English
Karnataka Board PUCPUC Science Class 11

The Equivalent Capacitance of the Combination Shown in Figure (31-q1) is

Advertisements
Advertisements

Question

The equivalent capacitance of the combination shown in the figure is _________ .

Options

  • C

  • 2 C

  • C/2

  • none of these.

MCQ
Advertisements

Solution

2C

Since the potential at point A is equal to the potential at point B, no current will flow along arm AB. Hence, the capacitor on arm AB will not contribute to the circuit. Also, because the remaining two capacitors are connected in parallel, the net capacitance of the circuit is given by

`C_"eq" = C + C = 2C`

shaalaa.com
  Is there an error in this question or solution?
Chapter 31: Capacitors - MCQ [Page 164]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 31 Capacitors
MCQ | Q 4 | Page 164

RELATED QUESTIONS

A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (b) The energy stored in the capacitor Justify your answer by writing the necessary expressions


Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now?

Also find the charge drawn from the battery in each case.


A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 μC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 μC. Calculate The potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?


As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?


A capacitor has capacitance C. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charges? If no, what other information is needed?


A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is


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.


Each capacitor shown in figure has a capacitance of 5⋅0 µF. The emf of the battery is 50 V. How much charge will flow through AB if the switch S is closed?


Convince yourself that parts (a), (b) and (c) figure are identical. Find the capacitance between the points A and B of the assembly.


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 capacitor of capacitance 2⋅0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4⋅0 µF as shown in figure . Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.


A 5⋅0 µF capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.


The figure show a network of five capacitors connected to a 10V battery. Calculate the charge acquired by the 5μF capacitor.


Define ‘capacitance’. Give its unit.


Obtain the expression for energy stored in the parallel plate capacitor.


The capacitance of a parallel plate capacitor is 60 µF. If the distance between the plates is tripled and area doubled then new capacitance will be ______.


The material filled between the plates of a parallel plate capacitor has a resistivity of 200Ωm. The value of the capacitance of the capacitor is 2 pF. If a potential difference of 40V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is ______.

(given the value of relative permittivity of a material is 50.)


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 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?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×