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?

Explanation:

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

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

    = 0.291

or,       = ln (0.291)

      = -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 =

                  =

                  = 2.259 sec

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

electrical to thermal

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