oe 1 primjeri rjesenih zadataka 1
DESCRIPTION
....TRANSCRIPT
00
3
43
21
4321
24
22πε
πε
ϕϕ
ϕϕ
ϕϕ
ϕϕϕϕϕ
a
Qa
QBB
BB
BB
BBBBB
=
⋅
⋅==
=
−=
+++=
5
24
220
2
2
0
3
43
21
4321
⋅=
+⋅
⋅==
=
−=
+++=
πεπε
ϕϕ
ϕϕ
ϕϕ
ϕϕϕϕϕ
a
Q
aa
QAA
AA
AA
AAAAA
( )
( ) 33
12
9
0000
1088025.010764.2
1085.82.05
764.21085.8
5
551
5
1
5
⋅−=⋅−=⋅⋅⋅
−⋅⋅=
⋅
−⋅=
−⋅=−
⋅=
−
−
ππ
πεπεπεπεϕ
AB
AB
U
a
Q
a
Q
a
Q
a
Q
VU AB 880−=
1. pismeni
?
_________________
12
7
101
101
3
242
=
=
=
==
==−
−
F
VU
mmmd
mcmS
rε
nNF
F
Ud
S
dI
WIF
Ud
SCUW
rel
rel
446
10
11441071085.8
2
1
2
1
2
1
2
1
6
412
2
20
2
0
2
=
⋅⋅⋅⋅⋅⋅=
=⋅
⋅=
==
−
−−
εε
εε
_______________
2.0
85.8
BAAB
ma
nCQ
ϕϕϕ −=
=
=
8
1
2
4
2
0
2
0 =
⋅
⋅
⋅=
a
Qa
Q
E
E
B
A
πε
πε
CQ
CQ
µ
µ
220
160
1
2
=
= ⇒
VVC
QU
VC
QU
67.263
80
6
160
33.73
2
22
1
11
====
==
VUUU 10067.2633.7321 =+=+=
2. pismeni
2
0
2
0
2
0
2
0
230sin
24
2
430sin
42
a
Q
a
QE
a
Q
a
QE
B
A
⋅
⋅=°⋅
⋅
⋅=
⋅=°⋅
⋅⋅=
πεπε
πεπε
66
21
1
10
2
2
2
1
210
2
2
1
1
21011021
5663
10
1110
1
2
11
10
2
1
8010
6
1
3
13
60100
11
60106106.31032106.0
22
−−
−−−−
⋅=⋅
+
−=
+
−
=
++
=+=
+=⇒−=+−
=⋅=⋅=⋅⋅⋅⋅=
⋅⋅=⇒⋅
=
CC
C
QU
Q
C
Q
C
C
Q
C
QU
QQQQQQ
CQ
CWQC
QW
µ
provjera:
3. pismeni
CCC
BBB
AAA
r
k
rE
r
k
rE
r
k
rE
=⋅
=
=⋅
=
=⋅
=
0
0
0
2
2
2
πε
λ
πε
λ
πε
λ
kako je BCAB = ⇒
AA
A
C
C
A
CAB
rEk
r
r
E
E
rrr
⋅=
=
+=
2
mV
E
E
E
E
Er
rE
E
Err
rErrrE
r
kE
C
A
A
C
AA
AA
C
AAA
AA
CA
AA
BB 20
21
60
1
2
1
22
2
=+
=
+
⋅=
+⋅
⋅⋅=
+
⋅⋅=
+
⋅==
( )
kVkVUC
QU
cQQQ
cUCQ
nFU
QC
QcUCQ
Auk
ABAuk
AA
B
B
BABAB
5.1361024
10180
18014436
1441061024
24105.1
1036
36105.16108
9
6
0
0
39
00
3
6
0
0
0
39
1
=+⋅
⋅=+=
=+=+=
=⋅⋅⋅=⋅=
=⋅
⋅==
==⋅−⋅⋅==
−
−
−
−
−
µ
µ
µ
4. pismeni
mnc
db
d
/10
2
10214.310185.3
2
22
2
36
0
0
00
=
⋅⋅⋅⋅==
=
−−
λ
ε
πελ
πε
λ
ε
σ
Fµ2 Fµ1
VU d 401
= VUd 362
=
cQd µ802401 =⋅= cQd µ362 =
VUU
VC
QU
VC
QUQQ
d
uk
ukdduk
36
182
36
54
3
12
1036
22
1
1
6
12
2
==
===
=⋅
⋅==⇒=
−
( )( )
( ) ( )Ra
R
Rb
R
ili
mc
mC
R
U
ba
RaRb
S
Q
AB
BA
+−
+=−
−=
⋅−=
⋅−=⋅⋅−
⋅⋅⋅⋅⋅=
⋅−
++==
−
−
−−
−−−
ε
σ
ε
σϕϕ
µσ
σ
σ
εσ
22
2
26
12
22
2312
2
0
/04.0
/10039825.0
103982510
75
1030
103010601085.8
5. pismeni
( )
( )( ) ( )( )( )( )
π
πε
πεπε
πεπεϕϕ
2
0
00
00
4
4
44
)(44
75
RS
Uba
RaRbQ
RaRb
baQ
RaRb
RbRaQU
Ra
Q
Rb
QU
VU
BA
BA
ABBA
BA
=
⋅⋅
++=
++
−=
++
−−+=
−−
+=−=
=
cmmdC
Dd
D
d
SC
pFC
cmd
6.181859.04
4
___________
12
2
0
2
00 ==⋅
⋅⋅=⇒
⋅⋅=⋅=
=
=
πε
πεε
[ ]
[ ]mVa
QE
mVa
QE
/3004
/5.3374
2
20
22
2
10
11
=⋅
=
=⋅
=
πε
πε
6. pismeni
AsQ
AsQ
cmd
cma
cma
11
2
11
1
2
1
103
106
5
3
4
−
−
⋅−=
⋅=
=
=
=
iz slike imamo
20coscos2
9025169
2
2
2
1
2
2
2
1
2
22
2
2
1
πααα
α
=⇒=⇒⋅−+=
°=⇒=+
=+
aaaad
daa
[ ]mVE
EEEEEEE
EEE
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cos2
22
2
2
2
1121
2
2
2
1
21
=+=
+=⋅−+=
+=
α
4.275.0
8.1'
'
'
00 ===⇒⋅⋅=⋅
=
d
d
d
S
d
S
CC
rr εεεε
7. pismeni
2
912
0
0
0
00
101505.11085.8130m
AsEE
SSE
SQsdE
ZZ
ZZZ
Z
−− ⋅=⋅⋅=⋅=⇒=
⋅=⋅
⋅==∫
εσε
σ
ε
σ
ε
σ
ε
rr
NSd
USEDESF
VU
mmd
dmS
77.1......2
1
2
1
2
1
____________
4000
2
10
2
2
0
2
0
2
=====
=
=
=
εε
privlačna sila
?4
1041000
10410100444
____________
100
1
0
2
0
1
0
3
00
0
=⇒⋅
=
⋅=⋅=
⋅=⋅⋅=⋅⋅=⇒⋅
=
=
=
−−
RR
Q
rQr
Q
V
mmr
kapi
kapikapi
kapljicekapi
kapljicekapljicekapljicekapljice
kapljicekapljice
πεϕ
πε
πεπεπεϕπε
ϕ
ϕ
kVVR
Q
mRrRRr
VV
kapi
kapikapi
kapikapljice
1010104
104
4
1010003
4
3
41000
1000
4
2
0
2
0
0
23333
==⋅
⋅=
⋅=
=⇒⋅=⇒⋅=⋅⋅
=⋅
−
−
πε
πε
πεϕ
ππ
8. pismeni
21 UUU += 1
2
122211
21
EEEE
DD
r
r ⋅=⇒=
=
ε
εεε
mkV
dd
UEdEdEU
r
rr
r 5........
2
2
11
121
2
111 ==
+
=⇒+=
ε
εε
ε
9. pismeni
VQ
A2
103
1063
3
=⋅
⋅==
−
−
ϕ
KVpF
nC
C
QU
pFd
sC
mKVD
E
mCs
QD
nCCUQ
pFd
sC
pFd
sC
novi
r
24.35
8.70
4.3510
104001085.8
/2001085.8
1077.1
/1077.110400
108.70
8.7010108.70
45.1908.7069.2
8.70105
104001085.8
'
'
2
412
'0
12
6
0
26
4
9
123
0
'
3
412
0
===
=⋅
⋅==
=⋅
⋅==
⋅=⋅
⋅==
=⋅⋅==
=⋅==
=⋅
⋅⋅==
−
−−
−
−
−
−
−
−
−
−−
ε
ε
εε
ε
10. pismeni
( )2121
2
1
2
UUCQQQ
CC
UCQ
UCQ
eq
eq
+⋅=+=
⋅=
⋅=
⋅=
( )
VUUU
C
UUC
C
QUUU
CUQ
eq
eq
eqeq
80
2
40120
2
21
2121
===
+=
⋅
+⋅====
⋅=
kVS
Fd
S
FdU
Ud
S
d
UC
d
UQSEDF
rr
r
106.01085.84
42104.22
22
2222
212
3
00
2
2
2
0
2
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⋅⋅⋅=
⋅
⋅⋅=
⋅
⋅=
⋅⋅
=⋅
=⋅
=⋅⋅
=
−
−
πεεεε
εε
CCeq 3=
11. pismeni
ddd
X
X
Q
d
Q
X
QE
d
QE
2
2
22
442
_____________
42
4
2
2
0
2
0
2
0
2
0
===
=
=
=
πεπε
πε
πε
VU 2000' =
12. pismeni
U
d
s
Ud
s
C
CU
C
QU
UCQ
CUQ
2
20
0
''
''
'''
'
====
=
=
=
ε
ε
CCCC
CCCAB 2.1
6
5
32
32==
+
⋅=
d
SC 01 ε=
d
SdS
C
d
SdS
C
r0
02
002
6
2
''
2
2
'
εεε
εε
⋅==
⋅==
3
2
2
3
2
3
2
3
8
12
'''
'''
1
1
2
1
100
2
0
22
222
==
==
⋅
⋅
=+
⋅=
C
C
C
C
Cd
S
d
Sd
S
CC
CCC ε
ε
ε
13. pismeni
JU
d
SUCW r µεε 7.17......
22
2
0
2
==⋅=⋅
=
mkVEEEE
mkVEEEE
B
A
37.42.....)(2
1
37.42.....)(2
1
321
0
321
321
0
321
−==+−⋅
=−+−=
==+−⋅
=+−=
σσσε
σσσε
0
33
0
22
0
11
2
2
2
ε
σ
ε
σ
ε
σ
⋅=
⋅=
⋅=
E
E
E
14. pismeni
( )NF
Nba
QQkF
Nb
QQkF
FFF
uk
uk
5.17.22.1
2.15.0
104108109
7.22.0
104103109
2
669
2
3131
2
669
2
3232
3231
−=−=
=⋅⋅⋅
⋅=+
=
=⋅⋅⋅
⋅==
−=
−−
−−
(desno)
(u lijevo)
CCd
SC
d
SC
UCW
CC2
1
2
2
0
0
2
0
=⇒⋅
=
=
⋅=
ε
ε0
2
2
1
2
2
1
WUC
Wc ⋅=⋅
=⇒
15. pismeni
( )
( )
( ) ( )bdb
dbQ
bdb
bdbQ
bbd
Q
bdbbbd
QU
bbdQ
bdbQ
r
Q
ABBA
B
A
−
−=
−
+−=
−
−=
−+−−
−=−=
−
−=
−−=
=
2
22
11
2
1111
4
4
1
4
1
4
1
4
1
4
0000
00
00
0
πεπεπεπεϕϕ
πεπεϕ
πεπεϕ
πεϕ
C
QUQUCW
⋅=
⋅=
⋅=
222
22
a) ( )
0
22
0 42
2
2W
UCW
UCW a ⋅=
⋅⋅=⇒
⋅=
b) ( )
0
22
0 42
2
2W
C
QW
C
QW b ⋅=
⋅
⋅=⇒
⋅=
c)
21
2
2
21
222
21
21
21
121
2
CC
CU
CC
CCUQ
CC
CCU
CC
CCUQ
CUQ
+⋅=
+⋅⋅=
+
⋅⋅=
+⋅⋅=
⋅=
2
12
2
2
12
2
121
2
2
1
1
21
21
1
CQ
C
CQQ
QC
CQQ
C
CQQ
C
Q
C
Q
UU
QQQ
+⋅=
+⋅=
⋅=⇒=
=
+=
16. pismeni
( )
( )
0
222
3
2
33
213
0
2323
0
2323
13
13
4
129
2
3223
4
129
2
3113
0
9.7092.103
300
489.31730092.103
30060300
605.012030sin
92.103866.012030cos
300
0
12010900
104030109
1030
30010225
103025109
1015
3030
15arcsin
===
=+=+=
=−=+=
−=⋅=⋅−=
=⋅=⋅=
=
=
=⋅
⋅⋅⋅=
⋅=
=⋅
⋅⋅⋅=
⋅=
==
−
−
−
−
−
−
arctgF
Farctg
NFFF
NFFF
NFF
NFF
NF
F
NQQ
kF
NQQ
kF
X
Y
YX
YYY
Y
X
Y
X
β
α
položaj 1 položaj 2
→
13F→
3F
→
23F
β
( )
220
31
819
89
4
6
4
69
21
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109.9106.1
/109.901.01.01091064
1064
104
1040109
smm
eE
m
Fa
maF
mVEEE
⋅=⋅
⋅⋅⋅===
=
⋅=+⋅=
⋅
⋅+
⋅
⋅⋅=+=
−
−
−
−
−
−
JUCW
mKVd
UE
VC
Q
C
QU
r
5492
000
3
00
0
0
0
107.171016102125.22
1
2
1
/100104
400
4004100
−−
−
⋅=⋅⋅⋅==
=⋅
==
=⋅===
=
ε
nFCC
JCUW
mKVd
UE
CCUQ
nFd
sC
r
r
2125.24
1085.81
10425.4101085.82
1
2
1
/25104
100
885.0101085.8
85.8104
141085.8
9
0
5492
3
29
3
12
0
=⋅
==
⋅=⋅⋅==
=⋅
==
=⋅⋅==
=⋅
⋅⋅==
−
−−−
−
−
−
−
ε
µ
εε
17. pismeni
2
2
22
2
1
11
19
31
1
106.1
101.9
642
40
r
QkE
r
QkE
Ce
kgm
CQ
CQ
c
=
=
⋅=
⋅=
=
−=
−
−
µ
µ
( )
( )
mVE
EEEEE
mVEE
a
kE
a
kE
EEE
EEE
a
QkE
EEEEE
BDDCBDAC
BDAC
BD
AC
BDBD
ACAC
DCBA
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22
/103610
1021092
1021042
1011032
2
2
5
22
5
2
69
66
2
66
2
2
⋅=
==+=
⋅=⋅⋅⋅⋅
==
⋅−⋅=
⋅−⋅=
−=
−=
=
+++=
−
−
−−
−−
rrrrr
18. pismeni
Ck
REQ
R
Qk
R
QE 11
2
22
0
1033.54
−⋅=⋅
=⇒==πε
negativni naboj ravnomjerno raspoređen na površini kugle
→
CE →
DE
→
AE→
BE
Cµ1 Cµ2
Cµ3Cµ4
→
E
→
AGE→
BDE
ukQCQ =⋅= 11 100( )
FC
C
CC
CCC
CCQuk
µ1
1066
105.140410
4040100
40
1
6
1
6
11
211
21
=
⋅=⋅
⋅⋅+⋅=⋅
⋅+⋅=⋅
⋅+=
−
−
19. pismeni
[ ]CN
mV
Q
am
Q
FE 1010256.8 −⋅=
⋅==
IIE
EE
R
EEI
T
T
T
r
1510
1058
1000
2
22
++⋅=
=
+
−==
1. pismeni
V
VU
VUU
A
RA
R
8
80
83
2
1
1
−=
−=−=
==
ϕ
ϕΩ=
Ω=
−==
6
3||
12
1
32
R
RR
VCB ϕϕ
+ +
1R
32 || RRV12V12
A
CB ≡
R
UR
R
UII
3
2
2
1
2
111 =
+
==
2. pismeni
sklopka u položaju 1
sklopka u položaju 2
mAI
mAI
mAK
I
32
6
27
620
41
610
3
2
1
==
=−
=
=−
=
Ω===
=−+=−+=
=−−+
KmAI
R
mAIIII
IIII
R
A
R
R
23
6
3324
0
321
321
ϕ
R
U
RR
UR
R
UII
54
22
2
1
2
2222 =
+=
+
==
2
1
2
1
2
1
3
5
5
3
3
2
U
U
R
UR
U
I
I=== 5
2
5
3
3
2
2
1 ==U
U
3. pismeni
Ω=
=+⇒+
⋅=⇒
+=
===
4
242168
832
8
8
38
24
8
R
RRR
II
AU
I
A
V
( )( )( ) 261436433
361526522
342514511
IRIRIRRRE
IRIRIRRRE
IRIRIRRRE
+−++=
++++=
−+++=
++−=
−−−=−
+−=
+++=
−+=
+−=
++=
−+=
213
312
213
312
321
213
312
321
38.2
3396.9
38.2
32.3
36.3
5:
551514
651516
551518
III
III
III
III
III
III
III
III
( )
AIII
AI
AI
II
II
II
II
III
II
I
5.17.12.332.3
07.17.1
7.1
24.37.1
7.127.1
27.1
4|:848.6
12
11
11
21
21
=−=−−=
=−=
=
+−−=
−−−=
−−=
−−=
12
21
21
7.1
7.14
86
8.644
II
II
II
−=
=⋅
=+
=+
4. pismeni
+
V
R
R
S zatvorena
II
R
UR
U
I
I
R
UI
AA
A
A
3
2
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2
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Jednadžbe čvorova su:
AIAIAIAIAIAI 5.1,7.1,2.0,5.1,0,7.1 654321 ======
12
2
===
=
R
UII
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R
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5. pismeni
a) u seriju
Ω=
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150
15.085.4585.4
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a
a
R
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c 49.3582.0
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582.5
582.5
82.5
82.5
82.5
82.5
6. pismeni
položaj 1
položaj 2
Točke 1 i 2 su na istom potencijalu
20400 =⇒= nn
RnR
R
R/3
R
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0
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312.1078.00039.0175.0
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7. pismeni
AII
R
U
RR
RU
RR
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R
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eq
92
3
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3
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12
2
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8. pismeni
AI
AI
II
II
B
A
AB
BA
515.1
545.4
15450
1527100
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AI
AI
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2
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9. pismeni
( ) ( ) RR
RRRRR
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3
2
2
3
2
2
2
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+
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3
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600
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400
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10. pismeni
+
uR
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11. pismeni
VE
UIRE u
98.109
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AII
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333.06
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128000128030057000016004800006201178000
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1900
1600300
1002
3002
3002500
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12. pismeni
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=
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228001754.0
2
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13. pismeni
( )
100
1
20
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2
max
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E
TR
14. pismeni
KNkWhKNkWh
trošak
kWhhkWtPW
satix
WUIP
1.895.02.172
:
2.17890980.1
330
19809220
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20
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15. pismeni
mVU
mm
R
mm
R
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310103
10104105.2
105.2
63
625
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R
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R
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4.01.04
07.2
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16. pismeni
Ptica neće stradati.
( )
VU
kkIU
mAI
I
V
AV
A
A
312.4
105.20
5.11910854.05.11||9
854.0
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9cm(3x3)
125x3=375cm
+ +
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TR
125 ćelija x 0.8V = 100V
17. pismeni
18. pismeni
29375
05.110503503
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AmAmAxI
x
uk
=
===
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AI
UI
VU
AB
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VRIU u 4.523.066 36 =⋅−=−=
AIAIAI
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I
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1.13.08.08.03
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46633
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II
IIIIII
IIIIII
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3.0;8.03.01.1;1.1
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19. pismeni
VEE
IRIE
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600121283
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28
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1. pismeni
?
______________
5.17
20
15
5
26.10
3
2
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?
__________
20
10
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3. pismeni
Φ
t
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Φ−=
4. pismeni
( )ππ
µ
π
µµ
π
µ
ππ
+=
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=
=+⇒==
242
11
242
4
1
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5. pismeni
Vdt
diLe
Hl
SN
R
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m
162.0
26.1
6.15.02
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6. pismeni
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100
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100
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7. pismeni
?
__________
1
1
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π
µ
π
µ
π
µ
π
µ
π
µ
2ln1
2ln2ln
2ln2
ln2
2
22
0
00
00
2
0
0
⋅⋅⋅−=⇒=
⋅⋅⋅⋅
=
⋅
⋅−=
Φ−=
⋅=
⋅=
⋅
⋅=Φ
⋅⋅
==Φ
∫
kaeN
kaN
kta
dt
dN
dt
dNe
aI
a
aaI
x
adxI
adxx
IBdsd
a
a
?
__________
14.3
5
10
=
=
=
=
H
NF
AI
cml
( )
mAlI
FH
HlIlIBF
BlIF
/1052
10
1041.05
14.3 67
7
0
0
⋅==⋅⋅⋅
=⋅⋅
=
⋅⋅⋅=⋅⋅=
×⋅=
−πµ
µ
rrr
8. pismeni
utjecaja ima samo promjena struje i=kt rotacijom petlje ne mijenja se tok obuhvaćen petljom
AN
I
Az
BH
BH
BH
BBTB
lHlHlH
2
1021032104
785.0
103106
106
1012
1012
104
785.0
104
785.0
104106
785.0
1041012
785.0
785.0
2
34
7
3
2
2
2
2
7
7
0
00
72
20
22
72
10
11
210
002211
=Θ
=
⋅=⋅⋅⋅
=Θ
⋅+
⋅
⋅+
⋅
⋅⋅
⋅=Θ
⋅==
⋅⋅⋅==
⋅⋅⋅==
===
++=Θ
−
−
−−
−
−
−
−
π
π
πµ
πµµ
πµµ
?
__________
1000
600
1200
5.1
6
12
875.0
2
1
0
2
1
0
=
=
=
=
=
=
=
=
I
N
mml
cml
cml
TB
µ
µ
9. pismeni
?
?
________________
500
50
4.0
2.0
100
10
2
1
2
1
=
=
=
=
=Φ
=Φ
=∆
=
e
L
mAI
mAI
mVs
mVs
mst
N
mVVt
Ne
mHI
NL
t
INe
2002.010100
102.010
44.410450
102.010
3
3
3
3
==⋅
⋅⋅=
∆
∆Φ=
=⋅
⋅⋅=
∆
∆Φ=
∆
∆−=
−
−
−
−
10. pismeni
?
_____________
10
2
/4
1
=
=
Ω=
=
=
B
AI
R
smv
cml
A
a ( )
Tlv
RiB
Rie
BvlBvle5.0
1014
210102
3
=⋅⋅
⋅⋅=
⋅
⋅=⇒
⋅=
⋅⋅=×=−
−rrr
TBBB
AIII
5.0
5
12
12
=−=∆
=−=∆
R
IH
π41 =
R
IH
24
32 ⋅=
03 =H
?
__________
1.1
6.0
6
1
100
20
2
1
2
1
2
=
=
=
=
=
=
=
L
TB
TB
AI
AI
N
cmS
mHHLI
BSNL
ILBSN
ILNt
IL
tNe
2002.0
5
5.01020100 4
==
⋅⋅⋅=
∆
∆⋅⋅=
∆⋅=∆⋅⋅
∆⋅=∆Φ⋅
∆
∆−=
∆
∆Φ−=
−
11. pismeni
+=
⋅
⋅+=
2
31
48
3
4 ππ R
I
R
I
R
IH uk
11
13
33
11
12
22
111
00
164
4
42
2
φµµφ
φµµφ
µφ
µµµµφ
=⋅⋅
⋅=⋅⋅
⋅=
=⋅⋅
⋅=⋅⋅
⋅=
⋅⋅
⋅=
⋅⋅
=⋅=⋅=
Sl
INS
l
IN
Sl
INS
l
IN
Sl
IN
Sl
INSHSB
srsr
srsr
sr
srrr
16:4:1:: 321 =φφφ
11lH
22lH
Φ
Θ
AI
lHlHIN
mAB
H
mAB
H
TS
B
TS
B
r
r
247.11000
2.07.31848.02.796
/7.3184500104
2
/2.7961000104
1
2105.12
105.2
11025
105.2
2211
7
20
22
7
10
11
4
3
2
22
4
3
1
11
21
=⋅+⋅
=
⋅+⋅=⋅=Θ
=⋅⋅
==
=⋅⋅
==
=⋅
⋅=
Φ=
=⋅
⋅=
Φ=
Φ=Φ=Φ
−
−
−
−
−
−
πµµ
πµµ
?
____________
500
1000
5.12
25
20
80
5.2
1000
2
1
2
2
2
1
2
1
=
=
=
=
=
=
=
=Φ
=
I
cmS
cmS
cml
cml
mVs
N
r
r
µ
µ
12. pismeni
13
12
23
12
321
4
2
2
2
?::
NN
NN
=
=
=
=
=
µµ
µµ
φφφ
⊗⊗
1I
2I
a
b
a
2I
ABFBCF
?
____________
10
20
10
30
2
1
=
=
=
=
=
F
AI
cma
cmb
AI
?
___________________
/10450
15
7
21
=
⋅=
==
−
d
mNl
F
AII
m
l
FI
d
d
I
d
I
d
I
l
F
110450
225102102
1022
104
2
7
727
27
272
0
=⋅
⋅⋅=
⋅⋅=
⋅⋅=⋅
⋅⋅=
⋅
⋅=
−
−−
−−
π
π
π
µ
13. pismeni
?
__________
4
6
20
2
1
=
=
=
=
x
AI
cmr
AI
cmI
Irx
r
I
x
I
HHHHHr
IH
x
IH
uk
55.94
06.020
2
2
22
0
2
2
2
121
2101
20
11
=⋅
=⋅
⋅⋅=⇒=
=⇒=−=
=
=
πππ
π
VvlBe 125080105 5 =⋅⋅⋅=⋅⋅= −
21
210
21
0
2
0
1
021
4
11
444 RR
RRI
RR
I
R
I
R
IBBBC
+⋅=
+⋅=+=+= µ
µµµ
BCADuk
BCAD
CDAB
FFF
FF
FF
−=
>
== 0
( ) ( )ba
aII
ba
IIaBaIF
b
aII
b
IIaBaIF
BC
AD
+⋅
⋅⋅⋅=
+⋅
⋅⋅⋅=⋅⋅=
⋅
⋅⋅⋅=
⋅
⋅⋅⋅=⋅⋅=
π
µ
π
µ
π
µ
π
µ
22
22
210102
''
12
210102
'
12
( ) ( )
NNNF
bba
a
ba
aII
bab
aIIF
uk
uk
µ
π
π
π
µ
π
µ
8010810800
103010102
104001030104
2
11
2
57
22
47210210
=⋅=⋅=
⋅⋅⋅⋅
⋅⋅⋅⋅⋅=
⋅+⋅
+⋅
⋅⋅⋅=
+−⋅
⋅⋅⋅=
−−
−−
−−
14. pismeni
1R
2R
CI
?
_____________
80
105
/250/900
5
=
=
⋅=
==−
e
md
TB
smhkmv
15. pismeni
Br
Fr
vrQ
uF vFII
0=B
a
mAR
eI
Vt
Ne
Wb
WbSB
aS
50100
5
51.0
105100
105
0
105101005.0
3
3
12
2
34
1
2
===
=⋅−
⋅−=∆
∆Φ−=
⋅−=Φ−Φ=∆Φ
=Φ
⋅=⋅⋅=⋅=Φ
=
−
−
−−
mWtRIW 251.0100102500 62 =⋅⋅⋅=⋅⋅= −
?
_____________
55.1
105
/120/432
5
=
=
⋅=
==−
F
mCQ
TB
smhkmv
( ) NBvQBvQF 6853 103.9109301051201044.1 −−−− ⋅=⋅=⋅⋅⋅⋅=⋅⋅=×⋅=rrr
_________
100
1.0
5.0
100
10
Ω=
=∆
=
=
=
R
st
TB
N
cma
smjer struje podržava opadajući tok. sile donje i gornje grane se poništavaju
( )HL
i
teL
dt
diLe
2.350
160
103020
102083
3
==⋅−−
⋅⋅−=
∆
∆⋅−=⇒−=
−
−
πρ
κκκ
π
ρε
πρεε
2
314
44
22
0
2
20
2
0
4
/10184
2
1
2
1
d
I
S
IEE
S
IEJ
mJd
I
d
IE
V
Ww el
el
=⋅
=⇒⋅=⇒⋅=
⋅==
=== −
π
µ
πµ
π
µ
πµ
µ
µ
d
IIB
mJd
I
d
IB
V
Ww
d
mm
0
2
0
3
22
2
0
22
0
22
0
0
2
2
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==
=====
sila na desni rub je 0 sila na lijevi rub iznosi NBlINF 25.05.010101050100 23 =⋅⋅⋅⋅⋅=⋅⋅⋅= −− 16. pismeni
?
__________
30
20
20
8
2
1
=
=
−=
=∆
=
L
mAI
mAI
mst
VU
__________
20
3
AI
mmdV
Ww
=
=
=
17. pismeni
petlja se giba
prema gore
5cm
5cm
10cm
1Fr
2Fr
3Fr
4Fr
Br
Br
1Ir
2Ir
d
lIIF
BlIF
⋅
⋅⋅⋅=
×⋅=
π
µ
2
)(
210
rrr
R
I
R
II
R
I
RBBB
R
IB
unuk8
216444
3
4
2
0000
00
µµµµ
µ
=⋅=⋅−⋅=−=
=
I4
3
I4
1
I IR
?
______________
1021
=
==
F
AII
31
42 0
FFF
FF
uk
rvr
rr
+=
=+
NF
NF
5
2
227
2
5
1
7
2
227
1
10210102
101010104
10410
10104
1052
101010104
−
−
−−
−
−
−
−
−−
⋅=⋅⋅
⋅⋅⋅⋅=
⋅=⋅
⋅⋅=
⋅⋅
⋅⋅⋅⋅=
π
π
π
π
π
π
NFFFuk5
31 102 −⋅=−= prema gore
( )SS
dS
dS
l
SNL
9´
4
3
42
2
0
2
=⇒
=′
=
⋅=
π
π
µ
18. pismeni
1 zavoj ⇒ dπ N zavoja ⇒ Ndπ (ukupna duljina žice) Ndπ zavoja sada je raspoređeno na Ndπ N´3dπ
Ndπ = Ndπ 3
'N
N =⇒ (namotano je tri puta manje zavoja)
duljina žice ostaje ista
Ll
SN
l
SNL 3
3
1
99
'
'''
00
2
=
⋅⋅
=⋅⋅
=
µµ
duljina ll3
1'= jer je namotano tri puta manje zavoja
TB
cmr
AI
Z41055.0
5
55
−⋅=
=
=
ZBB
Tr
IB
4
102.210221052
55104
2
45
2
7
0
=
⋅=⋅=⋅⋅
⋅⋅=
⋅= −−
−
−
π
π
π
µ
19. pismeni
rotirano za 90°
27.6386.4002'
4
10910104
3.01084.11'
''
437
3
0
0
2
==
⋅⋅⋅⋅
⋅⋅=
⋅⋅
⋅=⇒
⋅⋅⋅=
−−
−
N
S
lLN
l
SNL
r
r
ππ
µµ
µµ
AI
AI
cmd
15
20
20
2
1
=
=
=
TB
B
r
I
r
IBBB
uk
uk
uk
5
6
2
1
2
1
7
2
2
2
2
1
102
2
2
1
105
62510210
15
10
20
2
104
2
−
−
−−
−
⋅=
⋅⋅=
+
⋅=
+
=+=
π
π
π
µ
01000
?'
'
?
__________
3
30
2000
µµ =
=
=
=
=
=
=
r
N
LL
L
cmd
cml
N
mHHL
l
dN
l
SNL
84.111035.1184
3.0
4
109104104
4
5
467
2
0
2
0
2
=⋅=
⋅⋅⋅⋅⋅
=⋅⋅
=⋅⋅
=
−
−− π
ππ
µµ
N' = 64 puna zavoja
