. You charge a 6.0 F parallel-plate capacitor to a potential difference of 24 kV

Question

. You charge a 6.0 F parallel-plate capacitor to a potential difference of 24 kV across the plates. After the capacitor is fully charged, you connect this capacitor in parallel with a 10 M resistor.

a. Using Kirchhoff's loop rule, obtain (derive) the equation describing the behavior of the voltage across the capacitor as a function of time.

b. Show that the time constant RC has time units.

c. What is the initial current in the resistor when the capacitor discharge process begins?

d. How long will it take for the capacitor voltage to decrease to 0.37 of its original value?

e. Graph the current in the resistor as a function of time.

Explanation / Answer

a)From Kirchhoff's loop rule,

The currents flowing through each capacitor in the related to the voltage.

we know that, Q=CV

then we write it as, i=Q/t

it=CV

i=CV/t

b)

ohm=Volt/Ampere

       =Volt* second / Coulomb
Farad = Coulomb / Volt
substitute in above expression,

Ohm*Farad = Volt*second / Coulomb * Coulomb/Volt

                    = second

c)

V=I*R

I=V/R=(24 KV)/(10 M)

=2.4 mA

d)

time constant,t=RC

           t=10 M*6 F

            =60 s

0.37T=60 s*0.37

T=22.2 s

e)

The graph between current and time are inversly proportional to each other.

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