Answer fast
Asked by Varsneya Srinivas | 7th Jan, 2018, 09:54: PM
Expert Answer:
Charge as a function of time on the capacitor of capacitance C while charging is given by
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 
Solving for th we will get th = (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 
Hence 
To solve the above integration, let τ = (2t)/RC, then dt = (RC/2)dτ, limits t=0 -> τ=0 and t=th -> τ = ln4
Charge as a function of time on the capacitor of capacitance C while charging is given by
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
Solving for th we will get th = (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
Hence
To solve the above integration, let τ = (2t)/RC, then dt = (RC/2)dτ, limits t=0 -> τ=0 and t=th -> τ = ln4
Answered by Thiyagarajan K | 9th Jan, 2018, 02:13: PM
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