Physics M.C.Q (1 Marks)

$\therefore \frac{{R{R_2}}}{{R + {R_2}}} + {R_1} = R \Rightarrow 2R + R + 2 = {R^2} + 2R$

$ \Rightarrow {R^2} - R - 2 = 0 \Rightarrow R = - 1$ or $R = 2\,ohm$

Now $8\,\Omega $ is in parallel with $(1 + 1 + 4 + 1 + 1)\,\Omega $.

$\therefore R = 8\,\Omega ||8\,\Omega = \frac{8}{2} = 4\,\Omega $ $ \Rightarrow $ ${R_{AB}} = 4 + 2 + 2 = 8\,\Omega $

Now ${V_C} - {V_E} = \frac{4}{3} \times 1$ $ \Rightarrow $ ${V_E} = - \frac{4}{3}\,V$

Now such three resistance are joined in series,

hence total $R = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\,ohm$

Hence every new piece will have a resistance $\frac{R}{{10}}.$

If two pieces are connected in series, then their resistance $ = \frac{{2R}}{{10}} = \frac{R}{5}$

If $5$ such combinations are joined in parallel, then net resistance $ = \frac{R}{{5 \times 5}} = \frac{R}{{25}}$

Potential difference between $A$ and $B$ ${V_A} - {V_B} = 2 \times 12$

$ \Rightarrow $ ${V_A} - 0 = 24\,V$ $ \Rightarrow $ ${V_A} = 24\,V$

If all are in parallel then ${R_{eq}} = \frac{4}{3}\Omega = 1.33\,\Omega $

If two are in series then parallel with third, ${R_{eq}} = \frac{8}{3} = 2.6\,\Omega $

If two are in parallel then series with third, ${R_{eq}} = 6\,\Omega $

Current given by the cell $i = \frac{E}{{{R_{eq}} + r}} = \frac{{10}}{{(4 + 1)}} = 2\,A$

Hence, $({V_A} - {V_B}) = \frac{i}{2} \times ({R_2} - {R_1}) = \frac{2}{2}(2 - 4) = - 2\,V.$

$18=10+\frac{16 R}{16+R} \Rightarrow 8=\frac{16 R}{16+R}$

$128+8 R=16 R \Rightarrow 8 R=128$

$R=16 \Omega$

Now potential difference across the combination of $3$ $\Omega$ and $6$ $\Omega$, $V = 0.5 \times \left( {\frac{{3 \times 6}}{{3 + 6}}} \right) = 1\,Volt$

The same potential difference, also develops across $3$ $\Omega$ resistance.

Current $i = \frac{9}{1} = 9\,A$

Current passes through the ammeter $= 5\,A$.

$ \Rightarrow $  ${R_{AB}} = 5\,\Omega $.

${R_{AC}} = \frac{{200 \times 200}}{{200 + 200}} = 100\,\Omega $.

By using $R = \rho \frac{l}{A} \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{{A_2}}}{{{A_1}}} \Rightarrow \frac{{{R_1}}}{8} = \frac{1}{4} \Rightarrow {R_1} = 2\,\Omega $

Hence, ${R_{eq}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{2 \times 8}}{{(2 + 8)}} = \frac{8}{5}\,\Omega .$

$R = 3 + \frac{{10 \times (3 + R)}}{{10 + 3 + R}}$ $ = 3 + \frac{{30 + 10R}}{{13 + R}}$

$R = \frac{{39 + 3R + 30 + 10R}}{{13 + R}} = \frac{{69 + 13R}}{{13 + R}}$

$13R + {R^2} = 69 + 13R$ $ \Rightarrow $ $R = \sqrt {69}\, \Omega $.

Equivalent resistance $R = \frac{{3 \times (3 + 3)}}{{3 + (3 + 3)}} = 2\,\Omega $

Current $i = \frac{2}{2} = 1\,A.$ So,${i_1} = 1 \times \left( {\frac{3}{{3 + 6}}} \right) = \frac{1}{3}\,A$.

Potential difference between $A$ and $B =\frac{1}{3} \times 3 = 1\,volt.$

Hence, $\frac{6}{5} = \frac{{2 \times {R_2}}}{{2 + {R_2}}} \Rightarrow {R_2} = 3\,\Omega $.

$\Omega$, which is $V = i × 20$  and $i = \frac{{48}}{{(100 + 100 + 80 + 20)}} = 0.16\,A$

 $V = 0.16 \times 20 = 3.2\,V$.

So, $\frac{1}{x} + \frac{1}{{2x}} + \frac{1}{R} = 1 \Rightarrow x = \frac{3}{2}\left( {\frac{R}{{R - 1}}} \right)$

As, resistance is not fractional $ \Rightarrow $ $\frac{R}{{R - 1}} = 2$
$ \Rightarrow $ $x = 3,\,R = 2,\,2x = 6$

Hence, the value of largest resistance $= 6\,\Omega$.

 Resistance of each arm $ = \frac{{12}}{3} = 4\,\Omega $

 ${R_{effective}} = \frac{{4 \times 8}}{{4 + 8}} = \frac{8}{3}\Omega $

So $\frac{P}{Q} = \frac{R}{S}$

$==>$ $\frac{{21}}{{3 + \frac{{8X}}{{(8 + X)}}}} = \frac{{18}}{6}$

$==>$ $X = 8\,\Omega $

$ \Rightarrow $ ${R_{AB}} = \frac{{30 \times 60}}{{(30 + 60)}} = 20\,\Omega $

$R_{AB}=\frac {10}{3}\,\Omega $

Then  $i \times 25 = (2.1 - i)\,10$

$i = \frac{{10}}{{35}} \times 2.1 = 0.6\,A$

In balance condition, $\frac{P}{Q} = \frac{{Sr/(S + r)}}{R}$. This gives $r = 8\,\Omega $

Between $A$ and $C$ circuit becomes equivalent to balanced Wheatstone bridge so ${R_{AC}} = R$.

$P+Q>R+S$

$i=i_{1}+i_{2}$

$i>i_{2}>i_{1}>i_{g}$

When resistor is cut into two equal parts and connected in parallel, then

${R_{eq}} = \frac{{R/2}}{2}$ $ = \frac{R}{4} = \frac{6}{4} = 1.5\,\Omega $

$\frac{1}{{R'}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{3} = \frac{2}{3}$ or $R' = \frac{3}{2} = 1.5\,\Omega $

Now the equivalent resistance between $A$ and $B$ as $R' = 1.5\,\Omega $ and $2.5\,\Omega $ are connected in series, so

Now by ohm's law, potential difference between $A$ and $B$ is given by ${V_A} - {V_B} = iR = 2 \times 4.0 = 8\,\,volt$

When these resistance are connected in parallel.

$\frac{1}{{{R_P}}} = \frac{1}{4} + \frac{1}{{12}}$   $ \Rightarrow \,\,\,{R_P} = \frac{{4 \times 12}}{{4 + 12}} = \frac{{4 \times 12}}{{16}} = 3\,\Omega $

$10 = 5{i_1} + i$..... $(i)$

Applying in the second mesh

$5{i_1} = 4i - 4{i_1}$...... $(ii)$

Solving equation $(i)$ and $(ii)$, we get

${i_1} = \frac{{40}}{{29}}\,A$

current through the circuit $i = \frac{{2.4}}{{100}}\,A$

$P.D.$ across combination of voltmeter and $100\,\Omega$ resistance $ = \frac{{2.4}}{{100}} \times 50 = 1.2\,V$

Since the voltmeter and $100\,\Omega$ resistance are in parallel, so the voltmeter reads the same value i.e. $1.2\,V.$

$20 = (R + r)4$ $ \Rightarrow $ $R + r = 5$ $ \Rightarrow $ $R < 5\,\Omega $

$ \Rightarrow $ $R = 1\,\,\Omega $.