Advertisements
Advertisements
Question
Find the charge appearing on each of the three capacitors shown in figure .
Advertisements
Solution
Let us first find the equivalent capacitance. It can be observed from the circuit diagram that capacitors B and C are in parallel and are in series with capacitor A.
The equivalent capacitance can be calculated as follow :-
`1/C_(eq) = 1/C_A + 1/(C_B+C_C`
`1/C_(eq) = 1/8 + 1/(4+4) = 1/8+1/8`
`⇒ 1/C_(eq) = 2/8`
`⇒ C_(eq) = 4 "uF"`
Capacitors B and C are parallel and are in series with capacitor A. The equivalent capacitance of capacitors B and C is given by
(4 + 4) μF = 8 μF
It is the same as the capacitance of capacitor A. Therefore, equal potential difference will be there on capacitor A and the system of capacitors B and C.
Now,
Potential difference across capacitor A = 6 V
Thus,
Charge on capacitor A = (8 µF) × (6 V) = 48 µC
And,
Potential difference across capacitors B and C = 6 V
Charge on capacitor B = (4 µF) × (6 V) = 24 µF
Charge on capacitor C = (4 µF) × (6 V) = 24 µF
APPEARS IN
RELATED QUESTIONS
Define capacitor reactance. Write its S.I units.
(i) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 µF capacitance.
(ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?
Following operations can be performed on a capacitor:
X − connect the capacitor to a battery of emf ε.
Y − disconnect the battery.
Z − reconnect the battery with polarity reversed.
W − insert a dielectric slab in the capacitor.
(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.
(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYZ.
(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.
(d) The electric field in the capacitor after the action XW is the same as that after WX.
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.
Each of the capacitors shown in figure has a capacitance of 2 µF. find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.
A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm and of radii 2 mm and 4 mm. (a) Calculate the capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii 4 mm and 8 mm. Calculate the capacitance.
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.
The separation between the plates of a parallel-plate capacitor is 0⋅500 cm and its plate area is 100 cm2. A 0⋅400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.
A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.
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?
Define ‘capacitance’. Give its unit.
Derive the expression for resultant capacitance, when the capacitor is connected in parallel.
Calculate the resultant capacitances for each of the following combinations of capacitors.
Three capacitors 2µF, 3µF, and 6µF are joined in series with each other. The equivalent capacitance 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:
A 5µF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 µF capacitor If the energy change during the charge redistribution is `"X"/100`J then value of X to the 100 nearest integer is ______.
A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electric conductivity σ = 7.4 × 10-12 Ω-1 m-1. If the charge on the plate at the instant t = 0 is q = 8.85 µC, then the leakage current at the instant t = 12 s is ______ × 10-1 µA.
Two identical capacitors are connected as shown and have an initial charge of Q0. The separation between the plates of each capacitor is d0. Suddenly the left plate of the upper capacitor and right plate of the lower capacitor start moving with speed v towards the left while the other plate of each capacitor remains fixed. `("given" (Q_0V)/(2d_0) = 10 A)`. The value of current in the circuit is ______ A.
