NCERT Solutions | Current Electricity | Example 3.1, 3.2, 3.3, 3.4 | CBSE 12th #jee #neet - video Dailymotion

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00:00अगर आप होगा, क्योंकि आप दिख रहे हैं।

00:03हैलो स्टुड्रेंट से आगर वालोम मी और कोसी सहीचाच शुरुप से बहुत ब्राष्ट होकी हैं।

00:05ऐसा के साथ मैं एक अलग प्लेलिस्ट के आलक प्रकरण कर रहा हूँ।

00:09In this series, I am gonna start current electricity installed NCRT,

00:15पहला वीडियो होगा, फिर तो जकता खेल खेल खेल के लिए ही,

00:19Of course, it is for class 12 CDSC, NCRT exercises हमारा नेक्स वीडियो होने बाला होगा.

00:26आए चलते हैं, इंस्टेल्ट इंजांबल्स को आप अच्छे तरीके से आप मेंज कीजिए।

00:30फिर उसके आद कोई डाउट हो तो आप अगर कुछ विश्वाद नहीं भी होगी तो अगर एकसर्साइज़ करते करते आप काफी ज़्यादा खुले होगे और फिर यह एकसर्साइज़ और एकसर्साइज़ दोनों के लिए आप जी को भी ध्यान में रखना है तो बहुत नह

01:00होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी न

01:30हमें एन पहले मालूम करना होगा बिकोस ड्रिफ्ट स्पीड से रिलेटेट या तो

01:36हमें आई इस इकल तो एन एड़ी वीडी और वीडी इस माइनस इटाउ पाई एम

01:46हमारे लिए इस उप्शन इस गलोस और कौन-कौन से चीज गिवन हैं यहाँ पर डेंसिटी

01:58कौपर के लिए कौपर के लिए 9 के लिए 3 केजी पर यूबिक मीटर और इस एटाउमिक मास

02:08के लिए 63.5 यू यू यह एटाउमिक मास यूनिट है तो ड्रिफ्ट स्पीड से रिलेटेट या पहले मालूम करना होगा

02:20आप हमें इस से ही ये वर्क आउट करना पड़ेगा टेंड रेंज नम्बर डेंसिटी

02:25या अगर विशा बहुत लिए सीधा पुछना है नम्बर डेंसिटी मुकुल पर लिए बादकाम नम्बर पर लिए बादकाम

02:33पर यूनिट वालूम यहाँ पर मॉल कौन्सेप्ट इंबॉल हो रहा है और नमबर पर यूनिट वालूम

02:46अब नमबर के बारे में कुछ भी बात ही नहीं की गई है कुछ लेकिन उस से रिलेटेट एक चीज़ दे दी गई

02:53पर यूनिट वालूम यहाँ पर मॉल कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन

03:23कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और न

03:53मैं उनकिंगी का मात समय किसंगें पर तो हमको मात थो मालूं कर रही सकते था वन एटाम इपट का

04:02ఆ డ్ర్ థా ఆనంది లో లో పోలీ జూజ లో టగొన 2 కొ బ్రా అనూడ్ పున్న థ�ూండ్ పో టే కోలం౗ే ధినెను ప్రవా కౌన్ ఇంటర్ ఇట్ట్ న�

04:32एक परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद प�

05:02परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद

05:32गराम का मतलब

05:3463.5

05:36upon 1000

05:386 x 10

05:4023 x 9

05:4210 x 6

05:4463.5

05:4663.5

05:48नाव अफ़्टरवर्स

05:50आहव टू कैल्कुलेट

05:52आहव टू कैल्कुलेट

05:5429

05:569 x 6

05:5854 x

06:0063.5

06:028 x 5

06:040

06:0654

06:0863.5

06:100.8

06:125

06:140 x

06:1610 x

06:1829

06:208.5 x

06:2210 x

06:2428

06:26आव अगर वापस से

06:28हम बात करें

06:30from this equation

06:32Vd is equal to I upon N

06:34EA

06:36Vd is equal to

06:38directly

06:40I is 1.5 ampere

06:42N is

06:448.5 x

06:4610 x 28

06:48N is 1.6 x

06:5010 x

06:5219

06:54EA

06:56was 1 x

06:5810 x 7

07:00we need to multiply

07:02this only

07:041 upon

07:068.5 x

07:081.6

07:101.5

07:12upon

07:1413.6

07:1613.6 x

07:1810 x

07:207

07:2210 x

07:2426

07:2613.6

07:280.110

07:3010 x

07:322

07:341.1

07:3610 x

07:383

07:40meter

07:42per second

07:44it could be written as

07:461.1 x

07:48millimetre

07:50millimetre

07:52per second

07:54second part

07:56second part

07:58A part

08:00average speed

08:02average drift speed

08:04B part

08:06they are asking us

08:08to compare the drift speed obtained

08:10above with

08:12of copper atoms at

08:14ordinary temperatures

08:20first

08:22in part B

08:24we have to find thermal speed

08:26and

08:28one formula from the chapter

08:30kinetic theory of gases

08:32in eleventh

08:34and

08:36there we have to

08:38equate half mv square

08:40with kbt

08:42we got

08:44a formula

08:46that was

08:48thermal speed

08:50rms speed

08:52of electron

08:54having the kinetic energy

08:56half mv square

08:58half mv square

09:00is equal to

09:023 by 2 kbt

09:04where t was

09:06temperature

09:08mass v was

09:10speed

09:12same speed

09:14kb was

09:16Boltzmann constant

09:18you are right

09:20then we

09:22got v

09:24as kbt

09:26by m

09:28from this

09:30thermal speed of copper

09:32approximately

09:34this is not talked

09:36it is roughly

09:38taken

09:40room temperature

09:42and lastly

09:44calculation of m

09:463 kbt

09:481.38 into

09:5010 raise to minus 23

09:52300

09:54on

09:56raise to minus

09:583 on 6

10:00into 10 raise to

10:02it will come over here

10:04it will be directly cancelled

10:06this factor i am directly cancelling

10:08right

10:10then there after

10:12is

10:146 into

10:161.38

10:18into

10:20300 on

10:223.5

10:2410 raise to minus 3

10:26it should come

10:28342

10:30it should not come

10:32that is

10:34100

10:36354

10:38is equal to

10:40in

10:42comparison

10:44comparison

10:461.1 into

10:4810 raise to minus 3

10:50meter per second

10:52ratio required

10:54ratio would be

10:56of

10:58upon

11:00PRMS

11:02of per atom

11:04and

11:06342.57

11:08and after calculation

11:10right now we are

11:12into second part of this question

11:14and this is

11:16B part

11:18in B part it is first part

11:20in B part it is first part

11:22compare the drift speed

11:24obtained above with

11:26thermal speeds of copper atoms

11:28at ordinary temperature

11:30which we have found

11:32speed of propagation

11:34of electric field along the

11:36conductor which

11:38causes the drift motion

11:40B

11:42second part compare

11:44the speed of propagation

11:46electric field

11:48along the conductor

11:50electric field and this is very

11:52important very interesting point

11:54electric field propagates

11:56along a conductor

11:58with the speed of

12:00speed of EM waves

12:02EMW that is

12:04that means we are to compare

12:06required ratio

12:08would be Vdc

12:10that is equal to

12:12already

12:14we had this

12:161.1

12:18into

12:2010 raise to minus 3 upon

12:22upon 3 into

12:2410 raise to 8

12:26just final answer

12:28would be

12:30again approximately

12:3210 raise to 11

12:34final answer

12:36now for

12:38example 3.2

12:40this is appearing

12:42completely theoretical

12:44but still let's see

12:46there are 5 parts in that

12:48A part in example

12:503.1 that means previous question

12:52the electron drift speed

12:54is estimated to be only

12:56a few mm per second for current

12:58in the range of a few

13:00amperes how then

13:02is current established

13:04almost the instant

13:06a circuit is closed

13:08that means

13:10how current is generated so fast

13:12again

13:14for this

13:16factor is electric field

13:18the deciding factor is electric

13:20field and in previous

13:22question B part

13:24B2 second

13:26part we had this

13:28electric field in numerical

13:30format now

13:32same electric field we are

13:34to discuss for

13:36this first part

13:38so for this part

13:40for current generation

13:42they don't have to wait for electrons

13:44then second

13:46part the electron

13:48drift arises due to the

13:50force experienced by electrons

13:52in the electric field inside the conductor

13:54but force should

13:56cause acceleration

13:58why then do the electrons

14:00acquire a steady average drift

14:02this is basic fundamental

14:04for

14:06acquisition of drift velocity

14:08let's come to

14:10that means we need to explain

14:12the reason behind

14:14this average

14:16despite uniformly

14:18accelerated motion

14:20does accelerate

14:22drift speed collides

14:24positive ion

14:26and this

14:28collision only cause

14:30obstruction that is resistance

14:32inside the conductor

14:34it loses its drift speed

14:36after collision

14:38but starts to accelerate

14:40increasing its

14:42drift speed therefore

14:44electrons acquire

14:48and now

14:50third part

14:52if the electron drift

14:54speed is so small

14:56and the electrons charge is small

14:58how can we still

15:00obtain large amount

15:02of current in a conductor

15:04this is again very thoughtful

15:06question because the electron

15:08number density is very high

15:10in the order of 10 raise

15:12to 29 per cubic

15:14meter the reason

15:16further I am to

15:18give one reference

15:20I is equal to

15:22N E A

15:24V D that means

15:26current is depending

15:28upon N also

15:30and N

15:32is this much I

15:34then come to

15:36D part when electron

15:38drift in a metal from

15:40lower to higher potential

15:42does it mean that

15:44all the free electrons of the metal

15:46are moving in the same

15:48direction yes

15:50that is obvious

15:52it has to this was

15:54D part

15:56the drift velocity

15:58is superposed

16:00then come to E part

16:02are the parts of

16:04electrons straight lines

16:06between successive collisions

16:08positive ions of the metal

16:10in the first

16:12absence of electric field presence

16:14of electric field that means in both

16:16the situation it is

16:18it is been asked

16:20whether the movement

16:22of electrons path

16:24are a straight line

16:26or not

16:28in both the situation absence

16:30of electric field may be and

16:32presence may be E part

16:34electric field

16:36the paths are straight lines

16:38and hence this

16:40question number 3.2

16:42is over

16:443.3 tan electric

16:46toaster uses

16:48nichrome for its heating

16:50element when a negligibly

16:52small current passes

16:54through it its resistance

16:56at room temperature

16:58above that means

17:00R T

17:02equals 75

17:040.3 ohms

17:06and where

17:08T is

17:1027 degree Celsius

17:12and this is room temperature

17:14when the toaster is connected to a

17:16230 volt supply

17:18the current settles

17:20after a few seconds

17:22to a steady value

17:24of 2.68

17:262.68

17:28ampere then I is

17:302.68

17:32ampere what is the

17:34steady temperature of the nichrome

17:36element

17:38temperature

17:40the temperature coefficient of resistance

17:42of nichrome averaged

17:44over the temperature range

17:46involved is

17:48that means they are

17:50talking about alpha

17:52alpha is given

17:54temperature coefficient

17:56alpha

17:581.70

18:00into

18:0210 raise to minus 4

18:04degree per

18:06Celsius so this

18:08was alpha and one

18:10more thing what is the steady

18:12temperature of the nichrome

18:14element any final

18:16temperature then

18:18T1 then

18:20T2 is to be

18:22found but

18:24let's find batka

18:26resistance also from

18:28given data R2

18:30could be found R2

18:32equals 230

18:34volt

18:36on 2.68

18:38ampere

18:40that is

18:4285.82

18:44now for

18:46T2 how do we find

18:48now considering

18:50for your again

18:52I am to remind

18:54RT is equal to

18:56R0 1 plus

18:58alpha T

19:00but here

19:02alpha at

19:040 degree Celsius

19:06was not given

19:08alternative formula

19:10R2 is equal to R0

19:121 plus

19:14alpha delta T

19:16has to be used

19:18this is

19:20not to be used then

19:22coming back this

19:24equation R2

19:26is equal to R0

19:281 plus alpha

19:30delta T that means

19:32T2 minus

19:34T1

19:36T2 minus T1

19:38R2 minus R1

19:40upon

19:42R1 alpha

19:44R2 is

19:4685.8

19:48minus R1

19:5075.3

19:5275.3

19:54into

19:56alpha

19:58was 1.70

20:00into 10 raise to minus 4

20:02it implies that

20:04T2 minus T1

20:06is

20:08820 degree Celsius

20:10and already

20:12we had T1

20:1427 degree Celsius

20:16here you can notice

20:18and we are to find T2

20:20820

20:24plus

20:2627

20:28847 degree

20:30Celsius

20:32now for example 3.4

20:34again

20:36it is the question

20:38is related to

20:40the relation between resistance

20:42and temperature only

20:44let's see

20:46what is given the resistance

20:48of platinum wire

20:50of a

20:52platinum

20:54resistance thermometer

20:56at the ice point

20:58at ice point

21:00resistance is

21:025 ohm

21:04R0

21:065 ohm and

21:08at steam point

21:10that means R100

21:12equals

21:145.23

21:16when the thermometer is

21:18inserted in a hot bath

21:20the resistance of the

21:22platinum wire is

21:245.795

21:26RT let's assume

21:28795

21:30ohm calculate the

21:32temperature of the bath

21:34this time temperature

21:36is asked now again

21:38we would think for

21:40two relations

21:42first is

21:44RT is equal to

21:46R0 1 plus

21:48alpha

21:50T second

21:52is R2 is equal

21:54to R0

21:561 plus alpha delta

21:58T now let's see

22:00what is to be taken

22:02going to be noted no

22:04alpha is given

22:06applying this is equation

22:08number one second situation

22:10RT

22:12RT is equal to

22:14R0

22:161 plus alpha

22:20we have

22:22two equations

22:24finally

22:26equation number one and

22:28two in both the equations

22:30we have to

22:32get the value of T

22:34therefore T is

22:36equal to

22:38100 into

22:40what is RT

22:42RT was

22:445.795

22:46and what

22:48was R0

22:50R0 is

22:525 what is

22:54R100 5.23

22:56only

22:58calculation left

23:005.795

23:02minus

23:045

23:06that is

23:08795

23:10upon

23:120.23

23:14get me the final

23:16value

23:183.45 multiplied

23:20by

23:22345

23:24T is

23:26345

23:280.65 degree

23:30Celsius hence

23:32this is done see you

23:34in next video bye for now

23:36take care

NCERT Solutions | Current Electricity | Example 3.1, 3.2, 3.3, 3.4 | CBSE 12th #jee #neet

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