Problem 6 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

It takes about 92.10 seconds for the voltage to drop to 10% of its original value.

Step 1: Identify the type of differential equation

The given equation \( \frac{dV}{dt} = -\frac{1}{40}V \) is a first-order linear differential equation. It is a separable differential equation, which means we can rewrite it as \( \frac{dV}{V} = -\frac{1}{40} dt \).

Step 2: Integrate both sides

To solve for \(V\), we integrate both sides of the equation: \( \int \frac{dV}{V} = -\int \frac{1}{40} dt \). This results in \( \ln |V| = -\frac{1}{40}t + C \), where \(C\) is the constant of integration.

Step 3: Solve for \(V\)

To solve for \(V\), we exponentiate both sides: \( |V| = e^{C} e^{-\frac{t}{40}} \). Let \(e^C = V_0\), which is the initial voltage at \(t = 0\), giving us \( V = V_0 e^{-\frac{t}{40}} \).

Step 4: Calculate the time for voltage to drop to 10%

We want the voltage \(V\) to be 10% of its initial value \(V_0\), so set \( V = 0.1V_0 \). Substitute into the equation: \( 0.1V_0 = V_0 e^{-\frac{t}{40}} \). Simplifying, we get \( 0.1 = e^{-\frac{t}{40}} \).

Step 5: Solve for time \(t\)

To find \(t\), take the natural logarithm of both sides: \( \ln(0.1) = -\frac{t}{40} \). Solving for \(t\), we have \( t = -40\ln(0.1) \). Calculate this to find \( t \approx 92.10 \) seconds.

Key Concepts

Capacitor Discharge

In electronic circuits, a capacitor stores electrical energy temporarily. When a capacitor discharges, it releases this stored energy. The rate of this energy release is fundamental in many electronic applications, from camera flashes to electronic timing circuits.
A discharging capacitor can be modeled using a first-order differential equation. This captures the proportionality between the rate of voltage drop and the voltage itself. Mathematically, for a capacitor's voltage \(V\), we model it as \(\frac{dV}{dt} = -kV\), where \(k\) is a constant.

• This equation shows how the voltage decreases over time.
• It implies that as the voltage decreases, the rate of decrease slows down.

Understanding the discharge of capacitors helps in predicting how long a device will function under given power conditions.

Exponential Decay

Exponential decay describes processes where quantities reduce at a rate proportional to their current value. This concept is prevalent in natural processes and radioactive decay, in addition to the electrical context of capacitor discharge.
The differential equation representing exponential decay \(\frac{dV}{dt} = -kV\) demonstrates how a quantity, such as voltage \(V\) over time \(t\), decreases exponentially. Upon solving, this leads to the solution \(V = V_0e^{-kt}\), where:

• \(V_0\) is the initial value at \(t = 0\).
• \(e\) is the base of the natural logarithm.

This mathematical model reveals that exponential decay is characterized by a rapid drop initially, which slows over time, never quite reaching zero. Such an understanding is critical in designing circuits that rely on precise timing as they involve capacitor discharges.

Initial Value Problem

An initial value problem involves solving a differential equation with a specific condition given at the initial point, usually \(t = 0\). Here, we are given the equation \(\frac{dV}{dt} = -\frac{1}{40}V\) with the initial condition \(V = V_0\) at \(t = 0\).
Solving an initial value problem involves finding the specific solution that satisfies both the differential equation and the initial condition. In our case, the solution is found to be \(V = V_0e^{-\frac{t}{40}}\).

• This equation illustrates how the initial condition shapes the solution's exponential form.
• By substituting initial values, we can predict future behavior of the system, such as how long it takes for voltage to reduce to a particular fraction of its original value.

Initial value problems are crucial in physics and engineering because they allow for precise predictions of system behavior over time.