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Two capacitors with capacitances of 1.0 m F and 0.50 m F, respectively, are connected in series. The system is connected to a 100-V battery. What electrical potential energy is stored in the 1.0- m F capacitor?

Answer :

Answer:

Therefore energy is stored in the 1.0 mF capacitor is 5.56×10⁻⁹ J

Explanation:

Series capacitor: The ending point of a capacitor is the starting point of other capacitor.

If C₁ and C₂ are connected in series then the equivalent  capacitance is C.

where     [tex]\frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_2}[/tex]

Given that,

C₁ = 1.0 mF=1.0×10⁻³F  and  C₂ = 0.50mF=0.50×10⁻³F  

If C is equivalent capacitance.

Then    [tex]\frac{1}{C} =\frac{1}{1.0}+\frac{1}{0.5}[/tex]

[tex]\Rightarrow \frac{1}{C} =\frac{3}{1}[/tex]

[tex]\Rightarrow C=\frac{1}{3}[/tex] mF

Again given that the system is connected to a 100-v battery.

We know that

q=Cv

q= charge

C= capacitor

v= potential difference

Therefore

[tex]q=(\frac{1}{3} \times 10^{-3}\times10) C[/tex]

 [tex]=\frac{10^{-2}}{3} C[/tex]

The electrical potential energy stored in a capacitor can be expressed

[tex]U=\frac{q^2C}{2}[/tex]

q= charge

c=capacitance of a capacitor

Therefore energy is stored in the 1.0 mF capacitor is

[tex]U=\frac{q^2C_1}{2}[/tex]

[tex]\Rightarrow U=\frac{(\frac{10^{-2}}{3})^2\times 10^{-3} }{2}[/tex]

[tex]\Rightarrow U= 5.56\times 10^{-9}[/tex] J

 

The energy stored in the 1.0mF capacitor is 9.0*10^-5J

Data given;

  • c[tex]_1[/tex] = 1.0mF
  • c[tex]_2[/tex] = 0.50mF
  • V = 100V

Capacitor

Since the capacitor are in series, the total capacitance is;

[tex]C = \frac{1}{c_1} + \frac{1}{c_2}[/tex]

substitute in the values and solve.

[tex]C = \frac{1}{1} + \frac{1}{0.5}\\C = 3mF[/tex]

The capacitance in the capacitor is 3mF.

Potential Energy

This is the energy stored in the capacitor.

[tex]U= \frac{1}{2} q^2C[/tex]

U = potential energy

q = charge

C = capacitor.

But the charge of a capacitor is given as

[tex]q = cV\\q = (3*10^-^3)*100\\q = 0.3C[/tex]

The charge on the capacitor is 0.3C

We can proceed to use this formula and solve for the potential energy stored in the 1.0mF capacitor.

[tex]U = \frac{1}{2} q^2C\\U = \frac{1}{2}* 0.3^2*1.0^-^3\\U = 9.0*10^-^5J[/tex]

From the calculations above, the potential energy stored in the 1mF capacitor is [tex]9.0*10^-^5J[/tex].

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