Question

Discharging Capacitor Voltage Suppose that electricity is draining from a capacitor at a rate proportional to the voltage \(V\) across its terminals and that, if \(t\) is measured in seconds, $$\frac{d V}{d t}=-\frac{1}{40} V$$ (a) Solve this differential equation for \(V,\) using \(V_{0}\) to denote the value of \(V\) when \(t=0\). (b) How long will it take the voltage to drop to 10\(\%\) of its original value?

Short answer

The solution to the differential equation is \(V(t) = V_0 e^{-\frac{1}{40}t}\), and it will take about \(t = -40ln(0.1)\) seconds for the voltage to drop to 10% of its initial value.

Step-by-step solution

- Solving the differential equation

The differential equation we want to solve is \(\frac{d V}{d t} = -\frac{1}{40} V\). This is a simple first-order differential equation which can be solved using the method of separation of variables. By separating variables and integrating both sides, one would get: \( \int \frac{1}{V} dV = -\frac{1}{40} \int dt\). This would result in \( ln|V| = -\frac{1}{40}t + C\), where \(C\) is the constant of integration.

- Applying initial conditions

The resulting equation represents a family of solutions. To get the specific solution we use the initial condition that \(V=V_0\) when \(t=0\). By Substituting these values into the equation, we can solve for \(C\). Setting \(V=V_0\) and \(t=0\) yields that \(C = ln|V_0|\). Thus, the specific solution to the ODE is: \( ln|V| = -\frac{1}{40}t + ln|V_{0}|\). Exponentiating yields the voltage as a function of time: \(V(t) = V_0 e^{-\frac{1}{40}t}\)

- Find time for 10% voltage

The final step is to find the time it takes for the voltage to drop to 10% of its initial value. We set \(V(t) = 0.1V_0\) and solve for \(t\). This gives \(0.1V_{0} = V_0 e^{-\frac{1}{40}t}\). Simplify the equation and isolate for \(t\), one would get \(t = -40ln(0.1)\)