frac{E}{{2 + r}}$ ...... $(i)$

$0.25 = \frac{E}{{5 + r}}$ ..... $(ii)$

Dividing $(i)$ by $(ii)$,

we get $2 = \frac{{5 + r}}{{2 + r}}$ $ \Rightarrow $  $r = 1\,\Omega $

$0.5 = \frac{E}{{2 + 1}}$ $ \Rightarrow $ $E = 1.5\,V$

For cell  $A$   $E = V + ir$ $ \Rightarrow $  $V = 2 - \frac{4}{{3.8}} \times 1.9 = 0$.

$ \Rightarrow 0.5 = \frac{E}{{11 + r}}$ $ \Rightarrow $  $E = 5.5 + 0.5r$ ..… $( i)$

and $0.9 = \frac{E}{{5 + r}}$ $ \Rightarrow $ $E = 4.5 + 0.9r$ ..… $(ii)$

On solving these equation, we have $r = 2.5\,\Omega $

$r = \left( {\frac{E}{V} - 1} \right)R = \left( {\frac{{40}}{{30}} - 1} \right) \times 9$$ = \frac{{9 \times 10}}{{30}} = 3\,\Omega $

${E_{eq}} = \frac{{{E_1}{R_2} + {E_2}{R_1}}}{{{R_1} + {R_2}}} = \frac{{2 \times 4 + 2 \times 4}}{{4 + 4}} = 2\,V$ and

${R_{eq}} = \frac{4}{2} = 2\Omega $. Current $i = \frac{{2 + 2}}{2} = 2A$ from $A$ to $B$ through $E_2.$

$ = \frac{{1.5({r_1} + {r_2} + {r_3} + ...... + {r_n})}}{{({r_1} + {r_2} + {r_3} + ..... + {r_n})}} = 1.5\,A$.

Also $r = $ slope $=\frac{2}{5} = 0.4\,\Omega $

For maximum current in the external resistance $R$, the necessary condition is $R = \frac{{nr}}{m}$

$ \Rightarrow $   $12 = \frac{{n \times 2}}{m}$ $ \Rightarrow $ $n = 6\,m$ ..... $(i)$

Total cells $= 24 = n \times m$ ..... $(ii)$

On solving equations $(i)$ and $(ii)$

$n = 12$ and $m = 2$

i.e. $2$ rows of $12$ cells are connected in parallel.

$E_{1} r_{1}+E_{1} r_{2}+E_{1} R>E_{2} r_{1}+E_{2} R+E_{1} r_{1}+E_{1} R$

$E_{1} r_2 >E_{2}\left(r_{1}+R\right)$

Because $I$ is totally independent of $n$, hence it will remain constant.

Now $pd$ across $A$ or $B$ is

$I \varepsilon+I r$ (as they will be in charging state), brgt

$=\varepsilon=\frac{(n-4) \varepsilon}{n}=2 \varepsilon\left(1-\frac{2}{n}\right)=\frac{2 \varepsilon(n-2)}{n}$

$I=\frac{6-2}{1+3}=1 A m p$

$V_{2}=E_{2}+I r_{2}=2+1 \times 3=5 V$

Dividing $\varepsilon_{1}=\left(\frac{I_{1}+I_{2}}{I_{1}-I_{2}}\right) \varepsilon_{2}$

$A B$ is$V=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}+\frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}}=\frac{\frac{3}{1}+\frac{2}{1}+\frac{1}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=2 V$

$E=V+I r=120+10(1)=130 \mathrm{V}$

Hence, in circuit

$R=\frac{E-V}{I}$

$R=\frac{130-120}{10}=1 \Omega$

$R=\frac{V}{I}=1 \Omega$

When the point $A$ and $B$ are shorted by a conducting wire, the voltmeter measures $10 \mathrm{V}$ across the battery.

Hence the current is

$I=\frac{10}{1}=10 A$

Now, the voltage across the terminals of the battery is

$V=E-I r$

$10=12-10 r$

$10 r=2$

$r=0.2 \Omega$

$I=\frac{V-E}{I}$

$V=E+I r$

Therefore, when a cell is being charged the potential difference across its terminals is greater than emf of a cell. Also in charging the positive terminal is connected to anode of the cell and negative terminal to cathode.

Here, positive terminal of the battery is connected to terminal $A$ of charger and negative terminal of the battery is connected to terminal $B$ of charger.

Hence, Potential difference across points $A$ and $B$ must be more than $E$, $A$ must be at higher potential than $B$ and In battery, current flows from positive terminal to the negative terminal

$\mathrm{i}=\frac{\mathrm{E}_{\mathrm{eq}}}{10+\mathrm{r}_{\mathrm{eq}}}=0.03 \mathrm{\,A}$ from $\mathrm{P}_{2}$ to $ \mathrm{P}_{1}$

$\frac{6-0}{14-12}=\frac{E-0}{14-0}$

$\frac{6}{2}=\frac{E}{14} \quad \Rightarrow E=42$ $\mathrm{volt}$

$\mathrm{E}_{\mathrm{eq}}=42=10 \mathrm{\,E} \quad \Rightarrow \mathrm{E}=4.2 \mathrm{\,volt}$

$\Rightarrow \mathrm{r}+\mathrm{Rn}=\mathrm{R}+\mathrm{nr}$

$\Rightarrow \mathrm{r}=\mathrm{R}$

$\Rightarrow \Delta \mathrm{V}=\frac{6}{3} \times 0.5=1 \mathrm{\,V}$

$I_{1}=\frac{\varepsilon_{2}-V}{r_{1}} \Rightarrow I_{2}=\frac{\varepsilon_{2}-V}{r_{2}}$

Combining the last three equations

$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\varepsilon_{1}-\mathrm{V}}{\mathrm{r}_{1}}+\frac{\varepsilon_{2}-\mathrm{V}}{\mathrm{r}_{2}}$

$=\left(\frac{\varepsilon_{1}}{\mathrm{r}_{1}}+\frac{\varepsilon_{2}}{\mathrm{r}_{2}}\right)-\mathrm{V}\left(\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}\right)$

Hence, $\mathrm{V}$ is given by, $\mathrm{V}=\frac{\varepsilon_{1} \mathrm{r}_{2}+\varepsilon_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}-\mathrm{I} \frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$

If we want to replace the combination by a single cell, between $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ of emf $\varepsilon_{\mathrm{eq}}$ and internal resistance $\mathrm{r}_{\mathrm{eq}},$

we would have $\mathrm{V}=\varepsilon_{\mathrm{eq}}-\mathrm{Ir}_{\mathrm{eq}}$

$\mathrm{T.P.D.}=\mathrm{E}+\mathrm{Ir} \Rightarrow 15=\mathrm{E}+3 \mathrm{r} $          .........$(2)$

Solve ${\rm{e}}{{\rm{q}}^n}$ $(1)$ and $(2)$

Terminal potential $\quad \mathrm{V}=\mathrm{E}-\mathrm{Ir}$

$=\mathrm{E}-\frac{\mathrm{E}}{\mathrm{r}} \times \mathrm{r}=0$

$\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{S}}=7.5 \times 3=22.5$ $\mathrm{volt}$

but $\mathrm{V}_{\mathrm{S}}=0 \quad$ so $\mathrm{V}_{\mathrm{P}}=22.5$ $\mathrm{volt}$

$\mathrm{V}_{\mathrm{R}}-\mathrm{V}_{\mathrm{Q}}=1 \times 3$

$\mathrm{V}_{\mathrm{Q}}=-3 \mathrm{\,V}$

$2 \varepsilon=i(3+r)$

$r=1 \Omega$

$I_{2}=\frac{3-0}{30}=0.1 \,A$

$I=\frac{E_{1}-E_{2}}{500}$             ............$(1)$

$\mathrm{E}_{2}=\mathrm{IX}$           .........$(2)$

$2=\mathrm{i} \times \mathrm{x}$

$x=100$

$\mathrm{V}_{\mathrm{A}}-12-12-18+4-10=\mathrm{V}_{\mathrm{B}}$

$\Rightarrow \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=48 \mathrm{\,V}$

$\therefore $ Current in the circuit will be,

$\mathrm{i}=\frac{(\mathrm{n}-4) \mathrm{E}}{\mathrm{nr}}$

Hence, potential difference across $\mathrm{A}$ or $\mathrm{B}$ is,

${\mathrm{V}=\mathrm{E}+\mathrm{Ir}=\mathrm{E}+\frac{(\mathrm{n}-4) \mathrm{E}}{\mathrm{nr}} \cdot \mathrm{r}} $

${=2 \mathrm{E}\left(1-\frac{2}{\mathrm{n}}\right)}$

$I_{1}=\frac{E_{1}+E_{2}}{R+r_{1}+r_{2}}$

After short circuiting the battery of $\mathrm{emf}$ $\mathrm{E}_{2}$. current through resistance $\mathrm{R}$ would be,

$I_{2}=\frac{E_{1}}{R+r_{1}}$

Now, $\quad \mathrm{I}_{2}>\mathrm{I}_{1}$

$\therefore $ ${\frac{\mathrm{E}_{1}}{\mathrm{R}+\mathrm{r}_{1}}>\frac{\mathrm{E}_{1}+\mathrm{E}_{2}}{\mathrm{R}+\mathrm{r}_{1}+\mathrm{r}_{2}}} $

${\mathrm{E}_{1} \mathrm{r}_{2}>\mathrm{E}_{2}\left(\mathrm{R}+\mathrm{r}_{1}\right)}$

$I=\frac{2 V-V}{2 R+R}=\frac{V}{3 R}$

Potential differnce across $"\mathrm{R}"$ $=\frac{\mathrm{V}}{3 \mathrm{R}} \times \mathrm{R}=\frac{\mathrm{V}}{3}$

$\therefore \mathrm{V}_{\mathrm{AB}}=\mathrm{V}+\frac{\mathrm{V}}{3}$

and

So potential difference across $\mathrm{C}=\frac{\mathrm{V}}{3}$

In steady state capacitor is fully charged hence

No current will flow through line $(2)$ Hence potential difference across line $(1)$ is

$\mathrm{V}=\frac{\mathrm{E}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)} \times \mathrm{R}_{2},$ the same potential difference appears across the capacitor, so charge on capacitor

$\mathrm{Q}=\mathrm{C} \times \frac{\mathrm{ER}_{2}}{\left(\mathrm{R}_{2}+\mathrm{r}\right)}$

$\mathrm{V}_{\mathrm{p}}=7 \mathrm{\,V}$

$\mathrm{V}_{\mathrm{q}}=-1 \mathrm{\,V}$

$\mathrm{V}_{\mathrm{S}}=\mathrm{V}_{\mathrm{R}}=0 \mathrm{\,V}$

$\mathrm{I}_{1}$ is given by :

$I_{1}=\frac{n E}{R+n r}$   $. .(i)$

When the cells connected in parallel, current

$\mathrm{I}_{2}$ is given by :

$\mathrm{I}_{2}=\frac{\mathrm{E}}{\mathrm{R}+\frac{\mathrm{r}}{\mathrm{n}}}=\frac{\mathrm{nE}}{\mathrm{nR}+\mathrm{r}}$  $\ldots(\mathrm{ii})$

As $I_{1}=I_{2}$

So, $\frac{\mathrm{nE}}{\mathrm{R}+\mathrm{nr}}=\frac{\mathrm{nE}}{\mathrm{nR}+\mathrm{r}}$

$\therefore \quad \mathrm{R}+\mathrm{nr}=\mathrm{nR}+\mathrm{r}$

or $(n-1) r=(n-1) R$ or $r=R$

$V=E-I r=E-\frac{E}{r(n+1)} r=\frac{n E}{n+1} \Rightarrow \frac{V}{E}=\frac{n}{n+1}$

${{\rm{E}}_{{\rm{eq}}}} = \frac{{\Sigma \frac{{{{\rm{E}}_1}}}{{{{\rm{r}}_1}}}}}{{\Sigma \frac{1}{{{{\rm{r}}_1}}}}} \Rightarrow \frac{{\frac{{{{\rm{E}}_1}}}{{{{\rm{r}}_1}}} + \frac{{{{\rm{E}}_2}}}{{{{\rm{r}}_2}}}}}{{\frac{1}{{{{\rm{r}}_1}}} + \frac{1}{{{{\rm{r}}_2}}}}}{{\rm{E}}_{{\rm{eq}}}} = \frac{{{{\rm{E}}_1}{{\rm{r}}_2} + {{\rm{E}}_{2{\rm{r}}}}}}{{{{\rm{r}}_1} + {{\rm{r}}_2}}}$

$\frac{1}{{{r_{eq}}}} = \Sigma \frac{1}{{{r_1}}} = \frac{1}{{{r_1}}} + \frac{1}{{{r_2}}} \Rightarrow {r_{eq}} = \frac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}$

$R=1\, \Omega+2\, \Omega+3\, \Omega=6 \,\Omega$

Current, $\quad \mathrm{I}=\frac{10-4}{6}=1 \mathrm{\,amp}$

The direction of the current would be from $\mathrm{a}$ to $\mathrm{b}$ via $\mathrm{e}.$

$\Rightarrow \mathrm{E}=\frac{\frac{20}{4}+\frac{6}{3}}{\frac{1}{4}+\frac{1}{3}}=12$ $\mathrm{volt}$ $\Rightarrow \mathrm{r}=\frac{12}{7}$

$\Rightarrow \mathrm{V}=4\left[\frac{12}{4+\frac{12}{7}}\right]=8.4 \mathrm{\,volt}$