$(a)$ Explain how a cell maintains current in a circuit.
$(b)$ In the circuit given below,the resistance of the path $xTy = 2 \, \Omega$ and that of $xZy = 6 \, \Omega$.
$(i)$ Find the equivalent resistance between $x$ and $y$.
$(ii)$ Find the current in the main circuit.
$(iii)$ Calculate the current that flows through the path $xTy$ and $xZy$.

(N/A) The chemical action within a cell generates a potential difference across its terminals. This potential difference drives and maintains the flow of electric current in the circuit.
$(b)$ $(i)$ The paths $xTy$ and $xZy$ are connected in parallel between points $x$ and $y$. The equivalent resistance $R_e$ is given by:
$\frac{1}{R_e} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3} \, \Omega^{-1}$
$R_e = 1.5 \, \Omega$
$(ii)$ The total resistance of the circuit is the sum of the equivalent resistance $R_e$ and the series resistor of $1.5 \, \Omega$:
$R_{total} = 1.5 \, \Omega + 1.5 \, \Omega = 3 \, \Omega$
Using Ohm's law,the main current $I = \frac{V}{R_{total}} = \frac{6 \, V}{3 \, \Omega} = 2 \, A$
$(iii)$ The potential difference across the parallel combination of $xTy$ and $xZy$ is $V_{xy} = I \times R_e = 2 \, A \times 1.5 \, \Omega = 3 \, V$.
Current through path $xTy$ $(2 \, \Omega)$: $I_1 = \frac{V_{xy}}{R_1} = \frac{3 \, V}{2 \, \Omega} = 1.5 \, A$
Current through path $xZy$ $(6 \, \Omega)$: $I_2 = \frac{V_{xy}}{R_2} = \frac{3 \, V}{6 \, \Omega} = 0.5 \, A$