CBSE Class 12-science Answered
A 28microfard capacitor is charged to 100V and another 2microfard capacitor to 200V then they are connected in parallel determine the total initial and final energies account for the difference in two values
Asked by Devanshuagrawal36 | 25 Jun, 2019, 01:12: PM
Expert Answer
Charge on 28 μf capacitor = C×V = 28×10-6 × 100 = 2800 μC
Energy on 28 μf capacitor = (1/2)C V 2 = (1/2) ×28×10-6×100×100 = 0.14 J
Charge on 2 μf capacitor = C×V = 2×10-6 × 200 = 400 μC
Energy on 2 μf capacitor = (1/2)C V 2 = (1/2) ×2×10-6×200×200 = 0.08 J
Total energy on both capacitor = 0.22 J
when they are connected parallel , equivalent capacitance = 28+2 = 30 μf
total charge = 2800+400 = 3200 μC
potential difference across capacitors of parallel combination = 3200 / 30 ≈ 106.7 V
Energy on capacitors connected parallel = (1/2)×30×10-6 × 106.7×106.7 = 0.171 J
Energy loss while sharing charges between capacitors = 0.22-0.171 ≈ 0.05 J
This energy diference is due to loss of energy while sharing the charges between capacitors,
because it will generate heat and also spaking takes place
Answered by Thiyagarajan K | 25 Jun, 2019, 02:59: PM
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