DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

JEE: DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

The document DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

Introductory Exercise 22.1

Q.1. Find the dimensions of capacitance.

Sol. DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
= [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.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

Sol. Charge on outermost surfaces
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
Hence, charges are as shown below
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Sol. Charge on outermost surfaces.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
Hence charge on different faces are as shown below.

DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
Electric field and hence potential difference between the two plates is due to ±2.5 q
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

Introductory Exercise 22.2

Q.1. Find the charge stored in all the capacitors.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

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.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

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.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

Sol. In steady state current flows in lower loop of the circuit.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 800μC
In series, q remains same
q1 = q2 = 800 μC
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
and
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
(b) Now total charge will become 1600 μC. This will now distribute in direct ratio of capacity
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
They will have a common potential (in parallel) given by.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
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,
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
∴ Energy dissipated across R in the form of heat = E - U
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
q1 decreases exponentially from DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
While q2 increases exponentially from DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE Corresponding graphs and equation are given in the answer. Time constant of two exponential equations will be
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

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.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

Q.5. Determine the current through the battery in the circuit shown in figure
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

(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 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Q.6. For the circuit shown in the figure find
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
(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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
(b) In steady state no current flow through capacitor wire. PD across R1 is E.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
(c) In steady state potential difference across capacitor is E.
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE
(d) When switch is opened, capacitor is discharged through resistors R1 and R2
DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE

The document DC Pandey Solutions: Capacitors Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

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