Respuesta :
Answer:
[tex]1140.48\times 10^{-6}J[/tex]
Explanation:
We have given that the battery has an internal emf of 9 volt
So E = 9 volt
Capacitance [tex]C=44\mu F=44\times 10^{-6}F[/tex]
It is given that on the terminal voltage is only 80% of potential difference
So V = 0.8×9 = 7.2 volt
We know that energy stored in the capacitor is given by
[tex]E=\frac{1}{2}CV^2=\frac{1}{2}\times 44\times 10^{-6}\times 7.2^2=1140.48\times 10^{-6}J[/tex]
To answer the question we must know about the concept of capacitance.
What is capacitance?
The capacitance is the effect of a capacitor, while a capacitor is a device that stores the electrical energy into it. The energy stored in a capacitor can be calculated by the formula,
[tex]U = \dfrac{1 \times Q^2}{2 \times C} = \dfrac{1 \times Q}{2 \times V} = \dfrac{1}{2}CV^2[/tex]
The energy stored in a 44.0 μF capacitance when connected to a battery of 9 V is1.14x10⁻³ J.
Give to us
Internal EMF of battery = 9.00 V
Potential difference accross its terminal = 80%
Capacitance = 44.0 μF = 44 x 10⁻⁶ J
The voltage that will be given to the capacitor,
We know that the internal emf of 9.00 V but the potential difference across its terminals is only 80.0 % of that value. therefore,
V₁ = 9.00 x 80% = 7.2 V
The energy stored in the capacitor
We know the formula for the energy stored in a capacitor, therefore,
[tex]U = \dfrac{1}{2}CV_1^2\\\\U = \dfrac{1}{2} \times 44\times 10^{-6} \times (7.2)^2\\\\U = 1.14 \times 10^{-3} \rm\ J[/tex]
hence, the energy stored in a 44.0 μF capacitance when connected to a battery of 9 V is 1.14x10⁻³ J.
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