According to Faraday's first law of electrolysis
$$m = ZIt \Rightarrow m \propto It$$

Voltage across $$3\,\Omega $$  resistance is given by Ohm's law
i.e. $$V = IR = 3 \times {i_1}$$
$$ = 3 \times 0.8 = 2.4\,V\,\,\left( {{\text{as}}\,{i_1} = 0.8\,A} \right)$$
As $$3\,\Omega $$  and $$6\,\Omega $$  are in parallel, hence voltage across $$6\,\Omega $$  resistance will also be $$2.4\,V.$$
∴ Current through $$6\,\Omega $$  resistance
$${i_2} = \frac{V}{R} = \frac{{2.4}}{6} = 0.4\,A$$
∴ Total current in the circuit,
$$i = {i_1} + {i_2} = 0.8 + 0.4 = 1.2\,A$$
∴ Potential drop across $$4\,\Omega $$  resistance,
$$ = i \times 4 = 1.2 \times 4 = 4.8\,V$$

Input energy when the battery is charged
$$\eqalign{ & {E_{{\text{in}}}} = V\,it \cr & = 15 \times 10 \times 8 \cr & = 1200\,Wh \cr} $$
Energy released when the battery is discharged
$$\eqalign{ & {E_{{\text{out}}}} = 14 \times 5 \times 15 \cr & = 1050\,Wh \cr} $$
Hence, watt hour efficiency of battery is given by
$$\eqalign{ & = \frac{{{\text{Energy output}}}}{{{\text{Energy input}}}} = \frac{{1050}}{{1200}} \cr & = 0.875 = 87.5\% \cr} $$

When we measure the emf of a cell by the potentiometer then no current is drawn from the external circuit. Thus, in this condition the actual value of a cell is found. In this way potentiometer is equivalent to an ideal voltmeter of infinite resistance.
NOTE
The emf by the potentiometer is measured from null method in which zero deflection position is found on the wire.

Case 1: $${P_1} = \frac{{{V^2}}}{R}$$

Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $$\frac{R}{2}.$$ These are connected in parallel
$$\eqalign{ & \therefore {R_{eq}} = \frac{{\frac{R}{2}}}{2} = \frac{R}{4} \cr & \therefore {P_2} = \frac{{{V^2}}}{{\frac{R}{4}}} = 4\left( {\frac{{{V^2}}}{R}} \right) = 4{P_1} \cr} $$

$$I = \frac{E}{{R + r}},$$    Internal resistance $$\left( r \right)$$ is zero, $$I = \frac{E}{R} = {\text{constant}}{\text{.}}$$

$$H = \frac{{{V^2}t}}{R}$$
Resistance of half the coil = $$\frac{R}{2}$$
∴ As $$R$$ reduces to half, $$'H'$$ will be doubled.

From the principle of potentiometer, $$V \propto l$$
$$ \Rightarrow \frac{V}{E} = \frac{l}{L};\,{\text{where}}$$
$$V =$$  emf of battery, $$E =$$  emf of standard cell.
$$L =$$  length of potentiometer wire
$$V = \frac{{El}}{L} = \frac{{30E}}{{100}}$$

NOTE: In this arrangement, the internal resistance of the battery $$E$$ does not play any role as current is not passing through the battery.

$$\frac{P}{Q} = \frac{R}{S}\,{\text{where}}\,S = \frac{{{S_1}{S_2}}}{{{S_1} + {S_2}}}$$

$$r = \frac{E}{I} = \frac{{1.5}}{3} = 0.5\,{\text{ohm}}{\text{.}}$$