Question

Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the equivalent capacitance, and the energy contained in this equivalent capacitor.b. Find the energy stored in each individual capacitor.Show that the sum of these two energies is the same as the energy found in part (a). Will this equality always be true, or does it depend on the number of capacitors and their capacitances?c. If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part (a)? Which capacitor stores more energy in this situation?

Answer 1

Answer:

$13.83928\times 10^{-6}\ F$

$249.10704\times 10^{-6}\ J$

$137.89848\times 10^{-6}\ J$

$111.20842\times 10^{-6}\ J$

2.98273 V

Explanation:

$C_1=25\ mu F$

$C_2=31\ mu F$

V = Voltage = 6 V

Equivalent capacitance is given by

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F$

Equivalent capacitance is $13.83928\times 10^{-6}\ F$

Energy stored is given by

$E=\dfrac{1}{2}CV^2\\\Rightarrow E=\dfrac{1}{2}\times 13.83928\times 10^{-6}\times 6^2\\\Rightarrow E=249.10704\times 10^{-6}\ J$

Total energy stored is $249.10704\times 10^{-6}\ J$

Charge is given by

$Q=CV\\\Rightarrow Q=13.83928\times 10^{-6}\times 6\\\Rightarrow Q=83.03568\times 10^{-6}\ C$

Voltage is given by

$V_1=\dfrac{Q}{C_1}\\\Rightarrow V_1=\dfrac{83.03568\times 10^{-6}}{25\times 10^{-6}}\\\Rightarrow V_1=3.3214272\ V$

$E_1=\dfrac{1}{2}C_1V_1^2\\\Rightarrow E_1=\dfrac{1}{2}\times 25\times 10^{-6}\times 3.3214272^2\\\Rightarrow E_1=137.89848\times 10^{-6}\ J$

Energy strored in C1 is $137.89848\times 10^{-6}\ J$

$V_2=\dfrac{Q}{C_2}\\\Rightarrow V_2=\dfrac{83.03568\times 10^{-6}}{31\times 10^{-6}}\\\Rightarrow V_2=2.67857\ V$

$E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J$

Energy stored in C2 is $111.20842\times 10^{-6}\ J$

$E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J$

So, the energy is equivalent

Equivalent capacitance

$C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F$

$E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V$

The voltage would be 2.98273 V