Respuesta :
Answer:
The energy stored is [tex]E_s = 0.0064 \ J[/tex]
Explanation:
From the question we are told that
The capacitance is [tex]C = 8 \ \mu F = 8*10^{-6} \ F[/tex]
The resistance is R = 3.00-Ω
The emf is [tex]E_t = 70.0 V[/tex]
The power is P = 300 W
Generally the total emf is mathematically represented as
[tex]E_t = E_c + E_r[/tex]
Here [tex]E_c[/tex] is the emf across that capacitor which is mathematically represented as
[tex]E_c = \frac{q}{C}[/tex]
and [tex]E_r[/tex] is the emf across the resistor which is mathematically represented as
[tex]E_r = \sqrt{P R}[/tex]
So
[tex]E_t = \sqrt{PR} + \frac{q}{C}[/tex]
=> [tex]q = C[E_t - \sqrt{PR} ][/tex]
Generally the energy stored in a capacitor is mathematically represented as
[tex]E_s = \frac{q^2}{2C}[/tex]
=> [tex]E_s = \frac{[C [ E_t - \sqrt{PR} ]]^2}{2C}[/tex]
=> [tex]E_s = \frac{[8.0*10^{-6} [ 70 - \sqrt{300 * 3}}{2 *(8.0*10^{-6})}[/tex]
=> [tex]E_s = 0.0064 \ J[/tex]
The energy that has been stored in the capacitor will be 0.0064 J. The stored energy is utilized in the different electrical works.
What is a capacitor?
A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.
An electric insulator or vacuum, such as glass, paper, air, or a semi-conductor termed a dielectric, can be used as the non-conductive zone.
The given data in the problem is;
C is the capacitance = 8.00-μF = 8.00 ×10 ⁻⁶
R is the resistance = 3.00-Ω
E is the emf = 70.0 V
P is the power = 300 W
The total emf is the sum of the emf due to the capacitor and emf due to the resistor;
[tex]\rm E_t = E_C + E_R[/tex]
The emf generated by the capacitor is;
[tex]\rm E_C = \frac{q}{C}[/tex]
[tex]\RM E_R = \sqrt{PR}[/tex]
[tex]\rm E_t = \frac{q}{C} + \sqrt{PR} \\\\ \rm q= C [ E_t-\sqrt{PR} ]\\\\[/tex]
Generally, the energy stored in the capacitor is;
[tex]\rm E_S= \frac{q^2}{2C} \\\\ \rm E_S= \frac{\rm C [ E_t-\sqrt{PR} ]\\\\^2}{2C} \\\\ \rm E_S= \frac{\rm 8.0 \times 10^{-6}[ 70-\sqrt{300\times 3} ]\\\\^2}{2 \times (8.0 \times 10^{-6}} \\\\ \rm E_S= 0.0064 \ J[/tex]
Hence the energy that has been stored in the capacitor will be 0.0064 J.
To learn more about the capacitor refer to the link;
https://brainly.com/question/14048432
