Question

A capacitor with an initial potential difference of 100 V is discharged through a resistor when a switch between them is closed at t 0.At t 10.0 s, the potential difference across the capacitor is 1.00 V. (a) What is the time constant of the circuit

Answer 1

Answer:

τ = RC = 2.17 s

Explanation:

• The voltage through a capacitor can't change instantaneously, so immediately after the switch is closed, the potential difference will keep at 100 V.
• This voltage will produce a flow of charge (a current) from the capacitor to the resistor, which will be diminishing continuously, till the capacitor be totally discharged, and the current becomes zero.
• The voltage through the capacitor will follow an exponential function of time, as follows:

       $V_{C} =V_{o} * e^{-t/RC} (1)$

• Replacing by the givens in (1):

       $V_{C} = 1.00 V\\V_{o} = 100V\\t = 10.0 s$

      $\frac{1.00V}{100 V} = e^{-10s/RC} (2)$

• Taking ln on both sides in (2), and solving for RC, we have:

       $R*C= \frac{-10s}{ln 0.01} = 2.17s (3)$

• So, the time constant of the circuit (the product of R times C) is equal to 2.17s.