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I it possible to calculate the charge on a capacitor when only having the current, time and capacitance value?

I tried using this formula to first calculate the voltage and from there the charge. But in this formula I'm missing the initial voltage V0. would the resistance formula be \$V = \dfrac{1}{C} \cdot \int_{} idt+v0\$?

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  • \$\begingroup\$ The initial voltage determines the initial charge. Without it, you don't know the starting charge. You can still compute the change in charge without the initial voltage or charge, but not the absolute value. \$\endgroup\$
    – jonk
    Apr 18, 2019 at 19:02

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The direct answer to your question is no.

The missing piece of information is the initial voltage (or initial charge) as you correctly indicated in your formula for V.

Think about filling up a bucket with water. Your charge is the volume of water, your current is the flow of water, your voltage is the water level inside the bucket and the capacitance is a function of the geometry of the bucket (the bigger the area of the bucket, the bigger the C). If you don't know how much water you had in the beginning, you can't tell how much you'll have in the end.

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  • \$\begingroup\$ that makes sense. Could I calculate the charge if I had the total voltage and know the resistance? By using the current I guess I could determine if I use the charge or discharge function? As Q = Q_max (e^(-t/RC)) and Q_max = C * V_total \$\endgroup\$
    – user2246120
    Apr 19, 2019 at 4:21
  • \$\begingroup\$ yes, but if I only had a ammeter and no voltmeter available, I could still calculate the charge with the R*C formula, right? \$\endgroup\$
    – user2246120
    Apr 19, 2019 at 5:03
  • \$\begingroup\$ @user2246120 - I'm not sure I fully understand the proposed measurement circuit. I'm assuming you're talking about measuring the current as you discharge a capacitor through a resistor in parallel (please correct me if I'm wrong). Then, yes. If you take the Q(t) decay formula and do a time derivative you get the expected current waveform I(t) = I_max e^(-t/RC), where I_max = Q_max/RC. If you measure I_max and know R and C, you can calculate Q_max. \$\endgroup\$
    – joribama
    Apr 19, 2019 at 5:43
  • \$\begingroup\$ I don't think I was clear: I meant calculating the charge Q on the capacitor at time t. Not Q max. Q max is just C * V total, right? I have that information. \$\endgroup\$
    – user2246120
    Apr 20, 2019 at 13:22
  • \$\begingroup\$ I'm confused again - I thought you had no ability to measure voltage. In any case, if you measure voltage, you get Q(t) directly: Q(t)=C*v(t); if you measure current (i), and know Q(0), you can integrate the current waveform: Q(t)=Q(0)+integral(i(t),dt). \$\endgroup\$
    – joribama
    Apr 21, 2019 at 7:15
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If you want to know the charge, you can find it by integrating the current over time and adding the initial charge.

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