$(a)$ For the circuit shown in the diagram,calculate:
$(i)$ The value of current through the $30\, \Omega$ resistor.
$(ii)$ The total resistance of the circuit.
$(b)$ Give two advantages of connecting electrical devices in parallel with a battery.

(N/A) $(i)$ According to Ohm's law,the current $I$ through a resistor is given by $I = V / R$. Since the resistors are connected in parallel,the potential difference across each resistor is the same,i.e.,$V = 6\, V$.
For the $30\, \Omega$ resistor:
$I = 6\, V / 30\, \Omega = 0.2\, A$.
$(ii)$ For a parallel combination of resistors $R_1 = 5\, \Omega$,$R_2 = 10\, \Omega$,and $R_3 = 30\, \Omega$,the equivalent resistance $R_P$ is given by:
$1 / R_P = 1 / R_1 + 1 / R_2 + 1 / R_3$
$1 / R_P = 1 / 5 + 1 / 10 + 1 / 30$
$1 / R_P = (6 + 3 + 1) / 30 = 10 / 30 = 1 / 3$
Therefore,$R_P = 3\, \Omega$.
$(b)$ Two advantages of connecting electrical devices in parallel are:
$(i)$ If one device fails or is switched off,the others continue to work.
$(ii)$ Each device gets the full voltage of the power supply.