Physics M.C.Q (1 Marks)

$ \Rightarrow $ $12 = 0.1 (R + 2)$

$ \Rightarrow $ $R = 118\,\Omega$.

and $R = \pi \times 1 \times {10^{ - 6}}\,\Omega $

On solving, $R' = 0.88 \times {10^{ - 6}}\,\Omega $

${R_1} = \frac{{10}}{4} = 2.5\,\Omega $ and

Resistance of part $PMQ$;

${R_2} = \frac{3}{4} \times 10 = 7.5\,\Omega $

${R_{eq}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{2.5 \times 7.5}}{{(2.5 + 7.5)}} =\frac{{15}}{8}\,\Omega $.

Main Current $i =$ $\frac{3}{{\frac{{15}}{8} + 1}} = \frac{{24}}{{23}}A$

So, $i_1= i \times \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right) = \frac{{24}}{{23}} \times \left( {\frac{{7.5}}{{2.5 + 7.5}}} \right) = \frac{{18}}{{23}}A$

and ${i_2} = i - {i_1} = \frac{{24}}{{23}} - \frac{{18}}{{23}} = \frac{6}{{23}}A$.

Current $i = \frac{6}{{6 + 4 + 1}} = \frac{6}{{11}}A$

P.D. between $A$ and $B$,   $V = \frac{6}{{11}} \times 10 = \frac{{60}}{{11}}V$.

When only $S_1$ is closed ${V_1} = \frac{3}{4}E = 0.75\,E$

When only $ S_2$ is closed ${V_2} = \frac{6}{7}E = 0.86\,E$
and when both $S_1$ and $S_2$ are closed combined resistance of $6R$ and $3R$ is $2R$

${V_3} = \left( {\frac{2}{3}} \right)E = 0.67\,E$ $\Rightarrow $ ${V_2} > {V_1} > {V_3}$

Also $ - 2({I_1} - {I_{xy}}) + 4({I_2} + {I_{xy}}) = 0$

i.e. $2{I_1} - 4{I_2} = 6{I_{xy}}$ .... $(ii)$

Also ${V_{AB}} - 1 \times {i_1} - 2({i_1} - {i_{xy}}) = 0$  $\Rightarrow $ $50 = {i_1} + 2({i_1} - {i_{xy}})$

$ = 3{I_1} - 2{I_{xy}}$ .... $(iii)$

Solving $(i)$, $(ii)$ and $(iii)$, ${i_{xy}} = 2\,A$

$P.D.$  across $AC $$ = \frac{1}{5} \times 20 = 4\,V$

$P.D.$  across $AN = 4 + 4 = 8\,V$

Potential difference across $4\, \Omega$  resistance is $V = 2 \times 4 = 8\,volt$
Hence, potential difference across each capacitor is $4\,V$
So charge on each capacitor $Q = 3 \times 4 = 12 \,\mu C$.

Current through $3\,\Omega $ resistor $ = \frac{{24}}{{3 + 3}} = 4\;A$

Let potential of point $‘O’$ shown in fig. is ${V_O}$
then using ohm’s law

${V_O} - {V_a} = 3 \times 4 = 12\,V$  .... $(i)$

Now current through $5\,\Omega $ resistor $ = \frac{{24}}{{5 + 1}} = 4\,A$

So ${V_0} - {V_b} = 4 \times 1 = 4\,V$  ..... $(ii)$

From equation $(i)$ and $(ii)$ ${V_b} - {V_a} = 12 - 4 = 8\,V.$

$i = \frac{{10}}{{\frac{{1000}}{3} + 500}} = \frac{3}{{250}}\,A$

Now voltmeter reading $ = {i_v} \times {R_V}$ $ = \frac{2}{3} \times \frac{3}{{250}} \times 500 = 4\,V.$

$i = \frac{{20}}{{(3 + 2)}} = 4\,A$.

When switch is closed $i = \frac{{20}}{{3 + (2||2)}} = 5\,A$.

Now using ohm’s law $i = \frac{{25}}{{R + R'}}$$ \Rightarrow 0.5 = \frac{{25}}{{R + 4}}$

$ \Rightarrow R + 4 = \frac{{25}}{{0.5}} = 50$ $ \Rightarrow R = 50 - 4 = 46\;\Omega $

Current through $20\,\Omega $ resistor $ = \frac{{0.5 \times 5}}{{20 + 5}} = \frac{{2.5}}{{25}} = 0.1\,A$

Potential difference across middle resistor
$=$ Potential difference across $20\,\Omega $$ = 20 \times 0.1 = 2\,V$

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

$R=\frac{12.5}{0.5}=25 \Omega$

Now

$E=I R=\frac{V}{R+r}(R)$

$E=\frac{12.5}{25+1}(25)=(0.48)(25)=12 V$

the voltmeter with resistance $R_{v}=120$ ohm and resistor $R_{2}=60$

ohm are in parallel, hence equivalent resistance between them is

$R_{e}=\frac{R_{2} R_{v}}{R_{2}+R_{v}}$

$R_{e}=\frac{(60)(120)}{60+120}=40 \Omega$

Now the reading of voltmeter can be obtain by

$V_{v}=V\left(\frac{R_{e}}{R_{1}+R_{e}}\right)$

$V_{v}=120\left(\frac{40}{60+40}\right)$

$V_{v}=40\left(\frac{6}{5}\right)=48 \mathrm{V}$

Current through $20\, ohm$ resistor is $0.3 \mathrm{A},$ hence the voltage across it is

$V=0.3 \times 20=6 V$

Current through $15 \,ohm$ resistor is

$I=\frac{V}{R}=\frac{6}{15}=0.4 A$

Hence current through $\mathrm{R} 1$ is, $\mathrm{I}^{\prime}=0.8-(0.3+0.4)=0.1 \mathrm{A}$

Therefore,

$R 1=\frac{V}{I^{\prime}}=\frac{6}{0.1}=60 \Omega$

$V_{A}-V_{B}=\left(I-i_{g}\right) 0.1=i_{g}(9+0.9)$

or $I=\frac{10 \times 0.01}{0.1}=1 A$

$12=0.1(20+r) \Rightarrow r=100 \Omega$

The voltmeter reading will be $12 V$.

$I=\frac{28}{14}=2 A m p$

$V_{A}+3 \times 1-10 \times 1=V_{B}$

$V_{A}-V_{B}=7$ volt

$\Rightarrow x=4$

$(ii)$ $\mathrm{R}_{1}=6$

Now $\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{R}_{1} \times \mathrm{x}}{\mathrm{R}_{1}+\mathrm{x}}=\frac{6 \times 4}{10}=2.4$

$\Rightarrow \mathrm{R}=\frac{4 \mathrm{V}}{0.4 \mathrm{\,A}}=10\, \Omega$

$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{15}+\frac{1}{9}$ or $\quad \mathrm{R}_{\mathrm{eq}}=\frac{(15)(9)}{15+9}=\frac{135}{24}=5.62\, \Omega$

Current through circuit i $=\frac{10}{5.62}=1.78 \mathrm{\,A}$

If current through upper and lower branches of the network are $\mathrm{i}_{1}$ and $\mathrm{i}_{2}$, then according to Kirchoff's function law $-$

$\mathrm{i}_{1}+\mathrm{i}_{2}=1.78 \mathrm{\,A}$

and $\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)+\left(\mathrm{V}_{\mathrm{b}}-\mathrm{V}_{\mathrm{c}}\right)=\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{d}}\right)+\left(\mathrm{V}_{\mathrm{d}}-\mathrm{V}_{\mathrm{c}}\right)$

or $\quad \mathrm{i}_{1}(5+10)^{3 / 4}\, \mathrm{i}_{2}(3.0+6.0)$ or $(15) \mathrm{i}_{1}=(9.0) \mathrm{i}_{2}$

On solving, we get $-$ $\mathrm{i}_{1}=0.67 \mathrm{\,A}$ and $\mathrm{i}_{2}=1.1 \mathrm{\,A}$

$\frac{{\frac{{2{p^2}{l^2}}}{{3\pi {{\rm{R}}^4}}}}}{{\frac{{\rho l}}{{\pi {{\rm{R}}^2}}}\left( {1 + \frac{2}{3}} \right)}} = \frac{{2\rho {\rm{l}}}}{{3\pi {{\rm{R}}^2}}} \times \frac{3}{5}$

$=\frac{12+6+3+2}{12 R_{0}}$

$R=\frac{12 R_{0}}{23} \quad i=\frac{\varepsilon}{\left(\frac{2 R_{0}}{23}\right)}=\frac{23 \varepsilon}{12 R_{0}}$

$\mathrm{V}_{2}=3 \mathrm{\,r}$

$\mathrm{V}_{1}-\mathrm{V}_{2}=2 \mathrm{\,r}$

curront in middle one is $2 \mathrm{\,A}$ so reading should be $7 \mathrm{\,A}$

current in circuit $=\frac{12}{8}=\frac{3}{2} \mathrm{\,A}$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=9 \mathrm{V}$

$\mathrm{q}=\mathrm{C}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right)$

$=2 \times 9=18\, \mu \mathrm{C}$

$=\frac{85}{7}$

$\mathrm{V}=\mathrm{IR}_{\mathrm{eq}}=1 \times 20=20 \mathrm{\,V}$

$\Rightarrow \mathrm{x}=6 \mathrm{\,V}$

$\therefore \mathrm{I}_{20 \Omega}=\frac{6+6}{20}=0.6 \mathrm{\,Amp}$

Also $\mathrm{R}_{\mathrm{AB}}$ depends upon $\mathrm{R}_{1}, \mathrm{R}_{2}$ and $\mathrm{R}_{3}$