If in the figure `R_(1) = 10Omega, R_(2) = 40 Omega, R_(3)= 30Omega, R_(4) =20 Omega, R_(5)= 60Omega` and a 12 V battery is connected to the arrangement, calculate (i) the total resistance in the circuit and (ii) the total current flowing in the circuit.

Explanation

Let the equivalent resistance of parallel resistors `R_(1)` and `R_(2) = R.`.
The equivalent resistance of parallel resistors `R_(3), R_(4)` and `R_(2) = R"`.
Then, we have `(1)/(R.)= (1)/(R_(1))+ (1)/(R_(2))`

`implies (1)/ (R.)= (1)/(10)+ (1)/(40)= (4+1)/(40)= (5)/(40)`
`implies R.= 8Omega`
similarly, `(1)/(R")= (1)/(R_(3))+ (1)/(R_(4))+ (1)/(R_(5))`
`implies (1)/(R")= (1)/(30) +(1)/(20)+ (1)/(60)= (6)/(60)`
`implies R" = 10Omega`
Thus, the total resistance
`R= R.+ R"= 8Omega+ 10Omega= 18Omega`
According to Ohm.s law
`I= (V)/(R)= (12V)/(18Omega)= 0.67A`