DC Pandey Solutions: Capacitors

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

## Introductory Exercise 22.1

Q.1. Find the dimensions of capacitance.

Sol.

= [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. Charge does not flow if their potentials are same.

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.

Sol. Charge on outermost surfaces

Hence, charges are as shown below

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.

Sol. Charge on outermost surfaces.

Hence charge on different faces are as shown below.

Electric field and hence potential difference between the two plates is due to ±2.5 q

## Introductory Exercise 22.2

Q.1. Find the charge stored in all the capacitors.

Sol: All three capacitors are in parallel wit h the battery. PD across each of them is 10V. So, apply q = CV for all of them.

Q.2. Find the charge stored in the capacitor.

Sol. Capacitor and resistor both are in parallel with the battery. PD across the capacitor is 10V. Now apply q = CV.

Q.3. Find the charge stored in the capacitor.

Sol. In steady state current flows in lower loop of the circuit.

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.

= 800μC
In series, q remains same
q1 = q2 = 800 μC

and

(b) Now total charge will become 1600 μC. This will now distribute in direct ratio of capacity

They will have a common potential (in parallel) given by.

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

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= CV
Energy supplied by the battery,
E = qV =CV2
Energy stored in the capacitor,

∴ Energy dissipated across R in the form of heat = E - U

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

Sol.

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. Both capacitors have equal capacitance. Hence half-half charge distribute over both the capacitors.

q1 decreases exponentially from
While q2 increases exponentially from  Corresponding graphs and equation are given in the answer. Time constant of two exponential equations will be

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. qi = q0 qf = EC
Now charge on capacitor changes from qto qf exponentially.

Q.5. Determine the current through the battery in the circuit shown in figure

(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.

Q.6. For the circuit shown in the figure find

(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.

(b) In steady state no current flow through capacitor wire. PD across R1 is E.

(c) In steady state potential difference across capacitor is E.

(d) When switch is opened, capacitor is discharged through resistors R1 and R2

The document DC Pandey Solutions: Capacitors | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET

## Physics Class 12

105 videos|425 docs|114 tests

## FAQs on DC Pandey Solutions: Capacitors - Physics Class 12 - NEET

 1. What is the formula to calculate the capacitance of a capacitor?
Ans. The formula to calculate the capacitance of a capacitor is C = Q/V, where C is the capacitance in Farads, Q is the charge stored on the capacitor in Coulombs, and V is the voltage across the capacitor in Volts.
 2. How does a capacitor store electric charge?
Ans. A capacitor stores electric charge by accumulating opposite charges on its two plates. When a voltage is applied across the capacitor, electrons are attracted to the positive plate and repelled from the negative plate, creating an electric field between the plates. This accumulation of charge on the plates results in the storage of electric energy in the capacitor.
 3. What factors affect the capacitance of a capacitor?
Ans. The capacitance of a capacitor is affected by three main factors: the area of the plates, the distance between the plates, and the dielectric constant of the material between the plates. Increasing the plate area or decreasing the plate distance increases the capacitance, while using a material with a higher dielectric constant also increases the capacitance.
 4. How does a dielectric affect the capacitance of a capacitor?
Ans. A dielectric material placed between the plates of a capacitor increases the capacitance. This is because the dielectric reduces the electric field between the plates, allowing more charge to be stored for a given voltage. The dielectric constant of the material determines how much the capacitance is increased.
 5. What are some common applications of capacitors?
Ans. Capacitors have various applications in electronic circuits. Some common uses of capacitors include smoothing power supply voltages, filtering noise in audio circuits, timing in oscillators, storing and releasing energy in flash cameras, and providing backup power in uninterruptible power supplies (UPS).

## Physics Class 12

105 videos|425 docs|114 tests

### Up next

 Explore Courses for NEET exam

### Top Courses for NEET

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;