Question

Question: Switch S in Fig. 27-63 is closed at time t=0, to begin charging an initially uncharged capacitor of capacitance C= 150$\mu$F through a resistor of resistanceR =20.0$\Omega$. At what time is the potential across the capacitor equal to that across the resistor?

Short answer

Answer:

The potential across the capacitor is equal to that across the resistor at a time 0.21 ms

Step-by-step solution

The given data

The figure is closed at the time t = 0 .

The capacitance of the uncharged capacitor, $\text{C}=15.0\mu \text{F}\left(\frac{{10}^{-6}F}{1\mu \text{F}}\right)=1.50\times {10}^{-5}F$

The given resistance, $R=20.0\Omega$

Understanding the concept of time constant 

The RC time constant of an RC circuit is equal to the product of the circuit resistance (in ohms) and the circuit capacitance. Using the given relation of the voltage, we can get the potential equation across the capacitor and the resistor that will give the time value for this condition.

Formula:

The voltage equation for the resistor and capacitor connection, $V=V_0{e}^{-t/RC}$ (1)

Calculation of the time at which the potential across the capacitor and the resistor is the same

Now, using equation (1) and the given condition that the potential across the capacitor is equal to that across the resistor, the time at which the condition satisfies can be given as follows:

$$\begin{gathered} V_0{e}^{-t/RC}=V_0\left(1-e^{-t/RC}\right) \\ 2e^{-t/RC}=1 \\ t=RC\ln 2 \\ t=\left(20.0\right)\left(1.50\times {10}^{-5}\text{F}\right)\ln 2 \\ t=0.21\times {10}^{-3}\text{s}\left(\frac{1ms}{{10}^{-3}\text{s}}\right) \\ t=0.21\text{ms} \end{gathered}$$

Hence, the value of the time is 0.21 ms .