At instant t = 0 second, 4A current flows in inductor coil and this coil is parallelly connected with 6 Ω & 3 Ω resistance. Find total heat developed in 6Ω resistor during discharging ?
A
16 J
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B
32 J
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C
48 J
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D
8 J
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Solution
The correct option is A 16 J U=12Li2 & Heat(H)=V2R(dt) U=12Li2=12×6×4×4=48Joule Time for both the resistance will be same therefor neglecting it in ratio. H1H2=V2/R1V2/R2=R2R1=36=12 ..(i) H1+H2=48 from equation (i), H1+2H1=48 H1=16J