Two capacitors with capacitance values C1=(2000±10) pF and C2=(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 (rounded off to nearest integer)________

Equivalent capacity in case of series connection of Capacitors, 1Ceq=1C1+1C2 ⇒Ceq=C1C2C1+C2Ceq=C1C2C1+C2=2000×30002000+3000=1200μF Also, △CeqCeq2=△C1C12+△C2C22 ⇒△CeqCeq=(△C1C12+△C2C22).Ceq ⇒△CeqCeq={10(2000)2+15(3000)2}×1200=5×10−3Energy stored in this combination of capacitorsU=12CeqV2⇒△UU=△CeqCeq+2△VV ⇒△UU=5×10−3+2(0.02)5=13×10−3Hence % error = 1.3

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Two capacitors with capacitance values $C_{1} = (2000 \pm 10) p F$ and $C_{2} = (3000 \pm 15) p F$ are connected in series. The voltage applied across this combination is $V = (5.00 \pm 0.02) V$ . The percentage error in the calculation of the energy stored in this combination of capacitors is (rounded off to nearest integer)________

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Detailed Solution

Equivalent capacity in case of series connection of
Capacitors, $\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} \Rightarrow C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$
$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{2000 \times 3000}{2000 + 3000} = 1200 μ F A l s o, \frac{△ C_{e q}}{C_{e q}^{2}} = \frac{△ C_{1}}{C_{1}^{2}} + \frac{△ C_{2}}{C_{2}^{2}} \Rightarrow \frac{△ C_{e q}}{C_{e q}} = (\frac{△ C_{1}}{C_{1}^{2}} + \frac{△ C_{2}}{C_{2}^{2}}) . C_{e q} \Rightarrow \frac{△ C_{e q}}{C_{e q}} = {\frac{10}{{(2000)}^{2}} + \frac{15}{{(3000)}^{2}}} \times 1200 = 5 \times 10^{- 3}$

Energy stored in this combination of capacitors

$U = \frac{1}{2} C_{e q} V^{2} \Rightarrow \frac{△ U}{U} = \frac{△ C_{e q}}{C_{e q}} + \frac{2 △ V}{V} \Rightarrow \frac{△ U}{U} = 5 \times 10^{- 3} + \frac{2 (0.02)}{5} = 13 \times 10^{- 3}$