|4ート L2 A b b. a r a L+z, Figure 4: 4. (a) From the above figure aL2 20 = a tan a= (True, False) %3D La + 20 z = 2r- dz = 2dr (b) Assuming (a) is correct the resistance of the cone like part of the resistor shown above using Ohm's Law [J = oË] and the continuity equation Tin = lout = 1 = JJ. dà = JA = Jar? follows as: S Edz Sdz 1 %3D 2 dr Δν %3D dz L2 %3D 1 R2 = (True, False) %3D %3D 12 r2 παύσ (c) Since the current is common to both resistors the resistance of the cylinder of the radius a is R = Hence the total resistance is Taaa R + R2 (True, False)
|4ート L2 A b b. a r a L+z, Figure 4: 4. (a) From the above figure aL2 20 = a tan a= (True, False) %3D La + 20 z = 2r- dz = 2dr (b) Assuming (a) is correct the resistance of the cone like part of the resistor shown above using Ohm's Law [J = oË] and the continuity equation Tin = lout = 1 = JJ. dà = JA = Jar? follows as: S Edz Sdz 1 %3D 2 dr Δν %3D dz L2 %3D 1 R2 = (True, False) %3D %3D 12 r2 παύσ (c) Since the current is common to both resistors the resistance of the cylinder of the radius a is R = Hence the total resistance is Taaa R + R2 (True, False)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
answer if a and b true or false.
![TO Ja
- L2
r
A
В
b
b
r
a
L,+2,
Figure 4:
4.
(a) From the above figure
aL2
z0 = -a
tan a =
(True, False)
L2 + 20
L2
b-a
2 dr
r dz =
b-a
(b) Assuming (a) is correct the resistance of the cone like part of the resistor shown above using Ohm's Law [J = aĒ] and the continuity equation
Iin = Iout = I = [ J. dà = JA = Jar² follows as:
Δν
R2 =
( Edz
S dz
dr
a
dz
1
b
L2
(True, False)
-
%3D
I
I
r2
Tabo
(c) Since the current is common to both resistors the resistance of the cylinder of the radius a is R = L. Hence the total resistance is
L1
Taaa
R1 + R2 (True,False)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90ba196a-a4da-4c5b-97fd-f3d58be17337%2F29e76e36-53f6-4eec-ba7a-a713f6c2acb9%2Fq6t6ewi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:TO Ja
- L2
r
A
В
b
b
r
a
L,+2,
Figure 4:
4.
(a) From the above figure
aL2
z0 = -a
tan a =
(True, False)
L2 + 20
L2
b-a
2 dr
r dz =
b-a
(b) Assuming (a) is correct the resistance of the cone like part of the resistor shown above using Ohm's Law [J = aĒ] and the continuity equation
Iin = Iout = I = [ J. dà = JA = Jar² follows as:
Δν
R2 =
( Edz
S dz
dr
a
dz
1
b
L2
(True, False)
-
%3D
I
I
r2
Tabo
(c) Since the current is common to both resistors the resistance of the cylinder of the radius a is R = L. Hence the total resistance is
L1
Taaa
R1 + R2 (True,False)
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