Advertisements
Advertisements
प्रश्न
Find the charge supplied by the battery in the arrangement shown in figure.
Advertisements
उत्तर
The equivalent circuit for the given case can be drawn as :-
It can be observed that capacitors C1 and C2 are in parallel.
Therefore, the equivalent capacitance is given by
Ceq = C1 + C2
= 5 + 6 = 11 μF
Now ,
Charge supplied by the battery = (Ceq)(V) = (11 μF) × (10 V) = 110 μC
APPEARS IN
संबंधित प्रश्न
Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 μF. When the ends X and Y are connected to a 6 V battery, find out (i) the charge and (ii) the energy stored in the network.
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.
Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be
Find the charge appearing on each of the three capacitors shown in figure .
Two conducting spheres of radii R1 and R2 are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series−parallel connections.
Find the equivalent capacitance of the system shown in figure between the points a and b.
Find the capacitance of the combination shown in figure between A and B.
A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?
Define ‘capacitance’. Give its unit.
Obtain the expression for energy stored in the parallel plate capacitor.
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 ______.
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:
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 ______.
Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge on each capacitor.
Calculate equivalent capacitance of the circuit shown in the Figure given below:
