Chapter 28: Problem 4
$$ \begin{array}{l} \text { Show that the time for the voltage in the } R C \text { circuit to rise to } V_{\mathrm{o}} / 2 \text { ("half-max") }\\\ \text { is } t_{1 / 2}=\tau \ln 2 \end{array} $$
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:
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:
- \( V(t) = V_0(1 - e^{-t/\tau}) \)
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.
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.
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.
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 \( -t/\tau = \ln(\frac{1}{2}) \).
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, \( t_{1/2} = \tau \ln 2 \)