JEE Advanced2020PhysicsMathematics in PhysicsActual
Two capacitors with capacitance values C 1 = 2000 ± 10 pF and C 2 = 3000 ± 15 pF are connected in series. The voltage applied across this combination is V = 5 . 00 ± 0 . 02 V . The percentage error in the calculation of the energy stored in this combination of capacitors is __________.
Correct answer
0
Step-by-step solution
For the purpose of calculation of error, fundamental formula is considered 1 C = 1 C 1 + 1 C 2 ⇒ C = 1200   p F - d C C 1 2 = - d C 1 C 1 2 - d C 2 C 2 2 d C = 6   p F Equivalent capacitance = 1200 ± 6   pF E = 1 / 2   CV 2 ( d E / E = d C / C + 2 d V / V ) × 100 = 1 . 3 %