(i) See the attachment.
(ii) (a) 2.4 Ω, (b) 2.5 A, (c) 1.0 A
Explanation:
(ii) (a) If the resistance of the parallel combination is Rρ, then
[tex] \frac{1}{Rρ} = \frac{1}{4} + \frac{1}{6 } \\ = > \frac{3 + 2}{12} = \frac{5}{12 } \\ or \\ Rρ = \frac{12}{5} Ω = 2.4Ω[/tex]
(b) Let the current through the battery be I, then,
[tex]I = \frac{V}{Rρ} = \frac{6}{2.4} \\ I = 2.5 \: Α[/tex]
(c) The potential difference across each resister is V = 6 volt (same as of the battery) since they are in parallel.
Current through 4 Ω resister :-
[tex]I₁ = \frac{V}{R₁} = \frac{6}{4} \\ I₁ = 1.5 \: Α[/tex]
Current through 6 Ω resistor :-
[tex]I₂ = \frac{V}{R₂} = \frac{6}{6} \\ I₂ = 1.0 \: Α \\ \\ Note \: that \: I \: = I₁ \: + I₂.[/tex]
Hope it helps you!!
