CBSE Class 12-science Answered

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Asked by Varsneya Srinivas | 07 Jan, 2018, 09:54: PM

Expert Answer

Charge as a function of time on the capacitor of capacitance C while charging is given by

$begin mathsize 12px style q left parenthesis t right parenthesis space equals space xi C left parenthesis 1 minus e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent right parenthesis end style$

ξ is emf of the battery. After long time capacitor is charged its full charge, Q = ξC.

Let time taken for charging half of Q is th, then $begin mathsize 12px style Q over 2 space equals space Q open parentheses 1 minus e to the power of bevelled fraction numerator negative t subscript h space end subscript over denominator R C end fraction end exponent close parentheses end style$

Solving for t_h we will get t_h = (ln2)RC

Energy dissipated dE in resistor while charging for a time dt is given by dE = i²Rdt

Where i is the current as a function of time is given by $begin mathsize 12px style i left parenthesis t right parenthesis space equals space xi over R e to the power of bevelled fraction numerator negative t over denominator R C end fraction end exponent end style$

Hence $begin mathsize 12px style E space equals space integral d E space equals space xi squared over R integral subscript 0 superscript t subscript h end superscript space e to the power of bevelled fraction numerator negative 2 t over denominator R C end fraction end exponent d t end style$

To solve the above integration, let τ = (2t)/RC, then dt = (RC/2)dτ, limits t=0 -> τ=0 and t=t_h -> τ = ln4

$begin mathsize 12px style E space equals space fraction numerator xi squared C over denominator 2 end fraction integral subscript 0 superscript ln 4 end superscript e to the power of negative tau end exponent d tau space equals space fraction numerator 3 xi squared C over denominator 8 end fraction end style$