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×
= 5.10 J
Question 2 A 7.35 uF capacitor is charged to 1.23x103 V and is then connected in...
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