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…

Question

Two capacitors with capacitance values C 1 = 2000 &#177; 10 &#160; pF and C 2 = 3000 &#177; 15 &#160; pF are connected in series. The voltage applied across this combination is V = 5 . 00 &#177; 0 . 02 &#160; V . The percentage error in the calculation of the energy stored in this combination of capacitors is __________.

Aicharya · Accepted Answer

For the purpose of calculation of error, fundamental formula is considered

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} \Rightarrow C = 1200 p F$

$- \frac{d C}{C_{1}^{2}} = - \frac{d C_{1}}{C_{1}^{2}} - \frac{d C_{2}}{C_{2}^{2}}$

$d C = 6 p F$

Equivalent capacitance $= (1200 \pm 6) pF$

$E = 1 / 2 {CV}^{2}$

$(d E / E = d C / C + 2 d V / V) \times 100$

$= 1.3 %$

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Two capacitors with capacitance values C 1 = 2000 ± 10   pF and C 2 = 3000 ± 15   pF are…

Solution

Two capacitors with capacitance values C 1 = 2000 &#177; 10 &#160; pF and C 2 = 3000 &#177; 15 &#160; pF are…

Solution

Two capacitors with capacitance values C 1 = 2000 ± 10 pF and C 2 = 3000 ± 15 pF are…