Question

4.Three resistors of resistances 2 Ω,4 Ω and 4Ω are connected in series to a battery of 5V.​

Answer 1

Answer:

Eq resistance, of parallel combination 4,6 and 12 ohm resistors from relation 1/R = 1/R1 + 1/R2 + 1/R3, is

1/{1/4+1/6+1/12} = 1/(3/12+2/12+1/12) = 1/(3+2+1)/12 = 1/6/12 = 2 ohm.

Total resistance connected to battery = 2+2 = 4

battery current = 4/4 = 1 amp