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Asked by Prashant DIGHE | 09 Mar, 2020, 22:00: PM
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Voltage across capacitor, V = E [ 1 - e-t/(RC) ]
 
Current in RC circuit , I = (E/R) e-t/(RC)
 
Rate of energy stored , i.e., power delivered to capacitor, P  = V × I = (E2/R) [ 1 - e-t/(RC) ] e-t/(RC)
 
P = (E2/R) [ e-t/(RC) - e-2t/(RC) ]............................(1)
 
maximum power is obtained by equating (dP/dt) to zero
 
begin mathsize 14px style fraction numerator d P over denominator d t end fraction space equals space open parentheses E squared over R close parentheses space open square brackets open parentheses negative fraction numerator 1 over denominator R C end fraction close parentheses space e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent space plus space open parentheses fraction numerator 2 over denominator R C end fraction close parentheses space e to the power of bevelled fraction numerator negative 2 t over denominator R C end fraction end exponent close square brackets end style
 
begin mathsize 14px style fraction numerator d P over denominator d t end fraction space equals space open parentheses fraction numerator E squared over denominator R squared C end fraction close parentheses space e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent space space open square brackets 2 space e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent space minus space 1 close square brackets end style
hence we get maximum power delivered to capacitor, at a time t that satisfies the condition [ 2 e-t/(RC) - 1 ] = 0  
 
From above condition we get maximum power at  t = ln(2)RC
 
Maximum power is obtained by substituting t = ln(2)RC in eqn.(1)
 
we get Pmax =  E2 / ( 4R )
Answered by Thiyagarajan K | 10 Mar, 2020, 09:33: AM

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