HC Verma Physics Part 2 Solutions for Chapter 39 Alternating Current in PDF

HC Verma Physics Part 2 Solutions for Chapter 39 Alternating Current

5/21/2012

Download HC Verma Solutions for HC Verma Physics Part 2 Solutions for Chapter 39 Alternating Current solved by our expert teachers. We have curated solutions for all questions of chapter 39. You can download the Solution of HC Verma -Alternating Current and prepare for your upcoming competitive examinations.

Solution of HC Verma Concept of Physics -2- Chapter 39 Alternating Current

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27 Comments

Suman

11/1/2012 12:14:25 pm

please explain Ans 13

Admin Answer

11/2/2012 11:06:55 am

Dear Suman,
In this question we have just applied the formula V= IR.
Step 1
first we have calculated the resistance of bulb.
step 2
then we have calculated the current in the circuit.
step 3 now we multiply current with resistance of bulb and put it equal to power of bulb.
now we get answer

Admin

11/5/2012 07:52:25 pm

This is attempt to provide free education to children who are preparing for engineering entrance examination.
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DEEP

12/9/2012 09:21:27 pm

IN QUESTION 16 ....(C)....WHY WE HAVE NOT TAKEN AREA ENCLOSED THROUGH WHICH THE MAGNETIC FIELD IS PASSING AS 3*5=15 INSTEAD OF 2*5=10....PLS EXPLAIN

Admin

12/10/2012 12:08:02 am

Dear Deep,
We know that Electric field(E) depends on the rate of change of Area through which magnetic field is passing.
In part C, Initially the magnetic field was passing through the whole area i.e 25cm² but in the last 2 seconds 2 cm of the length(area =2*5 = 10 cm²) gets out of the magnetic field.
initial area A₁ = initial Area through which magnetic field is passing = 25cm²
Final Area through which magnetic field is passing A₂= 3*5 = 15 cm²
so Change in area = A₂-A₁ = 25 - 15 = 10 cm²
Rate of change of area = 10/2 = 5 cm²/s
Here we don't take the area through which the magnetic field is passing. We take the change in area through which magnetic field is passing as explained above.

Pranjal

2/20/2013 01:51:46 am

queries from objective-1
->question2
->question6
->question9 (solved it by integrating in d way rms current is derived..seeking a short method)

Pranjal

2/20/2013 05:13:57 am

some more genuine doubts from exercise:
->'an AC source of x Volts' in this statement the given voltage should be considered to be peak voltage or rms voltage? In Q.10 its considered rms voltage, while in Q.16(b) its considered peak voltage (max current is peak current). Why such differences?
->In Q.14 (in my book) the given 50 V is peak voltage (printed, Eo=50 V). However the rms current is calculated by peak voltage. Plz explain this.
-> are these topics a part of iit or aieee syllabus? Dip circle, tangent galvanometer, deflection magnetometer, oscillation magnetometer, hot wire ammeter and voltmeter?? The first four are present in chapter36 and rest are in chapter 39.
Please help back asap.

Pranjal

2/24/2013 07:52:22 am

Well thats a very good initiative.
Please do let us know when the service is enabled back again. Thanks.

Shubhangi-Modi

9/29/2013 11:13:52 pm

well in Q.9
since the inductor is connected with capacitor, won't we take X as (XL-Xc) and then calculate the peak current??????????????

theyatin

10/1/2013 03:53:29 am

dear,
capacitor wont effect peak current so we need to consider only inductor.

shivani sharma

2/9/2014 03:20:06 am

sir,are the following statements true?
1-an inductor always opposes the flow of current
2-a capacitor does not opposes the flow of current

theyatin

2/10/2014 01:55:54 am

dear,
1-an inductor always opposes the flow of current
yes true
inductor induces magnetic field so as it opposes electric current

2-a capacitor does not opposes the flow of current
false capacitor opposes flow of current.

shubhu

2/24/2014 02:05:16 pm

Need help in objective Q4 . And Q2
in 4 I thought average value in half time period is never zero !!! Need help!!

theyatin

3/18/2014 11:14:08 pm

dear shubhu,
In that question you can use the equation of average current/voltage.Considering average current equation,for less value of 't' the average value will increase 'i'.So for the 'r' value remaining constant the value of v will tend to decrease.
otherwise using general equation also when the value of t will be small,then the sine value will decrease or can be said that it might tend to 0 as angle value remain b/w -1 & +1.

Admin

6/22/2014 01:54:26 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

Pragya

11/26/2014 09:35:32 pm

I want solution of this que..plz solve it

A circuit consists of a resistance
of 10ohm and a capacitor of capacitance
0.1micro farad.If an alternating emf of 100volt
at 50herz is applied.calculate the current in the
circuit -

Shivani

8/20/2015 01:55:46 am

32.2 divided by 10⁴(not sure)

VARUN KUMAR

12/19/2014 02:36:09 pm

please explain Q.29,30 and Q.3

Oshin

1/29/2015 12:06:03 am

Please can you answer short answer questions of alternating current .

yugesh

4/17/2015 02:02:13 pm

sir can you please explain Q 7 in detail and also i dont understand why we have to use integration

mridul

8/7/2015 05:03:21 pm

Please explain me questn no 3...is the dc voltge required for buod the rms value of voltage and if yes y so?

Shivani

8/20/2015 01:32:28 am

If Sin with any even angle is there , say Sin4wt if integrated will be zero??? If yes plz See problem no 11.

Shivani

8/20/2015 01:51:12 am

Sir, will u please answer? I need it.

abhishek jha

9/6/2015 11:12:41 pm

Sir please explain question no 26 b of chapter no 31

Anshul

9/9/2015 09:59:02 am

i want to know the questions of ch 39

barbi

10/21/2015 06:01:44 am

physics is nature of science

Abhinav

1/20/2017 08:24:59 am

Question no 6 dielectric strength is given not the electric field

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