Question

A 0.754F capacitor is charged to 50 V . It is then connected in series with a 558 resistor and a 100 resistor and allowed to discharge completely. Part A How much energy is dissipated by the 5522 resistor? Express your answer with the appropriate units. HA Value Units Submit Previous Answers Request Answer * Incorrect; Try Again; 2 attempts remaining < Return to Assignment Provide Feedback

Answer 1

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