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

`13.83928* 10^(-6)\ F`

`249.10704* 10^(-6)\ J`

`137.89848* 10^(-6)\ J`

`111.20842* 10^(-6)\ J`

2.98273 V

Step-by-step explanation:

`C_1=25\ mu F`

`C_2=31\ mu F`

V = Voltage = 6 V

Equivalent capacitance is given by

`(1)/(C)=(1)/(C_1)+(1)/(C_2)\\\Rightarrow C=(C_1C_2)/(C_1+C_2)\\\Rightarrow C=(25* 10^(-6)* 31* 10^(-6))/((25+31)* 10^(-6))\\\Rightarrow C=13.83928* 10^(-6)\ F`

Equivalent capacitance is
`13.83928* 10^(-6)\ F`

Energy stored is given by

`E=(1)/(2)CV^2\\\Rightarrow E=(1)/(2)* 13.83928* 10^(-6)* 6^2\\\Rightarrow E=249.10704* 10^(-6)\ J`

Total energy stored is
`249.10704* 10^(-6)\ J`

Charge is given by

`Q=CV\\\Rightarrow Q=13.83928* 10^(-6)* 6\\\Rightarrow Q=83.03568* 10^(-6)\ C`

Voltage is given by

`V_1=(Q)/(C_1)\\\Rightarrow V_1=(83.03568* 10^(-6))/(25* 10^(-6))\\\Rightarrow V_1=3.3214272\ V`

`E_1=(1)/(2)C_1V_1^2\\\Rightarrow E_1=(1)/(2)* 25* 10^(-6)* 3.3214272^2\\\Rightarrow E_1=137.89848* 10^(-6)\ J`

Energy strored in C1 is
`137.89848* 10^(-6)\ J`

`V_2=(Q)/(C_2)\\\Rightarrow V_2=(83.03568* 10^(-6))/(31* 10^(-6))\\\Rightarrow V_2=2.67857\ V`

`E_2=(1)/(2)C_2V_2^2\\\Rightarrow E_2=(1)/(2)* 31* 10^(-6)*2.67857^2\\\Rightarrow E_2=111.20842* 10^(-6)\ J`

Energy stored in C2 is
`111.20842* 10^(-6)\ J`

`E=E_1+E_2\\\Rightarrow E=137.89848* 10^(-6)+111.20842* 10^(-6)\\\Rightarrow E=249.107* 10^(-6)\ J`

So, the energy is equivalent

Equivalent capacitance

`C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56* 10^(-6)\ F`

`E=(1)/(2)CV^2\\\Rightarrow V=\sqrt{(2E)/(C)}\\\Rightarrow V=\sqrt{(2* 249.107* 10^(-6))/(56* 10^(-6))}\\\Rightarrow V=2.98273\ V`

The voltage would be 2.98273 V