Question

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. How many time constants (a decimal number) must elapse before a capacitor in a series RC circuit is charged to 83.0% of its equilibrium charge?

Answer 1

Answer:

The number of time constants is $t = 1.171 time \ constants$

Explanation:

Let us denote the equilibrium charge with $q_e$

    From the question we are told that

       for the duration we are to obtain the final charge on the capacitor should be 83% of $q_e$

Generally in an RC circuit an equation that defines charge after a time t is  mathematically represented as

                  $q = q_e (1-e^{\frac{-t}{RC} })$

Where q is the final charge = 83% $q_e$  = 0.83$q_e$

           RC is the time constant

Substituting values

           $0.83 q_e = q_e (1 - e^{-\frac{t}{RC} })$

                $e^{-\frac{t}{RC} } = 0.17$

    Taking natural log of both sides

            $ln(e^{-\frac{t}{RC} }) = ln(0.17)$

                  $-\frac{t}{RC} = -1.171$

=>         $t = 1.171 RC$

Since RC is the time constant then the number of time constant is

         $t = 1.171 time constants$

                 

Answer 2

Answer:

The time constant is 1.772

Explanation:

Given that,

Charge,  $q{t}= 0.83 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$= time constant

Put the value into the formula

$0.83q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]1-e^{-(\dfrac{t}{RC})}=0.83$

$e^{-(\dfrac{t}{RC})}=0.17\\-\dfrac{t}{RC}=ln (0.17)\\-\dfrac{t}{RC}=-1.772\\\dfrac{t}{RC}=1.772$

Thus, The time constant is 1.772