(i) Three resistances $R_1, R_2$ and $R_3$ are connected in parallel to one another between the same two points. In this case, the potential difference across the ends of all the resistance will be the same.$∴ \;\; {V}= {V}_1= {V}_2= {V}_3$...(i)If the total current flowing through the circuit is $I$, then the current passing through $R_1$ will be $I_1$ through $R_2$ will be $I_2$ and through $R_3$ will be $I_3$.Then${I}= {I}_1+ {I}_2+ {I}_3$...(ii) If $R$ is the effective resistance of the circuit, connected across a battery of ${V}$ volts, through which I current flows, then$I=\;{V}/{R}$ Substituting the values in eq. (ii), we get$⇒ \;{V}/{R}=\;{V}/{R_1}+\;{V}/{R_2}+\;{V}/{R_3}$$⇒ \;{V}/{R}=V [\;{1}/{R_1}+\;{1}/{R_2}+\;{1}/{R_3}]$$⇒ \;{1}/{R}=\;{1}/{R_1}+\;{1}/{R_2}+\;{1}/{R_3}$ (ii) Given: ${R}_1=4 Ω, {R}_2=4 Ω, {R}_3=8 Ω$ Let, resultant resistance between $a$ and $c$ be ${R}^{'}$ Then, $\;\; R^{'}=R_1+R_2$(Series combination)${R}^{'}=4+4=8 Ω$If $R$ is the effective resistance between $A$ and $B$, then$\;{1}/{R}=\;{1}/{R^{'}}+\;{1}/{R_3}$( $R^{'}$ and $R_3$ are in parallel combination)$\;\;{1}/{R}\;=\;{1}/{8}+\;{1}/{8}=\;{2}/{8}$$⇒\;R\;=4 Ω$