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How to draw the curve of the charge Qc as a function of time for an RC circuit, whose capacitor is initially charged and without source, when closing the tap

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  • \$\begingroup\$ What do you have to use to draw it? Some program or pen&paper? What help do you need to get started? Or are you looking for a formula for it? \$\endgroup\$
    – Justme
    Commented Jul 19, 2023 at 16:15

2 Answers 2

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The equation for voltage output of an RC is this:

\$V(t)=V_0 e^{\frac{-t}{RC}} \$

V0 is the capacitor initial charge.

So plug in values for t, 0,1,2,3 and then take the output V(t) and graph it.

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  • \$\begingroup\$ The question asked for charge vs time, not voltage vs time. \$\endgroup\$
    – RussellH
    Commented Jul 19, 2023 at 19:29
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The relation between current and charge is given by:

$$\text{I}\left(t\right)=-\text{Q}'\left(t\right)\tag1$$

In a series RC circuit with an initially charged capacitor, the voltage across the capacitor and resistor is given by:

$$\text{V}\left(t\right)=\text{V}_0\exp\left(-\frac{t}{\text{RC}}\right)\tag2$$

So, the current is given by:

$$\text{I}\left(t\right)=\frac{\text{V}\left(t\right)}{\text{R}}=\frac{1}{\text{R}}\cdot\text{V}_0\exp\left(-\frac{t}{\text{RC}}\right)\tag3$$

So, we need to solve:

$$\frac{1}{\text{R}}\cdot\text{V}_0\exp\left(-\frac{t}{\text{RC}}\right)=-\text{Q}'\left(t\right)\space\Longleftrightarrow\space\text{Q}\left(t\right)=-\int\limits_0^t\frac{1}{\text{R}}\cdot\text{V}_0\exp\left(-\frac{t}{\text{RC}}\right)\space\text{d}t\tag4$$

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