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DC Pandey Solutions: Capacitors | Physics Class 12 - NEET PDF Download

Introductory Exercise 22.1

Q.1. Find the dimensions of capacitance.

Sol. DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
= [M-1L-2T4A2]

Q.2. No charge will flow when two conductors having the same charge are connected to each other. Is this statement true or false?

Sol.  False . Charge does not flow if their potentials are same. It does not depend on amount of charge present.

Q.3. Two metallic plates are kept parallel to one another and charges are given to them as shown in the figure. Find the charge on all four faces.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Sol. Consider the charge distribution shown in figure.

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Electric field at point P   

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

But P lies inside conductor

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Hence, the charge distribution is shown in figure.DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.4. Charges 2q and -3g are given to two identical metal plates of area of cross-section A. The distance between the plates is d. Find the capacitance and potential difference between the plates.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Sol. Charge on outermost surfaces.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
Hence charge on different faces are as shown below.

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
Electric field and hence potential difference between the two plates is due to ±2.5 q
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
DC Pandey Solutions: Capacitors | Physics Class 12 - NEETDC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Introductory Exercise 22.2

Q.1. Find the charge stored in all the capacitors.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Sol:All three capacitors are in parallel wit h the battery. Potential Difference across each of them is 10V. So, apply q = CV for all of them.DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.2. Find the charge stored in the capacitor.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Sol. Capacitor and resistor both are in parallel with the battery. PD across the capacitor is 10V. Now apply q = CV. DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.3. Find the charge stored in the capacitor.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Sol. In steady state current flows in lower loop of the circuit.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
Now, potential difference across capacitor = potential difference across 4Ω resistance.
= iR
= (3)(4)=12V
q = CV = (2μF) (12 V)
= 24μC

Q.4. A 2μF capacitor and a 2μF capacitor are connected in series across a 1200 V supply line.
(a) Find the charge on each capacitor and the voltage across them.
(b) The charged capacitors are disconnected from the line and from each other and reconnected with terminals of like sign together. Find the final charge on each and the voltage across them.

Sol. 
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
= 800μCDC Pandey Solutions: Capacitors | Physics Class 12 - NEET


(b) In series, q remains same
q1 = q2 = 800 μC
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
and
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
Now total charge will become 1600 μC. This will now distribute in direct ratio of capacity
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.5. A 100 μF capacitor is charged to 100 V. After the charging, the battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is 20 V. Calculate the capacity of the second capacitor.

Sol. Charge, q = CV = 104μC
In parallel common potential is given by
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
Solving this equation we get
C = 400μF

Introductory Exercise 22.3

Q.1. An uncharged capacitor C is connected to a battery through a resistance R. Show that by the time the capacitor gets fully charged, the energy dissipated in R is the same as the energy stored in C.

Sol. Charge supplied by the battery,
q= CVDC Pandey Solutions: Capacitors | Physics Class 12 - NEET
Energy supplied by the battery,
E = qV =CV2
Energy stored in the capacitor,
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
∴ Energy dissipated across R in the form of heat = E - U
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.2. How many time constants will elapse before the current in a charging RC circuit drops to half of its initial value?

Sol.

 DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.3. A capacitor of capacitance C is given a charge q0. At time t = 0 it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the first capacitor and the second capacitor as a function of time f. Also plot the corresponding q-t graphs.

Sol. Let capacitor C1 is initially charged and C2is uncharged.

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

At any instant, let charge on C2 be q, charge on C1 at that instant =  q0 - q 

By Kirchhoff’s voltage law,

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.4. A capacitor of capacitance C is given a charge q0. At time f = 0, it is connected to a battery of emf E through a resistance R. Find the charge on the capacitor at time t.

Sol. Let q be the charge on capacitor at any instant t DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

By Kirchhoff’s voltage law

DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.5. Determine the current through the battery in the circuit shown in figure
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

(a) immediately after the switch S is closed
(b) after a long time.

Sol. 

(a) Immediately after the switch is closed whole current passes through C
i = E/R1
(b) Long after switch is closed no current will pass through C1 and C2.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

Q.6. For the circuit shown in the figure find
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
(a) the initial current through each resistor 
(b) steady-state current through each resistor 
(c) final energy stored in the capacitor 
(d) the time constant of the circuit when the switch is opened.

Sol. 

(a) At t = 0 equivalent resistance of capacitor is zero. R1 and R2 are in parallel across the battery PD across each is E.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
(b) In steady state no current flow through capacitor wire. PD across R1 is E.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
(c) In steady state potential difference across capacitor is E.
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET
(d) When switch is opened, capacitor is discharged through resistors R1 and R2
DC Pandey Solutions: Capacitors | Physics Class 12 - NEET

The document DC Pandey Solutions: Capacitors | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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FAQs on DC Pandey Solutions: Capacitors - Physics Class 12 - NEET

1. What is the principle behind capacitors and how do they store energy?
Ans. Capacitors store energy in the form of an electric field created between two conductive plates separated by an insulator (dielectric). When a voltage is applied across the plates, positive and negative charges accumulate on the plates, resulting in energy storage. The amount of energy stored is proportional to the capacitance and the square of the voltage across the plates.
2. How do you calculate the equivalent capacitance in series and parallel combinations?
Ans. For capacitors in series, the equivalent capacitance (C_eq) is calculated using the formula: 1/C_eq = 1/C1 + 1/C2 + 1/C3 + ... For capacitors in parallel, the equivalent capacitance is simply the sum of the individual capacitances: C_eq = C1 + C2 + C3 + ...
3. What are the different types of capacitors and their applications?
Ans. There are several types of capacitors, including ceramic, electrolytic, tantalum, and film capacitors. Ceramic capacitors are commonly used in high-frequency applications, electrolytic capacitors are used in power supply circuits due to their high capacitance values, tantalum capacitors offer stability and reliability for precision applications, and film capacitors are used in audio and radio frequency applications due to their low loss characteristics.
4. How does the dielectric material affect the performance of a capacitor?
Ans. The dielectric material affects a capacitor's capacitance, voltage rating, and energy storage capacity. Different dielectrics have different dielectric constants, which determine how much electric field energy can be stored. A higher dielectric constant allows for more charge storage for a given size, while the type of dielectric can also influence the capacitor's temperature stability and voltage breakdown characteristics.
5. What are the common applications of capacitors in electronic circuits?
Ans. Capacitors are widely used in electronic circuits for various applications including energy storage, filtering (smoothing out voltage fluctuations), timing applications (in oscillators), coupling and decoupling signals in amplifiers, and in power factor correction in AC circuits. They play a crucial role in ensuring stable and efficient operation of electronic devices.
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