Here we have given that,
V = 50 V
C = 0.75uF
R1 = 55 ohm
R2 = 100 ohm
Since we know that,
Energy stored in the capacitor is given as,
E = ½CV² = ½QV J
So that here, Ratio of energy dissipated by resistor compared to
total resistance will be
R1/(R1+R2)
So that the portion of energy dissipated by R1 is
[R1/(R1+R2)] * ½CV²
E(55 ohm)= [55/(55+100)] (0.5) (0.75×10^-6) (50 )²
= (3.3266129) × 10^-4 J
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