Four resistors of `12 Omega` each are connected in parallel . Three such combinations are then connected in series. What is the total resistance ? If a battery of 9 V emf and negligible internal resistance is connected across the networks, find the current flowing through of each resistors.

Explanation

The network of resistors, connected to a battery of e.m.f. 9 V is shown in figure. Let I be the total current in the circuit and R' be the effective resistance of four resistors of `12 Omega` each in paralle, then
`1/R' = 1/12 +1/12 +1/12+1/12 = 4/12 or R'= 3 Omega`
The total resistance of the network of resistors is `R = 3 + 3 + 3= 9 Omega`
Current in the circuit is `I=epsilon/R =9/9 = 1A`
Since all the four resistors in parallel are of equal resistance, so same current will flow through each resistor. Therefore, current through each resistor `= I/4 = 1/4 = 0.25 A`