Question

Two capacitors with capacitance values \[{C_1} = 2000 \pm 10{\rm{ pF }}\] and \[{C_2} = 3000 \pm 15{\rm{ pF }}\] are connected in series. The voltage applied across this combination is \[V = 5.00 \pm 0.02{\rm{ V}}\]. The percentage error in the calculation of the energy stored in this combination of capacitors is _________.

Answer 1

Hint:In two capacitors which are connected in series, then the resultant capacitance is the sum of their inverse capacitances. If two capacitors are connected in parallel, then the resultant capacitance is the sum of their individual capacitances. Here also we use the formula for energy stored in a capacitor or capacitors.

Formula used:
Capacitor in series combination,
\[\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}......\]
Where C is an equivalent capacitor and \[{C_{1,}}{C_2}....\] are the capacitors connected in series combination.
Energy stored in this combination of capacitors is,
\[E = \dfrac{1}{2}C{V^2}\]
Where C is the capacitance and V is the potential difference.

Complete step by step solution:
Two capacitance values with error limits connected in series \[{C_1} = 2000 \pm 10{\rm{ pF }}\] and \[{C_2} = 3000 \pm 15{\rm{ pF }}\]
Voltage applied with error limits, \[V = 5.00 \pm 0.02{\rm{ V}}\]
As we know in capacitor in series combination,
\[\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\]
Substituting the given values
\[\begin{array}{l}\dfrac{1}{C} = \dfrac{1}{{2000}} + \dfrac{1}{{3000}}\\{\rm{ = }}\dfrac{{2000 + 3000}}{{2000 \times 3000}}\\\therefore {\rm{ C = 1200 pF}}\end{array}\]
Hence, we get capacitance, C = 1200 pF

Now given errors in capacitance as,
\[d{C_1} = 10{\rm{ pF}}\] and \[d{C_2} = 15{\rm{ pF}}\]
As we know,
\[\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\]
On integrating it, we get
\[ - \dfrac{{dC}}{{{C^2}}} = - \dfrac{1}{{{C_1}^2}} \times d{C_1} - \dfrac{1}{{{C_2}^2}} \times d{C_2}\]
\[\dfrac{{dC}}{{{C^2}}} = \dfrac{{d{C_1}}}{{{C_1}^2}} + \dfrac{{d{C_2}}}{{{C_2}^2}}\]
Substituting the given values
\[\dfrac{{dC}}{{{{1200}^2}}} = \dfrac{{10}}{{{{2000}^2}}} + \dfrac{{15}}{{{{3000}^2}}}\]
\[\Rightarrow dC = 5{\rm{ pF}}\]
Hence, error in net equivalent capacitance is, \[dC = 5{\rm{ pF}}\]

Now we have net capacitance, \[C = 1200 \pm 6pF\].
As we know Energy stored in this combination of capacitors is,
\[E = \dfrac{1}{2}C{V^2}\]
Error in energy can be as,
\[\dfrac{{dE}}{E} = \dfrac{{dC}}{C} + 2\dfrac{{dV}}{V}\]
Substituting the given values
\[\dfrac{{dE}}{E} = \dfrac{6}{{1200}} + 2\dfrac{{0.02}}{5}\]

To find the percentage error of the energy stored in this combination of capacitors is,
\[\dfrac{{dE}}{E} \times 100 = \left[ {\dfrac{6}{{1200}} \times 100 + 2\dfrac{{0.02}}{5} \times 100} \right]\% \]
\[\Rightarrow \dfrac{{dE}}{E} \times 100 = \left[ {0.5 + 0.8} \right]\% \]
\[\therefore \dfrac{{dE}}{E} \times 100 = 1.3\% \]

Therefore, the percentage error in the calculation of the energy stored in this combination of capacitors is \[1.3\% \].

Note:The capacitor is a device which stores electrical energy in the electrical field. A capacitor consists of two plates separated by a distance of equal and opposite charges. The area between the conductors may be filled by a vacuum or insulating material known as a dielectric.