I build this circuit in an experiment circuit

During the experiment I got these readings:

t (s)   0  10   20   30  40   50   60   70   80   90  100
U (V)   0 0,43 0,86 1,2 1,53 1,84 2,11 2,35 2,58 2,78 2,96

t (s) 110   120  130  140  150 160
U (V) 3,12 3,28 3,41 3,54 3,68 3,76

Can someone show me how to calculate the capacity of the capacitor c using these readings? The power supply is 5,34V.

  • 1
    \$\begingroup\$ It seems this is some kind of homework assignment. What have you done so far to figure this out? \$\endgroup\$
    – Justme
    Nov 17, 2019 at 13:52
  • \$\begingroup\$ Step 1: Plot the data. Step 2: determine the RC time constant of the circuit by reading of the time the output rises to x percent of the steady state value... \$\endgroup\$
    – sstobbe
    Nov 17, 2019 at 15:41
  • 1
    \$\begingroup\$ you need to know the value of R to calculate C \$\endgroup\$
    – vicatcu
    Nov 17, 2019 at 16:50
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    \$\begingroup\$ @nmq Since you agree that \$\tau\approx 125\:\text{s}\$, then you can say that \$R\cdot C\approx 125 \:\text{s}\$, or that \$C\approx \frac{125 \:\text{s}}{R}\$. Again, there is that missing \$R\$ value. No matter how you cut this, you need a quantity for \$R\$ to get a quantity for \$C\$. \$\endgroup\$
    – jonk
    Nov 17, 2019 at 18:47
  • 2
    \$\begingroup\$ @nmq The quantitative (as opposed to qualitative) value of \$C\$ isn't possible without \$R\$. Your data can provide information about the combined pairing. But that cannot be used to separate the one from the other. You need more information. There is something else that you haven't told us about and that you may not even realize in important. Or else the question itself is flawed. \$\endgroup\$
    – jonk
    Nov 18, 2019 at 6:52

2 Answers 2


Fit the curve to the known equation for the charging of a capacitor through a known resistance to a known voltage: V(t) = V * (1 - EXP(-t / (R * C))). Use Excel or similar and try different values of C until you find one that adequately fits your empirical data, then that must be your best guess at C.


Well, using Mathematica I got:

enter image description here



  • \$\begingroup\$ The curve doesn't match the data, so the excessive precision of your answer is inappropriate. \$\endgroup\$ Nov 19, 2019 at 2:10

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