Power of electric bulb, $$P = \frac{{{V^2}}}{R}$$
So, resistance of electric bulb, $$R = \frac{{{V^2}}}{P}$$
Given, $${P_1} = 25\,W,{P_2} = 100\,W,$$
$${V_1} = {V_2} = 220\,V$$
Therefore, for same Potential difference $$V,$$
$$R \propto \frac{1}{P}$$
Thus, we observe that for minimum power, resistance will be maximum and vice-versa.
Hence, resistance of $$25\,W$$ bulb is maximum and $$100\,W$$  bulb is minimum.

For various combinations equivalent resistance is maximum between $$P$$ and $$Q.$$

It is clear that the two cells oppose each other hence, the effective emf in closed circuit is $$18 - 12 = 6\,V$$   and net resistance is $$1 + 2 = 3\,\Omega $$   (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.

$$I = \frac{{{\text{Effective}}\,{\text{emf}}}}{{{\text{Total resistance}}}} = \frac{6}{3} = 2A$$
The potential difference across $$V$$ will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of $$18\,V,$$  hence,
$${V_1} = {E_1} - i{r_1} = 18 - \left( {2 \times 2} \right) = 18 - 4 = 14\,V$$
Similarly, current enters in the cell of $$12\,V,$$  hence,
$${V_2} = {E_2} + i{r_2} = 12 + 2 \times 1 = 12 + 2 = 14\,V$$
Hence, $$V = 14\,V$$

Volume of material remains same in stretching.
As volume remains same, $${A_1}{l_1} = {A_2}{l_2}$$
Now, given $${l_2} = n{l_1}$$
∴ New area $${A_2} = \frac{{{A_1}{l_1}}}{{{l_2}}} = \frac{{{A_1}}}{n}$$
Resistance of wire after stretching
$$\eqalign{ & {R_2} = \rho \frac{{{l_2}}}{{{A_2}}} = \rho \cdot \frac{{n{l_1}}}{{\frac{{{A_1}}}{n}}} \cr & = \left( {\rho \frac{{{l_1}}}{{{A_1}}}} \right) \cdot {n^2} = {n^2} \cdot R\,\,\left[ {\because R = \left( {\rho \frac{{{l_1}}}{{{A_1}}}} \right)} \right] \cr} $$

Energy loss in conductor, $$Q = {i^2}Rt$$
where, $$i =$$  current flowing through it
$$R =$$  resistance of conductor
$$t =$$  time for which current is passed
Heat developed $$ = ms\Delta \theta $$
$$\therefore ms\Delta \theta = {i^2}Rt$$
Since $$m,s,R,t$$   remains constant.
So, $$\Delta \theta \propto {i^2}$$
So, for two different cases
$$\eqalign{ & \frac{{\Delta {\theta _2}}}{{\Delta {\theta _1}}} = \frac{{i_2^2}}{{i_1^2}} \cr & {\text{or}}\,\,\Delta {\theta _2} = {\left( {\frac{{{i_2}}}{{{i_1}}}} \right)^2}\Delta {\theta _1}\,......\left( {\text{i}} \right) \cr} $$
Given, $${i_2} = 2{i_1},\Delta {\theta _1} = {5^ \circ }C$$
So, from Eq. (i),
$$\eqalign{ & \therefore \Delta {\theta _2} = {\left( {\frac{{2{i_1}}}{{{i_1}}}} \right)^2} \times 5 \cr & = 4 \times 5 \cr & = {20^ \circ }C \cr} $$

Given : $$emf\,\varepsilon = 2.1\,V$$
$$I = 0.2\,A,R = 10\Omega $$
Internal resistance $$r = ?$$
From formula.
$$\eqalign{ & \varepsilon - Ir = V = IR \cr & 2.1 - 0.2\,r = 0.2 \times 10 \cr & 2.1 - 0.2\,r = 2\,\,{\text{or}}\,\,0.2\,r = 0.1 \cr & \Rightarrow r = \frac{{0.1}}{{0.2}} = 0.5\,\Omega \cr} $$

Kirchhoff's junction law follows from the conservation of charge.
Kirchhoff's loop law follows from the conservation of energy.

$${R_A} = $$  resistance of ammeter
$$\eqalign{ & \frac{{4 - {V_1}}}{{100}} = \frac{{{V_1} - 1}}{{{R_A}}} + \frac{{{V_1} - 0}}{{100}}\,......\left( {\text{i}} \right) \cr & 1\,V - 0\,V = \left( {10mA} \right){R_A} \cr & {R_A} = 100\Omega \,......\left( {{\text{ii}}} \right) \cr} $$

$$\eqalign{ & \frac{{4 - {V_1}}}{{100}} = \frac{{{V_1} - 1}}{{100}} + \frac{{{V_1} - 0}}{{100}}\,\,\left[ {{\text{By using eq}}{\text{. }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right)} \right] \cr & {V_1} = \left( {\frac{5}{3}} \right)V \cr & \frac{{{V_1} - 1}}{{{R_A}}} = \left( {{\text{Current in ammeter }}\left( {{\text{II}}} \right)} \right) \cr & = \frac{{\left( {\frac{5}{3}} \right) - 1}}{{100}} = 6.67\,mA \cr} $$

Given $$X$$ is greater than $$2\Omega $$  when the bridge is balanced
$$\frac{R}{\ell } = \frac{X}{{100 - \ell }}$$

Case (i)
$$\eqalign{ & {\text{or,}}\,100R - R\ell = \ell X\,\,\,{\text{or,}}\,200 - 2\ell = \ell X \cr & {\text{or,}}\,\ell = \frac{{200}}{{X + 2}} \cr} $$
When the resistances are interchanged the jockey shifts $$20\,cm.$$  Therefore
$$\eqalign{ & \frac{X}{{\ell + 20}} = \frac{2}{{80 - \ell }} \cr & 80X - \ell X = 2\ell + 40 \cr & {\text{or,}}\,80X = \ell \left( {X + 2} \right) + 40 \cr & {\text{or,}}\,80X = \left( {\frac{{200}}{{X + 2}}} \right)\left( {X + 2} \right) + 40 \cr & {\text{or,}}\,X = \frac{{240}}{{80}} = 3\Omega . \cr} $$

The given circuit can be redrawn as shown.
From circuit, $$\frac{{FC}}{{CE}} = \frac{{FD}}{{DE}} = 1$$

Thus, it is balanced Wheatstone bridge, so, resistance in arm $$CD$$  is ineffective and so no current flows in this arm.
Net resistance of the circuit is
$$\eqalign{ & \frac{1}{{R'}} = \frac{1}{{\left( {R + R} \right)}} + \frac{1}{{\left( {R + R} \right)}} \cr & = \frac{1}{{2R}} + \frac{1}{{2R}} = \frac{2}{{2R}} = \frac{1}{R} \cr & \therefore R' = R \cr} $$
So, net current drawn from the battery is
$$i' = \frac{V}{{R'}} = \frac{V}{R}$$
As from symmetry, upper circuit $$AFCEB$$   is half of the whole circuit and is equal to $$AFDEB.$$   So, in both the halves half of the total current will flow.
Hence, in $$AFCEB,$$   the current flowing is
$$i = \frac{{i'}}{2} = \frac{V}{{2R}}$$