Problem 28 - 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 approximately 92.08 seconds for the voltage to drop to 10% of its original value.

Step 1: Write the General Solution

The given differential equation is \( \frac{dV}{dt} = -\frac{1}{40} V \). This is a first-order linear differential equation with the standard form \( \frac{dV}{dt} = kV \), where \( k = -\frac{1}{40} \). The general solution to this type of differential equation is \( V(t) = V_0 e^{kt} \).

Step 2: Substitute into the General Solution

Substitute \( k = -\frac{1}{40} \) into the general solution. This gives us \( V(t) = V_0 e^{-\frac{1}{40} t} \). This represents the voltage at any time \( t \).

Step 3: Determine Time for Voltage to Decrease to 10%

To find the time \( t \) it takes for the voltage to drop to 10% of its initial value, set \( V(t) = 0.1V_0 \). Substitute this into the equation: \( 0.1V_0 = V_0 e^{-\frac{1}{40} t} \).

Step 4: Solve for Time \( t \)

Divide both sides by \( V_0 \), giving \( 0.1 = e^{-\frac{1}{40} t} \). Take the natural logarithm of both sides to obtain: \( \ln(0.1) = -\frac{1}{40} t \).

Step 5: Calculate \( t \)

Solve for \( t \) by multiplying both sides by \(-40\): \( t = -40 \ln(0.1) \). Calculate \( \ln(0.1) \approx -2.302 \). Thus, \( t \approx -40(-2.302) = 92.08 \).

Key Concepts

Capacitor Discharge

A capacitor is an electrical component that stores energy in an electric field. When a capacitor discharges, it releases this energy over time. This release causes the voltage across the capacitor to decrease. Capacitor discharge can be modeled using a first-order differential equation where the rate of voltage decrease is proportional to its current value. In mathematical terms, this is expressed as \( \frac{dV}{dt} = -\frac{1}{40} V \), where \( V \) is the voltage, and \( t \) is time measured in seconds. The minus sign indicates that the voltage decreases with time. This principle is crucial in fields like electronics and signal processing, where understanding how capacitors discharge can inform circuit design and performance.

Exponential Decay

Exponential decay is a mathematical concept often used to describe processes where quantities decrease over time at a rate proportional to their current value. In the context of capacitor discharge, it describes how the voltage \( V(t) \) decreases as time progresses. The solution to the differential equation \( \frac{dV}{dt} = -\frac{1}{40} V \) is \( V(t) = V_0 e^{-\frac{1}{40} t} \). The exponent includes a negative sign, which signifies decay. The constant \( \frac{1}{40} \) determines the speed of this decay. A larger constant would mean faster decay, while a smaller one would slower it. Exponential decay is pivotal in understanding real-world phenomena such as radioactive decay, cooling of objects, and population decrease.

Initial Value Problem

An initial value problem in differential equations involves finding a particular solution to a differential equation, given an initial condition. The equation \( \frac{dV}{dt} = -\frac{1}{40} V \) with \( V(0) = V_0 \) is one such example. Here, \( V_0 \) represents the initial voltage across the capacitor when \( t = 0 \). To solve this problem, we first find the general solution of the differential equation, \( V(t) = V_0 e^{-\frac{1}{40} t} \). We then use the initial condition to determine the specific solution for the problem at hand. Initial value problems are foundational in physics and engineering, providing a way to model the behavior of dynamic systems based on initial conditions.

Natural Logarithm

The natural logarithm, denoted by \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.718 \). It is essential for solving exponential equations like \( V(t) = V_0 e^{-\frac{1}{40} t} \). In the capacitor discharge example, we find the time \( t \) it takes for the voltage to drop to 10% of \( V_0 \) by solving \( 0.1 = e^{-\frac{1}{40} t} \). Taking the natural logarithm of both sides helps us isolate the variable \( t \):

• \( \ln(0.1) = -\frac{1}{40} t \)
• \( t = -40 \ln(0.1) \)
• \( \ln(0.1) \approx -2.302 \)
• Therefore, \( t \approx 92.08 \) seconds.

Natural logarithms simplify calculations in mathematics and science when dealing with exponential growth or decay.