Answer to Question #159586 in Electric Circuits for wilder

Question

A 10.0 x 10^(-6) F capacitor is charged by a 10.0V battery through a resistance R. The capacitor reaches a potential difference of 4.00V in a time interval of 3.00 s after charging begins. Find R

Answer 1

Capacitance, $\text{C}= 10\times 10^{-6}\text{F}=10^{-5}\text{F}$

Voltage, $\text{V}=10 \text{volts}$

As the Capacitor reaches 4 volts in 3 sec

Then the charge stored in the capacitorin 3 sec is

$\text{Q=CV}\\Q=10^{-5}\times 4=4\times 10^{-5}\text{Columb}$

Total charge on the capacitor, $Q_o=CV=10^{-5}\times 10=10^{-4}\text{columb}$

Using the capacitor Charging equation,

$Q=Q_oe^{-\frac{t}{\tau}}$

$\Rightarrow 10^{-5}=10^{-4}e^{-\frac{3}{\tau}}$

$\Rightarrow 0.1=e^{-\frac{3}{\tau}}$

Taking log on Both sides and solving-

$\Rightarrow \dfrac{-3}{\tau}=log0.1$

$\Rightarrow \tau=\dfrac{-3}{-1}=3$

As we know Time constant, $\tau=RC$

$\Rightarrow R=\dfrac{\tau}{C}=\dfrac{3}{10^{-5}}=3\times 10^5 \Omega$