Two insulated metallic spheres of 3 μF and 5 μF capacitances are charged to 300 V and 500V respectively. The energy loss, when they are connected by a wire is?
Answers
Answer:
Hii
Explanation:
Capacitance of sphere one = C1 = 3 = 3 ×10-6F
Capacitance of sphere two = C2 = 5 = 5 ×10-6F
Voltage one = V1 = 300 volt
Voltage two = V2 = 500 volt
Initial Electrostatic energy of the capacitor -
Ui = 1/2 C1V1² + 1/2C2V2²
Ui = 1/2 × 3 ×10-6 × (300)2 + 1/2×5 ×10-6 × (500)2
= 76×10-2J
Total capacitance -
C = C1 + C2 = 3 ×10-6 + 5 ×10-6
= 8×10-6 F
Charge in the capacitors
Q = C1V1+C2V2
Q = 3 ×10-6×300 + 5 ×10-6×500
= 34×10-4 C
Let V be the common potential of the capacitors after connection
V = Q/C
V = 34 × 10-4/ 8 × 10-6
= 425 volt
Final electrostatic energy -
Uf = 1/2CV² = 1/2 × 8 × 10-6× (425)²
= 7225×10-4 J
Loss in energy in form of heat -
Ui - Uf = 76×10-2-7225 ×10-4 J
=7600×10-4-7225 ×10-4 J
= 375 ×10-4J
= .0375
I hope it helps you
Answer:
Explanation:
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