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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