In the arrangement shown below, two different metals $M_1$ and $M_2$ are short$-$circuited by an

external connection. Show that $P{D}_{M\frac{1}{S}}=P{D}_{M\frac{2}{S}}.$ Also write an expression to give $V$ in terms

of the quantity $P{D}_{re\frac{f}{S}}.$ Hint: Apply Kirchhoff's law to each of the three cycles a$,$ b$,$ and c shown

in the figure. For simplicity, assume that the potential difference across $M_1$ and its external

connecting wire can be neglected. Make the same assumption for $M_2$ and for "ref" and their

respective connecting wires. Also assume that the potential difference between two different

points in solution is negligible.