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?
Question
Answer:
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 .