Question

A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant for this circuit

Answer 1

Answer:

The  time constant is  $\tau = 1.265 s$

Explanation:

From the question we are told that

     the time take to charge is  $t = 2.4 \ s$

The mathematically representation for voltage potential of a capacitor at different time is

        $V = V_o - e^{-\frac{t}{\tau} }$

Where  $\tau$  is the time constant  

           $V_o$ is the potential of the capacitor when it is full

     So  the capacitor potential will be  100%  when it is full thus  $V_o =$100%  =  1  

and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

      V  = 0.85

Hence

     $0.85 = 1 - e^{-\frac{2.4}{\tau } }$

       $- {\frac{2.4}{\tau } } = ln0.15$

        $\tau = 1.265 s$