Question

Question #8801d

Answer 1

This is what I get

Explanation:

Voltage vs time relationship of a charging `RC` circuit is shown in the figure above and is given as

`V(t)=E_s(1-e^(-t/(tau)))` ......(1)
where `E_s` is the DC supply across `RC` circuit of time constant `tau`.

It is given that `E_s=V_C-0_"ref"=8.262-2.385=5.877\ V`

We also know that charge held in a capacitor can be written as

`Q=CV`.

Therefore (1) becomes

`Q(t)=CE_s(1-e^(-t/(tau)))` ......(2)

Total charge would be maximum at `t=oo`. From (2) we get

`Q_max=CE_s` ........(3)

Given condition is Charge on the capacitor must be `=15/16Q_max` `=>` at what time `t` or how many times `RC`.
Let the desired time be `=xtau`

`15/16Q_max=CE_s(1-e^(-(xtau)/(tau)))` .......(4)

From (3) and (4) we get

`15/16=(1-e^(-x))`

Inserting various values in (1) we get

`V(t)=5.877xx15/16=5.510\ V`