Question #154058

 Considering the LRC series with a charge of 𝛼 Henry, capacitance 0.000𝛽Farad, having the resistance of 50 times (𝛼 + 𝛽)ohms with emf of 100V. Suppose that no charge and current is present at time 𝑑 = 0 when the emf is applied. Determine the charge and current at any time 𝑑. Where 𝛼 and 𝛽 are the non zero digits of your registration number, 𝛼 being the tenth place digit and 𝛽 being the unit place digit. If 𝛼 or 𝛽 is zero then take any non-zero digit of your registration number. my reg no is 1 and 8


1
Expert's answer
2021-01-11T17:42:56-0500

q=∫0TcΟ΅(1βˆ’eβˆ’tRc)dt=cΟ΅t+Rceβˆ’tRc∣0T=cΟ΅T+Rceβˆ’TRc+Rcq=\int_0^T c\epsilon (1-e^{\frac{-t}{Rc}})dt=c\epsilon t+Rce^{\frac{-t}{Rc}}|_0^T=c\epsilon T+Rce^{\frac{-T}{Rc}}+Rc

c=0.0008 Farad

Ο΅=100V\epsilon =100V

R= 8.3145 J mol-1K-1

q=0.0008(100T+8.3145eβˆ’T8.3145x0.0008+8.3145)q=0.0008(100T+8.3145e^{\frac{-T}{8.3145 x 0.0008}} + 8.3145)

Ξ±=1Henry\alpha= 1 Henry 

r=50(1+8)=450 ohms

I=∫0Teβˆ’rtΞ±dt=βˆ’Ξ±reβˆ’rtα∣0T=βˆ’1450eβˆ’450Tβˆ’1450=βˆ’1450(eβˆ’450T+1)I=\int_0^Te^\frac{-rt }{\alpha}dt=\frac{-\alpha}{r}e^\frac{-rt }{\alpha}|_0^T=\frac{-1}{450}e^{-450T }-\frac{1}{450}=\frac{-1}{450}(e^{-450T }+1)



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