$$\eqalign{ & {i_g} \times G = \left( {i - {i_g}} \right)S \cr & \therefore S = \frac{{{i_g} \times G}}{{i - {i_g}}} = \frac{{1 \times 0.81}}{{10 - 1}} = 0.09\Omega \cr} $$

$$\eqalign{ & {\text{Here}}\,\,{R_{DA}} = {R_{AB}} = {R_{BC}} = \frac{R}{4} \cr & {\text{and }}{R_{DE}} = {R_{EC}} = \frac{R}{8} \cr & {\text{Now }}{R_{ED,}}{R_{DA,}}{R_{AB,}}{R_{BC}}{\text{ are in series}}{\text{.}} \cr & \therefore {R_s} = \frac{R}{8} + \frac{R}{4} + \frac{R}{4} + \frac{R}{4} = \frac{{R + 2R + 2R + 2R}}{8} = \frac{{7R}}{8} \cr & \therefore {R_{eq}} = \frac{{\left( {\frac{{7R}}{8}} \right)\left( {\frac{R}{8}} \right)}}{R} = \frac{{7R}}{{64}} \cr} $$

At cold junction, current flows from Antimony to Bismuth (because current flows from metal occurring later in the series to metal occurring earlier in the thermoelectric series).

$$\eqalign{ & I = \frac{{8 - 4}}{{1 + 2 + 9}} = \frac{4}{{12}} = \frac{1}{3}A; \cr & {V_P} - {V_Q} = 4 - \frac{1}{3} \times 3 = 3\,volt \cr} $$

If $$m$$ is the mass of a substance deposited or liberated on an electrode during electrolysis when a charge $$q$$ passes through electrolyte, then according to Faraday’s first law of electrolysis. Mass deposited is directly proportional to the charge flows.
i.e. $$m \propto q\,\,{\text{or}}\,\,m = Zq$$
where, $$Z$$ is a constant of proportionality and is called electrochemical equivalent (ECE) of the substance.
If an electric current $$i$$ flows through the electrolyte, then
$$m = Zit\,\,\left[ {q = it} \right]$$
Given, $$i = 1.5\,A,\,t = 10\,\min = 10 \times 60\,s,$$
$$Z = 30 \times {10^{ - 5}}g{C^{ - 1}}$$
Hence, mass of copper deposited on the electrode
$$\eqalign{ & m = 30 \times {10^{ - 5}} \times 1.5 \times 10 \times 60 \cr & = 27 \times {10^{ - 2}} = 0.27\,g \cr} $$

NOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoff’s second law is based on conservation of energy.

Current density $$J = \frac{I}{A}$$
$$\eqalign{ & = \frac{{50 \times {{16}^{ - 6}}}}{{50 \times {{10}^{ - 6}}}} \cr & = 1\,A{m^{ - 2}} \cr} $$

$${\text{hence}}\,{R_{eq}} = \frac{3}{2};\,\,\,\,\,\,\therefore I = \frac{6}{{\frac{3}{2}}} = 4A$$

As we know that,
$${\text{Power}}\,P = {i^2}R\,\left[ {_{R\, = \,{\text{resistance}}}^{{\text{where,}}\,i\, = \,\,{\text{current in circuit}}}} \right]$$
$$R = \frac{p}{{{i^2}}}$$
Given, $$P = 1\,W,\,i = 5\,A$$
$$\therefore R = \frac{1}{{{{\left( 5 \right)}^2}}} = 0.04\,\Omega $$

If a rated voltage and power are given, then
$${P_{{\text{rated}}}} = \frac{{V_{{\text{rated}}}^2}}{R}$$
$$\therefore $$ Current in the bulb, $$i = \frac{P}{V}\,\,\left( {\because P = Vi} \right)$$
$$i = \frac{{500}}{{100}} = 5\,A$$
$$\therefore $$ Resistance of bulb, $${R_b} = \frac{{100 \times 100}}{{500}} = 20\,\Omega $$
$$\because $$ Resistance $$R$$ is connected in series.
$$\eqalign{ & \therefore {\text{Current,}}\,\,i = \frac{E}{{{R_{{\text{net}}}}}} = \frac{{230}}{{R + {R_b}}} \cr & \Rightarrow R + 20 = \frac{{230}}{5} = 46 \cr & \therefore R = 26\,\Omega \cr} $$