Physics M.C.Q (1 Marks)

$I_{\text {net }}=\frac{9}{3}=3 \mathrm{\,A}$

$\mathrm{T.P.D.}$ of cell $=\mathrm{E}-$ $\mathrm{Ir}$ $=9-3 \times 1=6 \mathrm{\,V}$

reading of ameter $=\frac{6}{3}=2 \mathrm{\,A}$

$m n=40$

Current is maximum

$\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}} \Rightarrow 2.5=\frac{\mathrm{n}}{\mathrm{m}} 1$

$\mathrm{n}=\mathrm{m}(2.5)$

$m=4, n=10$

$I=\frac{10 E}{5}=4 \,A$

$0=10 \mathrm{\,E}-14(10 \mathrm{\,r})$               ..........$(2)$

So, $6=10 \mathrm{\,E}-12\left[10 \times \frac{\mathrm{E}}{14}\right]$

$E=\frac{42}{10}=4.2 \mathrm{\,V}$

$\frac{v_C-v_D}{3} =1$

$v_C=3+v_D=3+6$

$v_C=9\,V$

$\mathrm{I}=\frac{\mathrm{E}}{r}=\frac{10}{1}=10 \mathrm{A}$

$v=E-I r$

$V=10-10 \times 1$

$V=0$

from $\mathrm{E}_{1}$ is given by $\mathrm{V}_{\mathrm{P}}=\left(\frac{\mathrm{E}_{1}}{\mathrm{P}+\mathrm{Q}}\right) \mathrm{P}.$

Also, $\mathrm{V}_{\mathrm{P}}=\mathrm{E}_{2}$. Therefore,

$\mathrm{E}_{2}=\mathrm{E}_{1}\left(\frac{\mathrm{P}}{\mathrm{P}+\mathrm{Q}}\right) \Rightarrow \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{P}}{\mathrm{P}+\mathrm{Q}}$

$(n-2 \,m) \varepsilon=84$

$(32-2 \,m) 3=84$

$4=2 \,m \Rightarrow m=2$

$=\frac{(18 \times 1)+(12 \times 2)}{(2+1)}$

$ \mathrm{E}_{\mathrm{net}} =\frac{18+24}{3}=\frac{42}{3}=14 \mathrm{\,volt} $

$\mathrm{R}=20\, \Omega$

$=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}}=\frac{3}{1+5}=0.5 \mathrm{\,A}$

Potential cifference actioss the wire of $5 \Omega=0.5 \times 5=2.5 \mathrm{\,V}$

Length of wire $=1 \mathrm{\,m}$

$\therefore $ Potential gradient $=2.5 \mathrm{\,V} / \mathrm{m}$

then $I = \frac{{{E_{net}}}}{{{R_{net}}}} = \frac{{60 - 25}}{{15}} = \frac{{35}}{{15}} = \frac{7}{3}\,{\mkern 1mu} A$

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

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

$H_1=\frac{E^2 R}{\left(R+r_1\right)^2}$

$H_2=\frac{E^2 R}{\left(r+r_1\right)^2}$

$\frac{R}{\left(R+r_1\right)^2}=\frac{r}{\left(r+r_1\right)^2}$

$\frac{r+r_1}{R+r_1}=\frac{\sqrt{r}}{\sqrt{R}}$

$\sqrt{R} r_1+r \sqrt{R}=\sqrt{r} r_1+R \sqrt{r}$

$r_1(\sqrt{R}-\sqrt{r})=\sqrt{R r}(\sqrt{R}-\sqrt{r})$

$r_1=\sqrt{R r}$

Current through the circuit $i=\frac{1000 E}{1000 r}=\frac{E}{r}$

Potential drop across one cell $=E-i r=E-\left(\frac{E}{r}\right) r=0$

$\Rightarrow$ For $399$ cells, total potential drop is zero

$3=\frac{E_1+E_2}{R+r_1+r_2}$

$1=\frac{E_1-E_2}{R+r_1+r_2}$

$3=\frac{E_1+E_2}{E_1-E_2}$

$2=\frac{E_1}{E_2}$

$E=6 \,V$

$r=0.5 \,\Omega$

$R=0.75 \,\Omega$

$i=24 \,\Omega$

$S(0.5)=P(0.75)$

$2 \,s=3 p$

$i=\frac{P S E}{S r+P R}$

$24=\frac{P\left(\frac{3}{2} P\right) 6}{15 P}$

$P=4$ rows

$S=6$ cells

$E_{\text {net }}=\frac{\frac{10}{3}+\frac{10}{3}+\frac{10}{3}}{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=10$ volt

At junction $A$, ${i_{AB}} = 2 + 2 = 4\,A$

At junction $B$, ${i_{AB}} = {i_{BC}} - 1 = 3\,A$

At junction $C$, $i = {i_{BC}} - 1.3 = 3 - 1.3 = 1.7\,A$

$(2 + 2) = (0.1 + 0.3 + 0.2)i$ $ \Rightarrow $ $i = \frac{{20}}{3}\,A$

Hence potential difference across $A$

$ = 2 - 0.1 \times \frac{{20}}{3} = \frac{4}{3}\,V$ (less than $2\,V$)

Potential difference across $B = 2 - 0.3 \times \frac{{20}}{3} = 0$

${V_A} - {V_B} = 1 \times 1.5$

$ \Rightarrow \,\,\,{V_A} - 0 = 1.5\,V \Rightarrow {V_A} = 1.5\,V$

Potential difference between $B$ and $C$

${V_B} - {V_C} = 1 \times 2.5 = 2.5\,V$

$ \Rightarrow 0 - {V_C} = 2.5\,V \Rightarrow {V_C} = - 2.5\,V$

Potential difference between $C$ and $D$

${V_C} - {V_D} = - 2V$$ \Rightarrow \,\,\, - 2.5 - {V_D} = - 2 \Rightarrow \,{V_D} = - 0.5\,V.$

Applying $KVL$ along the loop $ABDA$, we get

$-6i_1 -3 i_2 + 15 = 0$   or   $2i_1 + i_2 = 5$  …..  $(i)$

Applying $KVL$ along the loop $BCDB$, we get

$-3(i_1 -i_2) -30 + 3i_2 = 0$   or   $-i_1 + 2i_2 = 10$  …..  $(ii)$

Solving equation $(i)$ and $(ii)$ for $i_2$, we get $i_2 = 5\, A$.

By $KVL,$ $15+3=(1+2+R) I$ or $I=\frac{18}{3+R}$

Here, $V_{a b}=0$

or $3-1 I=0$

or $3=I=\frac{18}{3+R}$

or $R=3 \Omega$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{AC}}=\frac{4}{40+60} \times 40=\frac{8}{5}$         .........$(ii)$

eq. $(i)$ $-$ eq. $(ii)$

$\Rightarrow \mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{B}}=\frac{9}{5}-\frac{8}{5}=\frac{1}{5}=0.2 \mathrm{\,V}$

Using Kirchhoff's first law, $ \mathrm{i}_{1}+\mathrm{i}_{2}=\mathrm{i}_{3}$

$\frac{10-V}{4}+\frac{5-V}{2}=\frac{V-0}{2}$

$\Rightarrow \mathrm{V}=4 \mathrm{\,V}$

$\therefore \mathrm{i}_{3}=\frac{\mathrm{V}-0}{2}=\frac{4-0}{2}=2 \mathrm{\,A}$

$\Rightarrow \frac{\mathrm{V}-20}{2}+\frac{\mathrm{V}-5}{4}+\frac{\mathrm{V}-0}{2}=0$

$\Rightarrow \mathrm{V}=9 \mathrm{\,Volt}$

Therefore $\mathrm{I}_{3}=\frac{\mathrm{V}}{2}=\frac{9}{2}=4.5 \mathrm{\,A}$

for loop $ABCDEA:$ $2 i_{1}+2\left(i_{1}+i_{2}\right)=2 \Rightarrow 2 i_{1}+i_{2}=1 \ldots . .(1)$

for loop $HCDFGH:$ $2\left(i_{1}+i_{2}\right)+2 i_{2}=2 \Rightarrow i_{1}+2 i_{2}=1 \ldots . .(2)$

From $(1)$ and $(2),$ $i_{1}+2\left(1-2 i_{1}\right)=1$ or $-3 i_{1}=-1 \Rightarrow i_{1}=1 / 3 A$

and $i_{2}=1-2(1 / 3)=1 / 3 A$

$\mathrm{V}_{\mathrm{G}}-\mathrm{V}_{\mathrm{H}}=4 \times 2-3+2 \times 2-3 \times 1$

$=8-3+4-3=6 \mathrm{\,V}$

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

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

$=-2+2+1=+1$ $\mathrm{volt}$

$-\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right) \mathrm{R}-\mathrm{i}_{1} \mathrm{r}_{1}+\mathrm{E}_{1}=0$

or $\mathrm{E}_{1}-\left(\mathrm{i}_{1}+\mathrm{i}_{2}\right) \mathrm{R}-\mathrm{i}_{1} \mathrm{r}_{1}=0$

$I_{y}=\frac{10-0}{2}=5 \,A$

$I_{2}=\frac{20-0}{5}=4 \,A$

applying $\mathrm{KCL}$ at $\mathrm{X}$

$I_{z}+I_{x}=I_{y}$

$4+I_{x}=I_{y}=5$

$ \Rightarrow \boxed{{{\text{r}}_{\text{x}}} = 1\,{\text{A}}}$

Applying Kirchhoffs law in mash $(I)$

$2 i_1+6 i_1-6 i_2+8 i_1+0.5 i_1-15=0$

$16.5 i_1-6 i_2=15 \ldots . \text { (I) }$

Applying Kirchhoff's law in mash $(II)$

$7 i _2+ i _2+10 i _2+6 i _2-6 i _1=0$

$24 i _2-6 i _1=0$

$4 i _2- i _1=0$

$i _1=4 i _2 \ldots \ldots \text { (II) }$

After solving of equation $(I)$ and $(II)$

Then,

$16.5 \times 4 i _2-6 i _2=15$

$66 i _2-6 i _2=15$

$60 i _2=15$

$i _2=\frac{1}{4} A$

Now put the value of $i_2$ in equation $(I)$

$i_1=4 i_2$

$i_1=4 \times \frac{1}{4}$

$i_1=1\,\,A$

Now, the current from the battery in the circuit is $1\,\,A$.

$\frac{x-10}{10}+\frac{x+6}{20}+\frac{x}{10}+\frac{x-10}{5}=0$

$\Rightarrow \mathrm{x}=6 \mathrm{\,V} \quad \therefore 1_{20\, \Omega}=\frac{6+6}{20}=0.6 \mathrm{\,A}$

$\mathrm{V}_{\mathrm{P}}+1-1=\mathrm{V}_{\mathrm{Q}}$

$\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}=0$

from kirchoff's rule

$\mathrm{V}_{\mathrm{x}}+6+2+2=0 \quad \Rightarrow \quad \mathrm{V}_{x}=-10$ $\mathrm{volt}$

${2-3 \mathrm{I}_{1}-2 \mathrm{I}_{1}=0} $

${\mathrm{I}_{1}=0.4 \mathrm{\,amp}}$

$\mathrm{K.V.L.}$ on right mesh

$4-3 I_{2}-5 I_{2}=0$

$\mathrm{I}_{2}=0.5 \mathrm{\,amp}$

Potential difference between $A $ and $ B$

${\mathrm{V}_{\mathrm{A}}+3 \mathrm{I}_{1}+4=\mathrm{V}_{\mathrm{B}}} $

${\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=-3(\mathrm{O} .4)-4=-5.2 \mathrm{\,V}}$

$6=6 i$

$i=1 A$

$\Delta V=6+4$

$\Delta V=10$

$V_A-4-6=V_B$

$V_A-V_B=w$

$10=i(2)$

$i=5 \,A$

$=\frac{R}{4}$

$i=\frac{4 E}{5 R}$

$\Delta V_1=\frac{E}{5}$

$\Delta V_2=\frac{4 E}{5}$

$\Delta V_1=\Delta V_2$

$=1: 4$

$\therefore 4 i _1+2\left( i _1+ i _2\right)-3+4 i _1=16\, V$

Using Kirchhoff's second law in the closed loop we have

$9-i_2-2\left(i_1+i_2\right)=0$

Solving equations $(i)$ and $(ii)$, we get $i _1=1.5 A ^{\text {and } i _2}=2 A$

$\therefore$ current through $2 W$ resistor $=2+1.5=3.5\,A$.

$ \Rightarrow $  $X = \frac{8}{4} = 2\,\Omega $

$\frac{A}{B} = \frac{D}{C} \Rightarrow \frac{{10}}{5} \ne \frac{4}{4}$ (Unbalanced)

$\frac{{A'}}{B} = \frac{D}{C} \Rightarrow \frac{{A'}}{5} = \frac{4}{4}$$ \Rightarrow $ $A' = 5\,\Omega $

$A'$ $(5\,\Omega )$ is obtained by connecting a $10\,\Omega $ resistance in parallel with $A$.