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CBSE Class 12-science Answered

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Asked by arup.isro | 25 Jun, 2019, 11:26: PM
answered-by-expert Expert Answer
Potential difference Vt at time t across capacitor of capacitance C discharging throuh resistance R , Vt = V0 e-t/RC  .........(1)
where V0 is initial voltage across capacitor
 
Potential difference after 1 sec,  V1 = V0 e-1/RC  or    40  = 50 e-1/RC .........................(2)
 
Potential difference after 2 sec,  V2 = V0 e-2/RC  or    V2  = 50 e-2/RC .........................(3)
 
using eqn.(3), we rewrite eqn.(3) as ,  V2 = 50×(40/50)×(40/50) = 32 V ...........................(4)
 
Energy stored in capacitor = (1/2) C V2   ...................(5)
 
as seen from eqn.(5), Energy stored in a given capacitor is directly proportional to Voltage across the capacitor plates.
 
Hence fraction of energy stored after 1 second  = (40×40) / (50×50) = 16/25
 
Hence only statements (A) and (B) are correct .
Answered by Thiyagarajan K | 26 Jun, 2019, 01:07: PM

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