Advertisements
Advertisements
प्रश्न
Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge on each capacitor.
उत्तर
 |
 |
 |
The equivalent capacitance = `200/3` pF
charge on C4 = `200/3 xx 10^-12 xx 300 = 2 xx 10^-8` C,
the potential difference across C4 = `(200 xx 10^-12 xx 300)/(3 xx 100 xx 10^-12)` = 200 V
potential difference across C1 = 300 - 200 = 100 V
charge on C1 = 100 × 10-12 × 100 = 1 × 10-8 C
potential difference across C2 and C3 series combination = 100 V
potential difference across C2 and C3 each = 50 V
charge on C2 and C3 each = 200 × 10-12 × 50 = 1 × 10-8 C
APPEARS IN
संबंधित प्रश्न
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge of the capacitor to become 12.6 μC. Find the resistance of the circuit.
A sphercial capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K up to a radius c as shown in figure . Calculate the capacitance.
An air-filled parallel-plate capacitor is to be constructed which can store 12 µC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is `3 xx 10^6 "Vm"^-1`
The variation of inductive reactance (XL) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.
(i) Calculate the self-inductance of the inductor.
(ii) When this inductor is used in series with a capacitor of unknown value and resistor of 10 Ω at 300 s–1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 106 Vm–1), lightning will occur.
- If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
- In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
The radius of a sphere of capacity 1 microfarad in the air is ______
The displacement current of 4.425 µA is developed in the space between the plates of the parallel plate capacitor when voltage is changing at a rate of 106 Vs-1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x × 10-3 m. The value of x is ______.
(Permittivity of free space, ε0 = 8.85 × 10-12C2N-1m-2).
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor.
The thickness of the dielectric slab is `3/4`d, where 'd' is the separation between the plate of the parallel plate capacitor.
The new capacitance (C') in terms of the original capacitance (C0) is given by the following relation:
