In the circuit shown below, calculate:
$(a)$ total resistance in arm $CE$,
$(b)$ total current drawn from the battery, and
$(c)$ current in each arm, i.e., $AB$ and $CE$ of the circuit.

(N/A) In arm $CE$, the resistors of value $2 \, \Omega$ and $4 \, \Omega$ are connected in series. Therefore, the total resistance in arm $CE = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega$.
$(b)$ The circuit consists of two parallel branches, arm $CE$ and arm $AB$, each having a resistance of $6 \, \Omega$. The total resistance of the circuit $R_P$ is given by:
$R_P = \frac{6 \, \Omega \times 6 \, \Omega}{6 \, \Omega + 6 \, \Omega} = 3 \, \Omega$.
Therefore, the total current $I$ drawn from the battery is:
$I = \frac{V}{R_P} = \frac{3 \, V}{3 \, \Omega} = 1 \, A$.
$(c)$ Since both arms $CE$ and $AB$ have the same resistance $(6 \, \Omega)$, the total current is split equally between them. Hence, the current in each arm ($AB$ and $CE$) is $0.5 \, A$.