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Asked by rbhatt16 12th June 2018, 4:14 PM
Answered by Expert
Answer:
Charge q(t) stored at a time on capacitor of capacitance C after connencting to a battery of EMF ξ is given by
 
begin mathsize 12px style q left parenthesis t right parenthesis space equals space C space xi left parenthesis space 1 minus e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent space right parenthesis
end style...................(1)
Energy E(t) stored in capacitor as a function of time, E(t)  begin mathsize 12px style equals space 1 half q squared over C space equals space fraction numerator 1 over denominator 2 C end fraction cross times C squared xi squared open parentheses 1 minus e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent close parentheses squared space equals space fraction numerator C xi squared over denominator 2 end fraction open parentheses 1 minus e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent close parentheses squared end style....................(2)
Rate of Energy storage, R(t) = begin mathsize 12px style fraction numerator d E over denominator d t end fraction equals space C xi squared open parentheses 1 minus e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent close parentheses e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent end style .......................................(3)
time of maximum rate will be obtained by differentiating eqn.(3) , equating it to zero and getting the time
 
begin mathsize 12px style fraction numerator d R over denominator d t end fraction space equals space xi squared e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent 1 over R open curly brackets 2 e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent space minus space 1 space close curly brackets space equals space 0 space end style..........................(4)
from (4) we get maximum rate at time t, if begin mathsize 12px style open curly brackets 2 e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent space minus space 1 space close curly brackets space equals space 0 space end style or t = ln2 × RC
maximum rate will be obtained by substituting t = ln2 × RC in (3)
 
maximum rate = (1/4) Cξ 
Answered by Expert 13th June 2018, 12:41 PM
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