# How can this differential Question about an RC circuit be solved? [closed]

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.

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

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• What have you tried? I think the most important thing to know is the physics equations behind the problem. – Matti P. Feb 22 at 13:46
• You posted two questions in a row of just copied-and-pasted homework problems. Where is the attempt? – Dylan Feb 22 at 14:07

## 1 Answer

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.