In the given electric circuit if the current flowing through `3 Omega` resistor is 1 A, find the voltage of the battery and the current I drawn from it.

Explanation

`R_(2) = 3 Omega" "I_(2) = 1A" "V_(2) = I_(2)R_(2) = 1 xx 3 = 3` volts
Resultant of `R_(2) and R_(3)` is R'
`(1)/(R')=(1)/(R_(2))+(1)/(R_(3))=(1)/(3)+(1)/(6)=(2+1)/(6)=(3)/(6)=(1)/(2)`
`R' = 2 Omega" "V' = V_(2) = 3 "volts"" "I' = ?`
`I' = (V')/(R')=(3)/(2)A = 1.5 A" "I'=I=1.5A`
Resultant resistance in the circuit = R
`R = R_(1) + R' = 2 + 2 = 4 Omega" "I=(3)/(2) A" "V = ?`
`V = "IR" = (3)/(2) xx 4 = 6 "volts"" "rArr" "E = 6 "volts"`