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Question 2 A 7.35 uF capacitor is charged to 1.23x103 V and is then connected in series with a 46.0 32 resistor and a 4.18 12

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Answer #1

Rate of energy dissipation from a resistor is given by

P = i2 R

Since resistors are in series hence current in each is equal. Therefore energy will be dissipated in the ratio of resistances.

Energy stored in capacitor

E = (1/2) C V2

=(1/2)×(7.35×10-6) ×(1.23×103)2

= 5.56 J

Energy dissipated from 46 ohm resistor is

= 5.56× 16 46 +4.18

= 5.10 J

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