Question

Pls help with this ques..​

Answer 1

(i) See the attachment.

(ii) (a) 2.4 Ω, (b) 2.5 A, (c) 1.0 A

Step-by-step explanation:

(ii) (a) If the resistance of the parallel combination is Rρ, then

`(1)/(Rρ) = (1)/(4) + (1)/(6 ) \\ = > (3 + 2)/(12) = (5)/(12 ) \\ or \\ Rρ = (12)/(5) Ω = 2.4Ω`

(b) Let the current through the battery be I, then,

`I = (V)/(Rρ) = (6)/(2.4) \\ I = 2.5 \: Α`

(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 :-

`I₁ = (V)/(R₁) = (6)/(4) \\ I₁ = 1.5 \: Α`

Current through 6 Ω resistor :-

`I₂ = (V)/(R₂) = (6)/(6) \\ I₂ = 1.0 \: Α \\ \\ Note \: that \: I \: = I₁ \: + I₂.`

Hope it helps you!!