this is a lecture No. 21 of a B 2 3 2 2
Today we will study how to determine the resistance 3
and capacitys of different types of resistors 4
and capacitys using boundary value problem 5
this will cover your section 6.5 of textbook 6
this is Academic Week No. 13 of this semester 7
the learning objectives of today's lecture would be 8
to determine the resistance 9
and capacity my solving be boundary value problem 10
you remember that in previous lecture 11
we have studied 12
how to solve these boundary value problems 13
using the Liplas and poisons equation 14
first of all the resistance 15
for conductors with uniform cross sectional area 16
the resistance it is related by this relationship 17
where to the 18
resistance is 19
directly proportional to the length of this conductor 20
and it is inversely proportional to the surface area 21
of that conductor 22
if the cross section of conductor is not uniform 23
then the resistance can 24
can still be obtained using this Homs law 25
but in that case your voltage is and current 26
they will be of some integral form 27
and here we also studied that 28
how to find out the voltages 29
using this integration of electrical 30
and how to find out this current 31
using the integration of this current density 32
okay surface current density here 33
so finding the resistance of a conductor with non 34
uniform cross sectionalia 35
it is also known as the boundary value problem 36
so we will involve this boundary value problem 37
to determine these two 38
these voltage and grant for the non uniform 39
cross sectionality of the resistors 40
these are the steps that we will follow 41
to find out the resistance of uh 42
any conducting media material 43
which is having non uniform surface area 44
first of all we will uh 45
choose a suitable coordinate system 46
that can be your cartision 47
slendicalant cervical coordinate system 48
thereafter we will assume uh 49
v not as the potential difference between the uh 50
conductor terminals 51
then will solve depending upon the situation 52
either it is a lap last equation case 53
or poison sequation case 54
so if it is a lap last equation case 55
then will involve this question 56
and uh we will uh 57
double integrate this equation to find out the v 58
in terms of here v 59
and then we will determine the E 60
electrically using this relationship between the uh 61
potential and the electrically 62
and then we will uh 63
use this integral equation to find out the current 64
in terms of in terms of we not right 65
so we will find out this in terms of we not 66
and then using this Onstar relationship 67
will find out the resistance 68
by simply dividing this assumed constant 69
potential 70
difference between the terminals of the conductor 71
divided by this current that we have determined 72
for the non uniform surface surface conductor 73
so it is also possible to operate in other ways 74
that will first resume the current is constant 75
okay and then we find out the potential 76
a corresponding potential defence in terms of I 77
not in terms of I not remember 78
so that this in the previous case 79
this we not and we not would be cancelled 80
in this case this I not and I not would be cancelled 81
and we are left with only be resistance 82
to have contestants we must have two conductors 83
and which carry that equal and opposite charges 84
the conductors may be separated by free space 85
or some insulating material 86
that is known as dialect material 87
so here's uh it is an example of your capactor 88
which is having two plates 89
and uh you can see over here 90
the charges are of uh same magnitude 91
but opposite polarity 92
and they are separated by some distance and then they 93
they can be this separation 94
it can be filled either by this free space 95
if this is free space 96
then your option is simply abslan not 97
and if it is other material 98
then you will have abslan is equal to abslan 99
not times D 100
relative permittivity 101
so the conductors are maintained at a potential 102
that is v in this case 103
and it is also 104
the potential difference between these two conductors 105
and it is uh related uh with this electric V E 106
using this relationship 107
we have already derived this relationship 108
that how 109
to find out 110
the potential difference between two conductors 111
by simply taking the line 112
integral of this electric feed 113
so you notice that this electric grid 114
it is always normal this electric cross lines 115
they are always normal to the surface 116
of these conductors so all flatlines 117
leaving one conductor 118
must necessarily terminate at the surface of 119
other conductor and also by a radical idea 120
at the normal orientation 121
to the surface of that originating conductor 122
and terminating conductor 123
so the capacitor saw 124
of the capacitor 125
is defined as the ratio of magnitude of the charge 126
on one of the plate 127
to the potential difference between them 128
so this is the basic relationship 129
that we have already utilized 130
in our early classes as well 131
in our early lectures as well 132
the only difference over here is 133
it is one non uniform surface area 134
so if it is the in non uniform surface areas 135
then obviously we are going to involve the integrals 136
and here uh from glazes law 137
we know that uh 138
this charge enclosed is equal to the uh 139
this uh surface integral of here 140
uh flux uh this is electric flux density right 141
this one 142
okay and uh 143
and this is the same relationship between the uh 144
that uh 145
voltage potential is equal to the integral of here 146
align integral of here electric feed 147
so there again 148
there are two methods of finding all the capacitys 149
for any two conductor capacitors 150
so 151
it is either by using the gauzesla or the liplactions 152
by solving the liplactions equation 153
so if we assume q the charge as a constant 154
and then we determine this V in terms of this q 155
then this will enroll your causes drop 156
and the vice versa if we assume V as a constant 157
and determine this q in terms of your V 158
then we will we need to solve this 159
the class integration we'll do these things 160
uh one by one in our uh 161
in our examples 162
examples as well and in our innovations as well 163
so magnet 1 involves the falling steps again 164
you need to choose the suitable corner system 165
according to your problem 166
and then let the conducting plates uh 167
let the let the two conducting plates carry a charges 168
uh equal 169
uh equal uh 170
magnitude charges but with 171
but with this opposite polarity vitamin E electrical 172
either using your coolants law or Gaza's Law 173
for symmetrical surface densities charges 174
charge distributions 175
and then if we know the electric field 176
then obviously 177
we can find out this voltage 178
using this relationship of uh 179
between this align integral of your electrically 180
I know yes 181
you know since you have found out the B in terms of Q 182
then you can uh 183
involve the specific relationship to find out 184
the capacitors of these two 185
uh conductor 186
play conductor uh 187
plate capacitors and hey remember 188
that negative sign may be ignored 189
as absolute value of V is required because 190
because we know that capacity 191
capacity is a skill of quantity 192
uh these are the uh 193
some of the examples of your capacitors as you can see 194
as you can see as you can see 195
that these are electrolytic types of the capacitors 196
right these 197
if you are you're seeing that these are 198
these are ceramic types of the conductors 199
so electrolytic types of the conductor 200
they involve the polarity 201
okay positive and negative like 202
whereas the ceramic types of the conductors they are 203
they are in the in nanos and fatter 204
nanoferribs and PICO ferbs of the capacitors 205
whereas these are you can as you can see here 206
these are in microfranger milli fat 207
whereas these are injured nano feathers and 208
and PICO feathers 209
so what is the purpose of your chair or this 210
this capacitor okay 211
a capacitor consists of two conductors of any shape 212
placed near one another without touching each other 213
without touching 214
remember this without touching each other 215
it is common to fill up 216
the region between these two conductors 217
with an insulating material 218
known as the dialectric 219
and we have seen that the capacity 220
the storage 221
is directly proportional to the dialectric value 222
of the medium 223
that is present in between these two conductor plates 224
right this this is the dialect 225
and uh we charge these plates with uh 226
with opposing charges to set up an electric field 227
so and uh so if we uh 228
if we uh connect these two these two plates 229
the better metallic connector plates with some uh 230
external potential then 231
then electric field is going to build up in between 232
these two 233
private plates 234
and then you're going to store the charge between 235
charge on this to pasture plates 236
uh this is very important 237
this slide is uh you can uh 238
you can uh 239
take this slide as a complete summary of your uh 240
these two lectures that 241
how you need to proceed to find out the capacity 242
and the resistance of conductors 243
so the first step if you keep this Q as a constant okay 244
q constant 245
and then you need to determine this V in terms of Q 246
and for this purpose 247
you will take help of this cooling stop 248
okay the second 249
the second one is your uh if 250
if you keep this V as a constant 251
v not as a constant between two plates 252
and then you can find out this Q using this um 253
definite definition of this uh flux 254
flux uh 255
this uh 256
that charge is equal to the uh 257
surface in general of your electric 258
uh electric uh flux 10 3 259
and then from here you can 260
you know that 261
this is the constitutive relationship between this 262
Indian electrical 263
and then if you if you have found out this electrical 264
from this equation 265
then you can use this relationship to find out 266
the electrical and hey 267
remember that we will 268
we will take help of the Prussians 269
or the poisons equation 270
according to this case to find out this 271
this v 272
like here okay 273
and the third thing is your resistance 274
so for the resistance if you again 275
keep this voltage across that resistance 276
at that constant 277
then you need to determine this current 278
flowing to that conductor 279
and that is your surface integral of your condensity 280
and where this conduct is equal to using the ONS 281
like it is equal to conductivity times e electrical 282
and again you can find out this electrical age 283
using this relationship and how to find out this V 284
again you need to take the help of this uh 285
solving this boundary problem or uh equently the uh 286
solution of finding out the solution of this leclasin 287
poison secretion 288
let's first find out the capacitors of our power plate 289
capacitor 290
in an ideal power plate capacitors 291
the plate separation this means the both conductors 292
they are separated by a distance or separation of D 293
it is very small 294
as compared to the dimensions of the plate 295
so this is the first Assumption 296
if this is an Assumption that the distance is very 297
very small as compared to the 298
the size of this place of the capacitor 299
then we can ignore the fringing fields 300
at the edge of place so this can be ignored 301
okay this can be ignored 302
okay the first thing 303
and then uh the second thing is the field 304
uh the field between them is considered to be constant 305
so they will be considered as constant 306
you can see over here they're 307
they're they're showing just like equal 308
equal magnitude lines so the space between the plates 309
it is filled up with a homogeneous 310
another resumption 311
homogeneous dialectic with permetuity absline 312
and we and this flux 313
fringing can be ignored by 314
using this 315
smaller separation between these two conductor plates 316
then what we can we can we can into that 317
the uh this electric trucks electric cross density 318
it is uh using either using this boundary conditions 319
it is equal to uh 320
this surface surface chart density time 321
this your vector country is your basically the normal 322
the unit normal vector to the surface of this 323
these two plates so we can uh uh 324
we can uh we can proceed in another 325
another way that this will derive in our chapter No. 4 326
that this is the uh 327
your electric field on the surface 328
on one conductor surface 329
uh which is having the surface on distribution of raw s 330
and if it is of negative here 331
if it is of negative charge 332
then you can you can see over here 333
you can see over here that uh 334
for this uh for these two things 335
uh if for these two two things 336
the uh the normal it is in the opposite direction right 337
so this is uh 338
this this was your end and equation that was uh uh 339
determine the electric field between two power plate 340
uh capactor and uh 341
it is in terms of the surface charge of uh 342
both capactors uh 343
and the unit normal between them 344
in the direction of your electric feed 345
so if you if we apply this equation on this case right 346
we have this negative q negative rules 347
you can also say okay 348
and this positive rules you can also say right 349
they are separated by a distance d and since uh 350
this positive positive rate is at top 351
so you can assume that this normal 352
normal direction is in minus x 353
right so 354
the electric field originates from the positive plate 355
and the terminates on the negative charge plate 356
so put in this this normal 357
uh you know normal factor 358
uh in the direction of your electric field 359
so this is your electric field 360
uh for this uh 361
particular uh example 362
okay 363
and uh 364
it is just the definition of your surface identity 365
that it is equal to your 366
what is you remember that it is your uh 367
that charge budget per unit alia 368
right so first charge then is your charge per unit alia 369
that is your Judi right 370
yes so that is your question 371
so suppose this is just again the reputation of that 372
suppose each plate has a surface area of s 373
the players are separated by a distance d 374
as in page 1 and 2 375
carry uniformly distributed charge plus Q and minus Q 376
and accordingly there will be 377
so this is the same as we have discarded now 378
previous try okay 379
but let's apply this relationship 380
the line into the relationship to find out the voltage 381
right so remember in this case we have assumed 382
we assume that the charge is constant 383
okay the first case 384
that the charge is constant on this surface of place 385
and we are 386
going to determine this V in terms of this Q 387
and for that we know this relationship 388
for this relationship 389
we need to find out this electrical 390
that we have no product 391
now take the integration of this thing 392
then uh you can find or that uh what is the variable 393
the variable is X this thing okay 394
and it is Jenny from it is Jenny from 0 2 d right 395
uh see your hair so uh here it is disciples that 0 396
x is equal 0 disciples that X is equal to 397
d 398
so this is that actually equal to zero 399
so that's why these integration limits are from 0 to D 400
okay and then uh 401
you just plug in uh this electrical lower here 402
you take uh 403
this uh center differential lenses in the X axis 404
so that's why it's uh unigratory is a a X 405
now you solve this equation 406
so this integral it is 407
you can say that it is it is like this okay 408
0 to d okay and then X integration of 1 409
you know that it is X in this case 410
okay so from here 411
if you move on and plug in this water to your hair 412
then this is your capacity okay 413
this is the capacity of your better grade capacity 414
capacity 415
that you have already started in your early classes 416
as well the capacity 417
it is directly proportional to the surface area of the 418
plates and it is inversely proportional to 419
the suppression between these two plates 420
right and 421
it is also directly proportional to the permittuity 422
of dialectic material 423
which has been placed in between these two 424
better plate capacitors 425
so if we want to find out this absent are 426
then what we need to do that the 427
the basically 428
this relative permativity of the medium 429
in between these two palpric pastor 430
then first we need to determine on major decapacitors 431
of pirate plate capacitor with space 432
with free space medium between this plate right 433
this absolute not case right 434
and then we need to 435
so with the redistrate of first of first cases sorry 436
the first first cases 437
the space between them 438
the space between them is 439
the space between players filled with the dialectric 440
right and we'll uh 441
we'll uh 442
we'll find out that capacity will level that uh 443
uh capacity as C okay 444
and the second one is see not for this case 445
we simply replace this dialect medium 446
medium with d free space Audi here 447
that is of perhaps not 448
so we just divide these two caps 449
tenses 450
to find out the relative promativity of the dialectic 451
that have have been placed in the previous 452
in the in the first case 453
so if we found we have found out the electric field 454
and we can also find out 455
this energy stored in a capacitor 456
so this was the relationship that we already studied 457
that the energy stored 458
that the energy stored can be related with this 459
electric field using this relationship 460
that it is half volume 461
integral of your square of this electrically 462
so now if we substitute this barrel plate capacity 463
electric field in this equation 464
and simplify this equation 465
like this one okay 466
and uh like this one okay 467
and uh what you can do with so if you 468
if you simplify this equation and if you simplify 469
let's see what 470
what simplification we are going to involve further uh 471
so this uh is going to cancel out okay 472
this option is going to cancel with this one 473
so you are left over with this one 474
and the second thing is very important thing is 475
you are changing this volume material 476
that is the severe volume is equal to your 477
is equal to your surface area 478
and the tire dimension of your 479
of your this coordinate system so here it was d okay 480
and so that's why this SND is coming over here 481
the integration of this volume 482
the volume trigger okay 483
and so if you put in order to simplify this 484
so this is the remember this equation 485
this is the relationship to find out the 486
the energy stored in the power plate 487
capacitor in terms of the charge and the voltage 488
and if you if you know the capacity then you can 489
you can you can just simplify 490
you can simply involve that relationship between your Q 491
and bring the charge and reward case 492
to find out further this relationship 493
so just take a take help of this relationship that you 494
C is equal to q divided by v okay 495
plug in this over here and uh 496
you can uh find out the different forms of the same 497
this energy stored you know 498
my parallel blade capacitor 499
so it is only for the pilot parallel blade capacity 500
remember this thing 501
so let's let's discuss in other case 502
uh for which we need to find out the uh 503
capestons that is your co xial capacitor 504
so consider that it is comprising of lent L and uh 505
it is consisting of two co xial conductors 506
the space between them is filled up with homogeneous 507
dialectic material of permittivity 508
permittivity action here 509
assumed that the conductor 1 and the conductor 2 510
that carry uniformly distributed charges 511
of same magnitude cube 512
but with opposite polarity 513
now we need to apply this uh 514
if we if we apply this Gaza's law 515
using this to the arbitrary guardian 516
uh 517
cylindrical surface of this radius row with this row 518
with this row it is changing from here a to B 519
okay so apply the Gaza 520
Gaza's larger 521
arbitrary gazian slendic surface of radius row 522
changing between this a and B to 523
this is your a and this is your B 524
okay 525
so uh we have already seen that 526
if we involve this Guardian's Law 527
and then uh this 528
if you solve this interior 529
and so the things are in your uh 530
cylindrical coordinate system because it is 531
it is exhibiting your cylindrical symmetry 532
remember this thing and uh 533
so this was electrical that would drive for this 534
for this this type of tea 535
for this type of your uh pyroplex okay 536
and if if you see that you can simplify this thing okay 537
you can simplify this thing that this is your 538
if you solve this integral 539
so you will find out this 2 Pyro L 540
and then if you know this electric field 541
you just plug in this electric field 542
in this relationship to find out this 543
again this DL 544
it is your 545
your this things are changing in your radial direction 546
remember 547
the things that changing in your radial direction 548
so that's why you're this DL 549
d L is to do that treated as d row 550
okay so you take the dark product 551
then uh what is left over here 552
so 553
you just need to solve the integration of one of our 554
row with this back to row 555
and head limits they're changing from B to a again 556
and uh you know that the integral of this 557
it is Eleanor Row okay 558
and then you need to simply plug in this uh 559
these ranges a and B to eliminate this negative sign 560
you can just reverse this ratio okay 561
so that is your lover to make drew 562
so now you have to 563
you have found out this V in terms of this Q 564
not involve your basic relationship 565
so q q will be cancelled 566
so here we have assume that your Q is constant okay 567
so this is your capacitors for your parallel blade 568
capacitor for your 569
this parallel coaxial coaxial conductors okay 570
so it is a coaxial capacitor example 571
or coexist cynentical capacity 572
or this is also you can assume like your cables 573
the cables the TV cables 574
a wire so in the center you are having the signal okay 575
in between you're having the dialective insulator 576
outer is your against shield 577
and then it is your plastic insulation right 578
so this is equirant thing 579
the third case uh 580
no in this case you are having to concentrate uh 581
spherical conductors here 582
so it is your third uh 583
uh uh 584
partner system 585
and here only the difference is the industry is uh 586
with the radius a and the outer sphere is uh 587
which radius b 588
they are separated again by dialectic medium 589
homogeneous dialect medium 590
medium with permetuity abslon 591
and again we are resuming that 592
the inner grade is having positive q charge 593
and the outer is having minus q charge 594
and now 595
if we apply the causing law to arbitrary guardian 596
spherical surface 597
so symmetry is spherical in this case and radiuses are 598
so for these cases we have already studied that 599
if we solve the secretion in the radio direction 600
so this is your 601
remember that this is your surface area of your sphere 602
so this was the electric vehicle that would 603
derived for this vertical corner system 604
in our chapter number four 605
now you know 606
we have found out this electrical 607
not involve this relationship 608
solve this thing here remember that this is only 609
the variations are 610
with respect to your radial direction that your Dr 611
and uh so you need to solve this integral 1 or R square 612
okay solve this integral with this spectral R plug 613
needs these values B to a 614
so this is your voltage in terms of share 615
this charge I know you can involve this relationship 616
that is your capacity it is 617
it is a ratio of charge 618
and devot it across that capacity 619
so this is very useful formula 620
and if if if we let the letter case tag this outer 621
this outer plate 622
it is a seriously approaching infinity 623
then it's uh resistance it will simply approach uh 624
4 by action on times the radius of your inner plate 625
which is the capacitors of a spherical capacitor 626
whose outer plate is infinitely large okay 627
infinity large 628
so just you need to put in this thing as infinity 629
anything divided by infinity it is equal to zero right 630
so this should go in the numerator 631
so then this is your very important example as 632
uh these are the just you can uh see that uh 633
your predact of your physics that uh 634
if the two capacitors with the capacitor 635
C1 and C2 are in Cds 636
have same this me 637
if they are in series then definitely 638
they're going to have the same charge on them 639
and uh for this case uh they're resistant capacitys 640
the total capacitys it can be found out by using this 641
case by using this uh 642
uh just the addressing program 643
of the things that we have done 644
with the parallel barrel resistors 645
that for for Cds capacitors 646
we would need to involve this 647
uh this uh if if if you call this 648
this uh this your uh harmonic additive 649
uh something like uh 650
so if you remember that 651
it is just like your harmonic mean 652
things like this 653
and if the two capacitors they are in parallel 654
so this means indeed this is the case for this case 655
the voltages are same across them 656
so just just need to add up these two uh 657
uh different capacitys 658
to find out the total capacitys of this circuit 659
okay so uh if you know this 660
this where I already discussed that uh uh 661
this is your homes law to 662
how to find out the resistance 663
by using this relationship of uh 664
voltage and current for this non uniform surface alias 665
circus alia plates and this is uh 666
for your capacitors 667
that it is the ratio of your charge on this uh 668
uh place and the what is difference between them 669
and if we take the product of them 670
okay let's take the product of them 671
and what you will see that these two things 672
what is going to cancel out 673
this thing is going to going to cancel out 674
okay 675
and this thing is going to cancel out 676
so what is what is your left out with 677
that 678
is the ratio of your permitivity and the conductivity 679
and this is known as the relaxation time of the medium 680
separating the two conductors okay 681
and this it is only 682
only when the medium is homogeneous 683
so it is only one single type 684
so the medium 685
which is residing in between these two place 686
so that is your time cost 687
and we also say that we also there said 688
we also say that it is relaxation time of the medium 689
separating them 690
so this is the property of your medium so uh 691
it is uh narrated it is narrated in terms of sure 692
this uh using this uh uh electrical 693
electrical uh properties of this medium 694
that is your conductivity and the permitivity 695
the ratio of these two uh quantities 696
it is equal to the relaxation time of the medium 697
and we have always studied in our previous lectures 698
that what is the significance of this relaxation time 699
that how the charges are going to decay 700
or how the electric fees are going to decay 701
and depending upon the relaxation time of this medium 702
so if if you remember 703
I guess we have left out with around 33% of the chart 704
okay left out 705
laptop okay 706
rodency or something like charge or things like 707
things like this 708
so if uh if uh 709
we assume that the it is a homogeneous medium and uh 710
if we need need to find out the resistance of uh 711
uh resistance of uh various capacitors 712
uh mentioned earlier can be obtained using this uh 713
relationship like this one 714
uh and if we have already derived that 715
you remember that 716
we have already derived these capacitansies 717
for the battle plate capacitor 718
slendrical capacitor and spherical capacitor 719
and if we involve this relationship then 720
then we can find out the resistance of the medium 721
remember that resistance of the medium 722
this resistance are in this equation 723
is not the resistance of the capacitor 724
remember that it is the just know that 725
it is not the resistance of the capacitor plate 726
but the leakage resistance between the players 727
therefore 728
stigma is the conductivity of dialect to medium 729
separating them 730
and abslon is the permituity of that medium 731
so this is regarding your medium related thing okay 732
the leakage resistance of your medium basically 733
some here today's lecture 734
uh today we uh for 735
for just for uh for generalized case uh case 736
uh uh this uh generalized cases 737
we determine that uh how to find out the resistance 738
and the capacity of these uh 739
capasters 740
for standard that is your barrel blade capaster 741
your work Z Capaster your spherical capaster 742
that how we can find out the capasters of these 743
different types of this gender capacitors 744
and then we found out how to 745
how we can determine this resistance of the medium 746
in between these two plates 747
next time we're going to solve the examples 748
by taking help of this boundary problem solution 749
and then we will find out the resistance 750
and the capacity of the conductors 751
here I thank you all 752
if you have any questions uh 753
they would be entertained uh 754
to your emails or online synchronous sessions 755
that we have arranged at the department