Question #154058

 Considering the LRC series with a charge of š›¼ Henry, capacitance 0.000š›½Farad, having the resistance of 50 times (š›¼ + š›½)ohms with emf of 100V. Suppose that no charge and current is present at time š‘” = 0 when the emf is applied. Determine the charge and current at any time š‘”. Where š›¼ and š›½ are the non zero digits of your registration number, š›¼ being the tenth place digit and š›½ being the unit place digit. If š›¼ or š›½ is zero then take any non-zero digit of your registration number. my reg no is 1 and 8


Expert's answer

q=∫0Tcϵ(1āˆ’eāˆ’tRc)dt=cϵt+Rceāˆ’tRc∣0T=cϵT+Rceāˆ’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

ϵ=100V\epsilon =100V

R= 8.3145 J mol-1K-1

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

α=1Henry\alpha= 1 Henry 

r=50(1+8)=450 ohms

I=∫0Teāˆ’rtαdt=āˆ’Ī±reāˆ’rtα∣0T=āˆ’1450eāˆ’450Tāˆ’1450=āˆ’1450(eāˆ’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)



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS