Using the superposition method, calculate the current through r5 in figure 8-71

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

now the current flowing through the battery is

now this will divide into R3 and R2 so current flowing in R3 will be

now this will again divide in R4 and R5

so current in R5 will be

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

also after its series with R3

now it is in parallel with R5 on other side

now current through the battery will be given as

now it is divide in r2 and R5

so current in R5 is given as

now the total current in R5 will be given by super position which is

so there is 0.85 mA current through R5 resistance

okay i have so many questions. but the big one here is: um wt..f is this mess? ?