You set up the circuit shown in Fig. 26.20, where C=5.00 ×10^-6 F. At time t=0, you close the s

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You set up the circuit shown in Fig. 26.20, where C=5.00...

You set up the circuit shown in Fig. 26.20, where $C=5.00 \times 10^{-6} \mathrm{~F}$. At time $t=0,$ you close the switch and then measure the charge $q$ on the capacitor as a function of the current $i$ in the resistor. Your results are given in the table: $$ \begin{array}{l|lllll} i(\mathrm{~mA}) & 56.0 & 48.0 & 40.0 & 32.0 & 24.0 \\ \hline q(\mu \mathrm{C}) & 10.1 & 19.8 & 30.2 & 40.0 & 49.9 \end{array} $$ (a) Graph $q$ as a function of $i .$ Explain why the data points, when plotted this way, fall close to a straight line. Find the slope and $y$ -intercept of the straight line that gives the best fit to the data. (b) Use your results from part (a) to calculate the resistance $R$ of the resistor and the emf $\mathcal{E}$ of the battery. (c) At what time $t$ after the switch is closed is the voltage across the capacitor equal to $10.0 \mathrm{~V} ?$ (d) When the voltage across the capacitor is $4.00 \mathrm{~V},$ what is the voltage across the resistor?

You set up the circuit shown in Fig. 26.20, where $C=5.00 \times 10^{-6} \mathrm{~F}$. At time $t=0,$ you close the switch and then measure the charge $q$ on the capacitor as a function of the current $i$ in the resistor. Your results are given in the table:
$$
\begin{array}{l|lllll}
i(\mathrm{~mA}) & 56.0 & 48.0 & 40.0 & 32.0 & 24.0 \\
\hline q(\mu \mathrm{C}) & 10.1 & 19.8 & 30.2 & 40.0 & 49.9
\end{array}
$$
(a) Graph $q$ as a function of $i .$ Explain why the data points, when plotted this way, fall close to a straight line. Find the slope and $y$ -intercept of the straight line that gives the best fit to the data. (b) Use your results from part (a) to calculate the resistance $R$ of the resistor and the emf $\mathcal{E}$ of the battery. (c) At what time $t$ after the switch is closed is the voltage across the capacitor equal to $10.0 \mathrm{~V} ?$ (d) When the voltage across the capacitor is $4.00 \mathrm{~V},$ what is the voltage across the resistor?

Key Concepts

Linearization of Transient Response

By rearranging the KVL equation for an RC circuit, one obtains a linear relationship of the form q = -CR·i + C?. Plotting the capacitor charge (q) versus the resistor current (i) yields a straight line whose slope and y-intercept correspond to the product of resistance and capacitance and the product of capacitance and the battery emf, respectively. This technique simplifies the analysis of transient data.

RC Time Constant and Exponential Charging

The RC time constant, given by ? = RC, characterizes the rate at which the capacitor charges. The voltage across the capacitor as a function of time is described by the exponential equation V(t) = ?(1 - e^(-t/RC)). This concept is fundamental in determining the time required for the capacitor to reach a specified voltage during the charging process.

RC Charging Circuit

An RC charging circuit consists of a resistor and a capacitor connected to a voltage source, where the capacitor charges over time. The process is governed by an exponential relation as the capacitor accumulates charge, reflecting the dynamic balance between the charging current through the resistor and the increasing voltage across the capacitor.

Kirchhoff's Voltage Law (KVL)

KVL states that the sum of potential differences around any closed circuit loop is zero. In an RC charging circuit, this principle leads to the equation ? = iR + q/C, where ? is the battery emf, i is the resistor current, R is the resistor's resistance, q is the charge on the capacitor, and C is the capacitance. This law is critical for relating the electrical quantities in the circuit.

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