A series combination of 12 Ω and 3 Ω is connected in parallel with another series combination of 6 Ω and 3 Ω. If a potential difference of 4 V is applied across it find the i) current drawn from the battery ii) current through 12 Ω resistor (iii)potential difference across 6 Ω resistor?

Explanation:

Resultant of 12 and 3 ohm in series = 15 ohm

Resultant of 6 and 3 ohm = 9 ohm

Total resultant resistance of circuit = 15 x 9 / (15 + 9)

= 5.625 ohm

current drawn from battery = 4 / 5.625

= .711 A

ii )current through 12 ohm = 4 / (12 + 3 )   , because potential  diff over 12 and 6 ohm will be 4 V .

current through 12 ohm = .267 A

iii )

current through 6 ohm

= .711 - .267

= .444 A

potential difference

= .444 x 6

= 2.664 V .

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