Advertisements
Advertisements
प्रश्न
Obtain the equivalent capacitance of the network in Figure. For a 300 V supply, determine the charge and voltage across each capacitor.
Advertisements
उत्तर
Capacitance of capacitor C1 is 100 pF.
Capacitance of capacitor C2 is 200 pF.
Capacitance of capacitor C3 is 200 pF.
Capacitance of capacitor C4 is 100 pF.
Supply potential, V = 300 V
Capacitors C2 and C3 are connected in series.
Let their equivalent capacitance be C'
∴ `1/"C'" = 1/200 + 1/200 = 2/200`
∴ C' = 100 pF
Capacitors C1 and C' are in parallel. Let their equivalent capacitance be C''.
∴ `"C''" = "C'" + "C"_1`
= 100 + 100
= 200 pF
C'' and C4 are connected in series. Let their equivalent capacitance be C.
∴ `1/C = 1/("C''") + 1/("C"_4)`
= `1/200 + 1/100`
= `(2 + 1)/200`
C = `200/3` pF
Hence, the equivalent capacitance of the circuit is `200/3` pF.
Potential difference across C" = V"
Potential difference across C4 = V4
∴ `"V''" + "V"_4 = "V" = 300 "V"`
Charge on C4 is given by
Q4 = CV
= `200/3 xx 10^-12 xx 300`
= `2 xx 10^-8 "C"`
∴ `"V"_4 = "Q"_4/"C"_4`
= `(2 xx 10^-8)/(100 xx 10^-12)` = 200 V
∴ Voltage across C1 is given by
`"V"_1 = "V" - "V"_4`
= `300 - 200 = 100 "V"`
Hence, potential difference, V1, across C1 is 100 V.
Charge on C1 is given by,
`"Q"_1 = "C"_1"V"_1`
= `100 xx 10^-12 xx 100`
= `10^-8 "C"`
C2 and C3 have the same capacitances have a potential difference of 100 V together. Since C2 and C3 are in series, the potential difference across C2 and C3 is given by,
V2 = V3 = 50 V
Therefore, charge on C2 is given by,
`"Q"_2 = "C"_2"V"_2`
= `200 xx 10^-12 xx 50`
= `10^-8 "C"`
And charge on C3 is given by,
`"Q"_3 = "C"_3"V"_3`
= `200 xx 10^-12 xx 50`
= `10^-8 "C"`
Hence, the equivalent capacitance of the given circuit is `200/3` pF with
Q1 = 10−8 C, V1 = 100 V
Q2 = 10−8 C, V2 = 50 V
Q3 = 10−8 C, V3 = 50 V
Q4 = 2 × 10−8 C, V4 = 200 V
संबंधित प्रश्न
A bulb is connected in series with a variable capacitor and an AC source as shown. What happens to the brightness of the bulb when the key is plugged in and capacitance of the capacitor is gradually reduced?
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC.
Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V?
A capacitor of capacitance ‘C’ is being charged by connecting it across a dc source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so, how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor.
As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?
The capacitance of a capacitor does not depend on
The plates of a parallel-plate capacitor are made of circular discs of radii 5⋅0 cm each. If the separation between the plates is 1⋅0 mm, what is the capacitance?
Suppose, one wishes to construct a 1⋅0 farad capacitor using circular discs. If the separation between the discs be kept at 1⋅0 mm, what would be the radius of the discs?
Find the charge supplied by the battery in the arrangement shown in figure.
The outer cylinders of two cylindrical capacitors of capacitance 2⋅2 µF each, are kept in contact and the inner cylinders are connected through a wire. A battery of emf 10 V is connected as shown in figure . Find the total charge supplied by the battery to the inner cylinders.
Consider the situation shown in the figure. The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery.(c) Find the change in energy stored in the capacitors.(d) Find the heat developed in the system.
A parallel-plate capacitor with the plate area 100 cm2 and the separation between the plates 1⋅0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.
You are provided with 8 μF capacitors. Show with the help of a diagram how you will arrange minimum number of them to get a resultant capacitance of 20 μF.
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?
Two spherical conductors A and B of radii a and b(b > a) are placed concentrically in the air. B is given a charge +Q and A is earthed. The equivalent capacitance of the system is ______.
Between the plates of parallel plate condenser there is 1 mm thick medium shoot of dielectric constant 4. It is charged at 100 volt. The electric field in volt/meter between the plates of capacitor is ______.
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:
Two plates A and B of a parallel plate capacitor are arranged in such a way, that the area of each plate is S = 5 × 10-3 m 2 and distance between them is d = 8.85 mm. Plate A has a positive charge q1 = 10-10 C and Plate B has charge q2 = + 2 × 10-10 C. Then the charge induced on the plate B due to the plate A be - (....... × 10-11 )C
A capacitor has charge 50 µC. When the gap between the plate is filled with glass wool, then 120 µC charge flows through the battery to capacitor. The dielectric constant of glass wool is ______.
A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?
