Physics M.C.Q (1 Marks)

$I=\frac{v}{4}=\frac{2}{4}=\frac{1}{2}$

$\mathrm{V}_{\mathrm{J}}=\frac{32}{(32+64)} \times 60=20 \mathrm{\,V}$

$\mathrm{V}$ $=\frac{6 \times 24}{40+24}$ $ = \boxed{2.25\,{\text{V}}}$

$R_{e q}=\frac{(20-x) x}{(20-x)+x}=1.8$

Solving, we get; $\mathrm{x}=2\, \Omega$

The resistance per unit length is $1\, \Omega / \mathrm{m} .$ So, length of shorter part $=2 \mathrm{\,m}.$

$\therefore $ $\frac{12}{500+R}=\frac{2}{R}$

or $\mathrm{R}=100\, \Omega$

then $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{R}_{\mathrm{wire}}}=\frac{2}{\mathrm{R}+3}$

$\phi  = \frac{{{V_{{\rm{wire }}}}}}{\ell } = \frac{{I{R_{{\rm{wire }}}}}}{\ell }$

$\frac{10^{-3}}{10^{-2}}=\left(\frac{2}{\mathrm{R}+3}\right) \frac{3}{(1)}$ solving this, we get

$\mathrm{R}=57 \,\Omega$

voltage reading:

$\mathrm{V}=\frac{2}{2+2} \times 10=5 \mathrm{\,Volt}$

$10 \times 1 =\mathrm{i}_{1} \times(2+\mathrm{x}) $

$\mathrm{i}_{1} =\frac{10}{2+\mathrm{x}}$       .......$(i)$

$\mathrm{i}_{2}=\frac{10}{\mathrm{x}}$

$\therefore i_{2}=2 i_{1}$

$\frac{10}{x}=2\left(\frac{10}{2+x}\right)$

$x=2\, \Omega$

$\mathrm{R}_{\mathrm{eff}}=5 \,\Omega$

Hence $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{eff}}}=\frac{60}{5}=12 \mathrm{\,A}$

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

$T.P.D.$ of cell $=E-I r=9-3 \times 1=6 \,V$

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

$\frac{R_1 R_2}{R_1+R_2}=\frac{6}{5}$

$R_1=2$

$\frac{2 R_2}{2+R_2}=\frac{6}{5}$

$5 R_2=6+3 R_2$

$2 R_2=6$

$R_2=3 \,\Omega$

$\frac{R_1 R_2}{R_1+R_2}=3$

$R_1+R_2=16$

$R_1 R_2=48$

$R_1\left(16-R_1\right)=48$

$R_1^2-16 R_1+48=0$

$R_1=4,12$

$\frac{E}{16 R} R_3=12 R_1$

$R_{\text {net }}=4 R_1$

$i=\frac{E}{4 R_1}$

$P=\frac{E^2 R_1}{16 R_1^2}=\frac{E^2}{16 R_1}$

$P_2=\frac{9 E^2}{256 R_1{ }^2} \cdot 4 R_1=\frac{9 P}{4}$

$P_3=\frac{E^2}{256 R_1{ }^2} \cdot 12 R_1=\frac{3 E^2}{64 R_1}=\frac{3 P}{4}$

$P+\frac{9 P}{4}+\frac{3 P}{4}=\frac{16 P}{4}=4P$

$\frac{i}{11-i}=\frac{1.2}{1}$

$i=13.2-1.2 i$

$i=\frac{13.2}{2.2}=\frac{132}{22}$

$i=6 \,A$

$R_{\text {net }}=2 \,\Omega$

$0.4=i(2)$

$i=0.2 \,A$

$i=\frac{12}{4}=3 \,A$

$\Delta V_4=6 \,V$

For $R=4 r$, the sequence repeats itself.

$R_{ net }=3 \,\Omega$

$2 \,\Omega$ net resistance

$R_{\text {net }}=9$

$V=9 V$

$i=1 A$

$R_2$ and $R_3 \Rightarrow$ Parallel

$\Delta V=\frac{50}{5}=10 V$

$10=i\left(2 \times 10^6\right)$

$i=5 \,\mu A$

$R_{ net }=8 \,\Omega$

$V=24 \,V$

$24=i(8)$

$i=3 \,A$

$i=\frac{180}{90}=2\,A$

As no current flows through arm $E B$ then $V _{ D }=0 V ; \quad V _{ E _{ D }}=0 V ; \quad V _{ B }=-4 V ; \quad V _{ A }=5 V$ So, potential difference between the points $A$ and $D$

$V_A-V_D=5\,V$

Thus in each row,

$q=6\, \mu C.$

$\mathrm{emf}$ $\mathrm{E}$ is shared between

$\mathrm{r} $ and $ \mathrm{R}_{2}=$ in the ratio of their resistance voltage across $\mathrm{R}_{2}$ is $\frac{\mathrm{ER}_{2}}{\mathrm{R}_{2}+\mathrm{r}}=2 \mathrm{\,Volts}$

$=$ Voltage across capacitor.

$\therefore $ $\mathrm{Q}=\mathrm{CV}_{\mathrm{c}}=2 \mu \mathrm{C}$

So current in the circuit will be clockwise.

Applying Kirchoff's voltage law

$ - \,1 \times i + 10 - 4 - 2 \times i - 3i = 0$ $ \Rightarrow $ $i = 1\,A$ $(a$ to $b$ via $e)$

Current $ = \frac{V}{R} = \frac{{10 - 4}}{6} = 1.0\,ampere$

Across the cell, $E = V + ir$ $ \Rightarrow $ $r = \frac{{E - V}}{i} = \frac{{E - V}}{{V/R}} = \left( {\frac{{E - V}}{V}} \right)\,R$

Then $0.5 = \frac{E}{{(r + 2)}}$ and $0.25 = \frac{E}{{(r + 5)}}$

On dividing, $2 = \frac{{5 + r}}{{2 + r}}$ $ \Rightarrow $  $r = 1\,\Omega $

and $4i = 2$ $ \Rightarrow $ $i = \frac{1}{2}$

Putting the value of $i$ in $(i)$, we get $r = 0.4\, ohm$.

Total resistance $R + nr$

$ \Rightarrow $  $i = \frac{{nE}}{{R + nr}}$

In parallel, ${I_2} = \frac{E}{{2 + \frac{r}{2}}} = \frac{{2E}}{{4 + r}}$

Since ${i_1} = {i_2}$ $ \Rightarrow $ $\frac{{2E}}{{4 + r}} = \frac{{2E}}{{2 + 2r}}$ $ \Rightarrow $ $r = 2\,\Omega $

Therefore total emf $=$ $2E$.

Total resistance $=$ $R + 2r$

 $i = \frac{{2E}}{{R + 2r}} = \frac{{2 \times 1.45}}{{1.5 + 2 \times 0.15}} = \frac{{2.9}}{{1.8}} = \frac{{29}}{{18}} = 1.611\,amp$

$ \Rightarrow $ ${V_2} = 4 + {V_1} = 4 + 10 = 14\,volt$