Capacitors C₁=2000±10 pF and C₂=3000±15 pF are in series across V=5.00±0.02 V. Find the percentage error in the calculated stored energy.
Physics · Unit dimension and error
JEE Advanced 2020

Question

Capacitors C₁=2000±10 pF and C₂=3000±15 pF are in series across V=5.00±0.02 V. Find the percentage error in the calculated stored energy.

G1 Physics · Accepted Answer

Series capacitance is C=C₁C₂/(C₁+C₂)=1200 pF. Differential propagation gives ΔC/C≈0.5%. Stored energy U=(1/2)CV², so ΔU/U=ΔC/C+2ΔV/V≈0.5%+2(0.02/5×100%)=1.30%.