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The voltage, $5e^{-t} \cos(100\pi t)$, is applied to a circuit with a capacitance, $0.9\times 10^{-6} \mathrm{F}$, in series with a resistance, $13\times 10^3\mathrm{\Omega}$.

Show that $\displaystyle \frac{dV}{dt}=427.35e^{-t}\cos(100\pi t)-85.47V$

where $V$ is the voltage across the capacitor.

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closed as off-topic by Dylan, Shuhao Cao, Cesareo, Leucippus, stressed out Feb 23 at 3:36

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  • $\begingroup$ What have you tried? I think the most important thing to know is the physics equations behind the problem. $\endgroup$ – Matti P. Feb 22 at 13:46
  • $\begingroup$ You posted two questions in a row of just copied-and-pasted homework problems. Where is the attempt? $\endgroup$ – Dylan Feb 22 at 14:07
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The instantaneous charge $Q$ on the plates of the capacitor is given by $CV$, where $V$ is the voltage across the capacitor (and $C$ is the capacitance).

The current passing through the capacitor is the first derivative wrt time of the charge, i.e. $I = \frac{dQ}{dt} = C\frac{dV}{dt}$

The capacitor and the resistor are in series so the current through the resistor is also $I$. By Ohm's Law, the voltage across the resistor is given by $RI = RC\frac{dV}{dt}$ where $R$ is the resistance.

But the voltage across the resistor can also be computed (by Kirchoff's Voltage Law) as $V_s - V$, where $V_s$ is the source voltage.

Hence $V_s - V = RC\frac{dV}{dt}$

Now put in what they've given you and see if you can't get the required result.

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