
The amount of energy needed in charging a condenser of capacitance $5\mu F$to 20 kV is-
A. 1 kJ
B. 10 kJ
C. 100 kJ
D. 5kJ
Answer
571.5k+ views
Hint: A capacitor (or condenser) is a circuit element which is used to store energy in the circuit. It stores energy in the form of an electric field. Capacitor is a passive component of the circuit. More such examples of passive components are resistor and inductor. It is used in places where we need a high supply of charge for a smaller interval of time, for example ceiling fans.
Formula used:
$U = \dfrac{1}{2}CV^2$
Complete answer:
When a capacitor is charged, it accumulates charge on its plates. This charge accumulation develops some potential difference in between the places and hence results in an electric field. Hence we can say that in a capacitor, energy is stored in the form of an electric field.
Given, $C = 5\mu F,\ V = 20kV$, thus using the formula $U = \dfrac{1}{2}CV^2$, putting the values in the equation, we get;
$U = \dfrac{1}{2}{(5\mu )}{(20 k)}^2 = \dfrac12 {5\times 10^{-3} (20 \times 10^3)^2}$
$\implies U = \dfrac52 (400 \times 10^3) = 10^5 J = 100 kJ$
So, the correct answer is “Option C”.
Additional Information:
Dielectric material is a medium having special property and having some dielectric constant value. Dielectric constant is a value which may be defined as the ratio of intensity of electric field in vacuum to that in the dielectric medium. In order to meet our required value of charge stored, we use dielectric material, which when inserted in a capacitor, its capacitance gets altered.
Note:
Capacitor is an important element of a circuit. It stores energy in the form of electrical energy. There is one more element in the circuit which is used to store energy but in the form of a magnetic field. This is known as inductor. These two circuit elements have various applications in electronic items.
Formula used:
$U = \dfrac{1}{2}CV^2$
Complete answer:
When a capacitor is charged, it accumulates charge on its plates. This charge accumulation develops some potential difference in between the places and hence results in an electric field. Hence we can say that in a capacitor, energy is stored in the form of an electric field.
Given, $C = 5\mu F,\ V = 20kV$, thus using the formula $U = \dfrac{1}{2}CV^2$, putting the values in the equation, we get;
$U = \dfrac{1}{2}{(5\mu )}{(20 k)}^2 = \dfrac12 {5\times 10^{-3} (20 \times 10^3)^2}$
$\implies U = \dfrac52 (400 \times 10^3) = 10^5 J = 100 kJ$
So, the correct answer is “Option C”.
Additional Information:
Dielectric material is a medium having special property and having some dielectric constant value. Dielectric constant is a value which may be defined as the ratio of intensity of electric field in vacuum to that in the dielectric medium. In order to meet our required value of charge stored, we use dielectric material, which when inserted in a capacitor, its capacitance gets altered.
Note:
Capacitor is an important element of a circuit. It stores energy in the form of electrical energy. There is one more element in the circuit which is used to store energy but in the form of a magnetic field. This is known as inductor. These two circuit elements have various applications in electronic items.
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