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$$ \begin{array}{l} \text { Show that the time for the voltage in the } R C \text { circuit to rise to } V_{\mathrm{o}} / 2 \text { ("half-max") }\\\ \text { is } t_{1 / 2}=\tau \ln 2 \end{array} $$

Short Answer

Expert verified
The time is \( \tau \ln 2 \).

Step by step solution

01

Understand the RC Circuit Charging Equation

The voltage across a charging capacitor in an RC circuit is given by the equation \( V(t) = V_0(1 - e^{-t/\tau}) \), where \( V_0 \) is the maximum voltage, \( \tau = RC \) is the time constant, \( R \) is the resistance, and \( C \) is the capacitance. This formula describes how the voltage changes with time.
02

Set the Voltage to Half-Maximum

To find the time \( t_{1/2} \) when the voltage reaches half of \( V_0 \) (i.e., \( V_0/2 \)), we substitute \( V(t) = V_0/2 \) into the charging equation: \( V_0/2 = V_0(1 - e^{-t_{1/2}/\tau}) \).
03

Simplify the Equation

Cancel \( V_0 \) from both sides of the equation, which gives \( 1/2 = 1 - e^{-t_{1/2}/\tau} \). Rearrange this equation to isolate the exponential term: \( e^{-t_{1/2}/\tau} = 1 - 1/2 = 1/2 \).
04

Solve for \( t_{1/2} \)

Take the natural logarithm of both sides to solve for \( t_{1/2} \). We have \( -t_{1/2}/\tau = \ln(1/2) \). Solving this yields \( t_{1/2} = -\tau \ln(1/2) \).
05

Simplify the Expression Using Logarithm Rules

Use the logarithmic identity \( \ln(1/2) = -\ln 2 \) to simplify the expression for \( t_{1/2} \). Substitute this back into the equation: \( t_{1/2} = -\tau (-\ln 2) = \tau \ln 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charging Capacitor
In an RC circuit, a capacitor stores and releases electrical energy. When a voltage is applied, the capacitor begins to "charge." This process involves the accumulation of electric charge, and with time, it approaches the maximum voltage (\( V_0 \)).
Unlike instantly reaching this voltage, the charging happens gradually. The rate at which a capacitor charges depends on resistance (\( R \)) and capacitance (\( C \)).
When initially powered, the capacitor voltage increases as given by the formula:
  • \( V(t) = V_0(1 - e^{-t/\tau}) \)
Here, \( \tau \), known as the time constant, is crucial in predicting how quickly the capacitor charges.
Time Constant
A key concept in understanding RC circuits is the time constant, denoted by \( \tau \). The time constant is the product of resistance and capacitance: \( \tau = R \times C \).
It represents how fast the capacitor charges or discharges and is measured in seconds.
The time constant indicates the time required for the voltage to reach about 63% of its maximum value when charging. This gives a practical sense of timing for the capacitor's behavior.
  • After \( 1\tau \), the voltage reaches 63% of \( V_0 \).
  • After \( 5\tau \), the capacitor is nearly fully charged.
Knowing \( \tau \) can thus help in predicting how quickly a circuit responds to changes in voltage.
Exponential Equation
The equation governing the charging of a capacitor in an RC circuit is an exponential equation. In equations like \( V(t) = V_0(1 - e^{-t/\tau}) \), the term \( e^{-t/\tau} \) defines the exponential change.
Initially, when \( t = 0 \), the capacitor holds no charge, and the exponential term is 1.
As time progresses, \( e^{-t/\tau} \) decreases, causing the voltage \( V(t) \) across the capacitor to approach its maximum. Exponential equations like these beautifully illustrate how the voltage changes over time in an RC circuit.
  • Exponential decay dominates the charging process.
  • Each time constant \( \tau \) moves the charging closer to maximum.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a key tool in solving exponential equations in RC circuits. For instance, in solving for the time \( t_{1/2} \) when the capacitor reaches half its voltage, the equation involves \( \ln \).
To isolate \( t \), one must apply the natural logarithm to both sides of the equation:
\( e^{-t/\tau} = \frac{1}{2} \) becomes \( -t/\tau = \ln(\frac{1}{2}) \).
  • Using log properties, \( \ln(1/2) \) is \( -\ln(2) \).
  • This step simplifies the solution for \( t \).
  • Thus, \( t_{1/2} = \tau \ln 2 \)
Understanding and using \( \ln \) is crucial for working through problems that involve exponential changes in RC circuits.

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