C1 = 56uF
C2 = 12uF
R = 12kohm
E = 13 V
I have this circuit and I want help to find out how long in takes for the voltage Uc2 to achieve the value 7 V?
I have calculated the time constant to 117 ms, and I tried to find the time through t = 0.117 * ln(13/13-7) but that doesn't seem to be right.
Could someone help me to go to the right direction?
Thank you!
Best Answer
In your circuit:
Since the current in a series circuit is everywhere the same at any instant, when S1 closes, the accumulation of charge in C1, in time, will be equal to the accumulation of charge in C2, but since their capacitances are different the voltages across them will be different. .
Then, since \$ Q=CV\$, and E3 is given as 7 volts,
for C2, $$ Q = C2 \times E3 =12\mu F \times 7V =84\mu C $$
and for C1, $$ V=\frac {Q}{C} = \frac{84\mu C}{56\mu F} = 1.5\text{ volts.}$$
The voltage across C2 is given as 7 volts, so E2 will be equal to 7 volts plus the voltage across C1, 1.5 volts, so $$ E2 = E3+1.5V = 8.5V $$
C1 and C2 are in series, so their equivalent capacitance is:
$$C_T = \frac{C1 \times C2}{C1+C2} \approx 9.9 \text{ microfarads.}$$
Our circuit now looks like this:
and we can get the time taken to charge Ct to 8.5 volts (which is the time taken to charge C2 to 7 volts) by solving:
$$\tau = k \ RC, $$
where
$$ k = ln\frac{V}{V-V_T} = ln\frac{13V}{13V-8.5V} =1.06$$
which works out to:
$$\tau = 1.06\times\ 1.2\cdot10^{4}\Omega\times 9.9\cdot10^{-6}F \approx 126\text{ milliseconds}. $$
The proof:
and the LTspice circuit list: