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A 10-8-Fcapacitor (10 nano-farads) is charged to 50Vand then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is51013; on a humid day, the resistance is7106. How long will it take the capacitor voltage to dissipate to half its original value on each day?

Short Answer

Expert verified

The value of time on each day is 96.26h and on the humid days0.048517 sec .

Step by step solution

01

Important formula.

The equation isRdqdt+qC=0.

02

Evaluate the value of q

Here, C=108, q = 50V,r=7106

The equation is

Rdqdt+qC=0Rdqdt=-qCdqq=-dtCRlnq=-tCR+Kq=Ke-tCR

When t=0,q=50V,thenK=50

role="math" localid="1664228869574" q=50e-tC(R+r)(Where,ristheresistanceoftheairgap)

03

Find the value of time on each day

The equation for the capacitor voltage

VC=qCVC=-50e-tC(R+r)10-8VC=5109e-t10-8(R+51013)

Since the capacitor voltage to dissipate to half its original value each day.

So, the equation will be

VC=12VCo5109e-t10-8(R+51013)=1250e-0C(R+r)10-85109e-t10-8(R+51013)=12

When R = 0 then

e-t10-8(0+51013)=12lne-t5105=ln12t=0.69315105t=346560sect=96.26h

04

Calculate the value of time in humid days.

Here, r=7106, and R = 0 then

e-t10-8(0+7106)=12lne-t710-2=ln12t=0.6931710-2t=0.048517sec

Therefore, the value of time on each day is 96.26h and on the humid days 0.048517 sec .

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