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 _______.
Answers
It has given that, capacitance of capacitors C₁ and C₂ are (2000 ± 10) pF and (3000 ± 15) pF connected in series. The voltage applied across this combination is V = (5 ± 0.02) V
To find : The percentage error in the calculation of the energy stored in this combination of capacitors is....
solution : first find equivalent capacitance,
1/Ceq = 1/C₁ + 1/C₂
differentiating both sides we get,
dCeq/Ceq² = dC₁/C₁² + dC₂/C₂²
Ceq = (2000 × 3000)/(2000 + 3000) = 1200 pF
so, dCeq/(1200)² = (10)/(2000)² + (15)/(3000)²
⇒dCeq = (1200)² [1/4 × 10^5 + 15/90 × 10^5]
= 144 [1/40 + 1/60]
= 144 [ 100/2400]
= 144/24
= 6
therefore Ceq = (1200 ± 6) pF
now E = 1/2 CV²
dE/E = dC/C + 2dV/V
⇒percentage error in E = percentage error in C + 2 × percentage error in V
= 6/1200 × 100 + 2 × 0.02/5 × 100
= 0.5 + 0.8
= 1.3 %
Therefore percentage error in energy stored in this combination of capacitors is 1.3 %