Question

A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.90 MΩ. After a time of 3.50 s the voltmeter reads 3.5 V. After a time of 4.00 s the voltmeter reads 3.0 V. What are (a) the capacitance and (b) the time constant of the circuit?

Answer 1

Explanation:

(a)  Formula to calculate the capacitance is as follows.

           $V_{c} = V_{o} (e^(\frac{-t}{RC}))$

Now, putting the given values into the above formula as follows.

          $V_{c} = V_{o} (e^(\frac{-t}{RC}))$

          $3.5 V = 12 V (e^(\frac{-3.50 sec}{3.90 \times C}))$

    $e^(\frac{-3.50 sec}{3.90 \times C})$ = 0.291

or,       $\frac{-3.50 sec}{3.90 \times C}$ = ln (0.291)

      $\frac{-3.50 sec}{3.90 \times C}$ = -1.23

              C = 0.729 F

Hence, the value of capacitance is 0.729 F.

(b)   Formula to calculate the constant of circuit is as follows.

               T = $R \times C$

                  = $3.1 \times 0.729$

                  = 2.259 sec

Therefore, the time constant of the circuit is 2.259 sec.