Question

A 10 µF capacitor is charged by a 10.0 volt battery through a resistance R. The capacitor reaches a potential difference of 4.0 V in a time 3.0 s after charging begins. Determine R.

Answer 1

Let \(\mathcal E\) be the battery voltage, \(C\) be the capacitance, \(Q\) be the charge on one capacitor plate, \(I\) be the current in the circuit, \(V_f\) be the described potential difference across the capacitor, and \(t_f\) be the described time to reach that state (\(t\) being time in general). Start with Kirchoff's voltage law about our circuit. The contributions below are the voltage provided by the battery, the voltage drop from the resistance, and the voltage drop from the capacitor.\[\mathcal E - IR - \frac{Q}{C} = 0\]Recall that \(I=\frac{dQ}{dt}\) Substitute this in and rearrange slightly.\[\frac{dQ}{dt}=\frac{\mathcal E}{R}-\frac{Q}{RC}\]\[\frac{dQ}{\frac{\mathcal E}{R} - \frac{Q}{RC}}=dt\]Integrate both sides. Let \(K_i\) be constants of integration.\[\int\frac{dQ}{\mathcal E - \frac{Q}{RC}}=\int dt\]\[-RC\cdot \ln\left\vert \frac{\mathcal E}{R} - \frac{Q}{RC} \right\vert = t+K_1\]\[\ln\left\vert \frac{\mathcal E}{R} - \frac{Q}{RC} \right\vert = -\frac{t}{RC}+K_2\]\[\frac{\mathcal E}{R} - \frac{Q}{RC} = e^{-\frac{t}{RC}+K_2}\]\[Q=C\mathcal E-e^{-\frac{t}{RC}+K_2}\]\[Q=C\mathcal E-K_3e^{-\frac{t}{RC}}\], now since we are to assume that the capacitor was originally uncharged, it follows that \(K_3=C\mathcal E\) as to force \(Q=0\) when \(t=0\).\[Q=C\mathcal E \left(1-e^{-\frac{t}{RC}}\right)\]We can determine an expression for the voltage across the capacitor by dividing both sides by \(C\).\[V=\mathcal E \left(1-e^{-\frac{t}{RC}}\right)\]Plug in given values for the situation.\[V_f=\mathcal E \left(1-e^{-\frac{t_f}{RC}}\right)\]Solve for resistance.\[\boxed{\displaystyle R=-\frac{t_f}{C\cdot \ln\left(1-\frac{V_f}{\mathcal E}\right)}}\]Obviously, plug in your found values and evaluate with a calculator, rounding to significant figures as appropriate.