Equivalent capacitance problem

[gracy]

Similarly here
$V_i$=$I$$R_i$=$V_{in}$$\frac{R_i}{R_1+R_2+...+R_n}$

 

[epenguin]

Read #40. I have added to the "Relevant equations". When you take these into account you can solve this problem in 10 seconds... unless you prefer to be a pedantic student, but I see no virtue in taking days you could use for other study or something else over a 10 s problem.

 

[ehild]

Should not it be $V_i$=Q/$C_i$=$V_{in}$$\frac{1/C_i}{1/C_1+1/C_2+...+1/C_n}$

Yes, thank you . I corrected it. You see how easy it is to make mistakes!
Yes i denotes any random element.
You can also notice that in case all n components of the chain are equivalent, the voltage across one of them is V_in/N . That was epenguin suggested you in #40.

 

[gneill]

in the picture why $V_{Rn}$ =$I$$R_n$ and $V_{Cn}$=$\frac{Q}{C_n}$ is not shown

Good catch! That's down to me getting sloppy with the cut & paste to duplicate the expressions, forgetting to "touch up" those entries.

I've updated the picture.