Page # 14 Solutions Slot – 2 (Physics)

EXERCISE – III SUBJECTIVE QUESTIONS

1. Net charge per second = i = icu  + iSo4 2 X b
icu  = iSo  = i1 4V
4
+
i = 2i1 = 0.5 – 3 3 5
–
0 .5 + 2V 4
 i1 = i 4V i
7.
2 8
i1t 0.5 5 2 a y
So total Cu++ =  10 = × 1019 i=
=
ze 2e 3.2 5

e i V 2 4
2. No. of = = Vx + 3 × +4–3× = Vy
sec e Re 5 8
20 2 6 3
= 19 =  1017 Vx + +4– = Vy
1000  1.6  10 1.6 5 2
Vx – Vy = 3.7 V
i V 20 No current in branch ab so there is no change
J= = 2 =
A Rr 1000    10 8 in Vx – Vy.

2
=  10 6 Amp/m2 4 10V 6V

3. V = IR = 10 × 5 = 50 Volt
2A 6

V 50
E= = = 25 V/m 4V 10
 2 8.
4. (a) V = E – ir = 12 – 90 × 5 × 10–2 10
V = 7.5 volt O O O
E
(b) i =
r R
For imax  R = 0 100
A
E 12
So imax = = = 24 mA
r 500
(c) greater then 12 V
R 50

x 9.
+ –
300
3 6 10 1.5 V

5. 4.5 3 Open
4.5V 1.5
3V i=
450
O
1
i= Amp.
x  4.5 x 0 x 3 300
+ + =0 Closed
3 6 10
x =3 1
volt A O
3 1/300
33
So i10 = = OA
10 1 R
volt O
6. Using KVL 3
30 – 2 – 1 – V1 – 3 – 5 = 0 1
 V1 = 19V volt O
3 300 1.5V
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 Solutions Slot – 2 (Physics) Page # 15

1 1 60  20
×R=  R = 600 I= = 2.5 Amp
6  300 3 16
1 1
= 1.5 – i × 300 I1 = × 2.5 = 0.5 Amp.
3 5
V = 7 × 0.5 = 3.5 Volt
7 x
i=
6  300 i4 i3 i2 i1
10. For imin Rmax . So S1 is open, 10 2 1
5 5A
S2 is open & S3 is open 10A
13.
6 3 2A
6 i1 + i2 + i3 + i4 + 5 = 10
1A 1A
1 9 x x x x
1 1 + + + + 5 = 10
2A 1 2 10 5
24V 50 25  50  1 5
x= = V, i10 =   = A
O O O 18 9  18  10 18
500 2V 2V
G
5.5V i i 2V
12V
R
14.
A
O O
11.
12  2 1
i= = A
r 500 50
10  20 20
 2 1
20  10 3  =  R = 100
R 50
E
imax =
20 A
 rmin
3
2
5.5
= = 0.825 A
20
0 1
O 1
3 1 1

E 5 .5 15. x x
imin = = = 0.15 A
20 20 1
 rmax  30 2 2
3 3
y 1 y
5 7
60 A

2
I 0.4
41 7
12. 20 O 1
A 1
4 8
1 1  1
2 10 O
12  48 2
Req. = + 0.4 + 6 = 16  2
12  48

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 Page # 16 Solutions Slot – 2 (Physics)

16. 22. (a) P = VI

 1000 = 220 × I
50
 I= A = 4.55 A
11
V2 220  220 22  11
(b) R = = =  = 48.4
P 1000 5
B (c) P = 1 kWatt
A
P 1000 cal 240cal
(d) Joule/sec. = _
~
Folding symmetry 4.2 4.2 sec . sec
dm
(e) P = L
dt
dm dm 12
240 = 540 ×  = gm / sec
17. FD & EC will be removed due to symmetry dt dt 27
because D & F are symmetric to E & C.
12 80
2 1 × 60 gm/minute = gm / min ute
18. 27 3
 Req = 3 iR
23. 60 = × 100  3 (R + r) = 5R
i(R  r )
E 6
i= R = = 2A 3r = 2R
eq 3
i(6R) 6R
1 1 1 now  =
i(6R  r )
× 100 = × 100= 90
i1 : i2 : i3 :: : :  1 : 2 : 1 2R
8 4 8 6R 
3
1 24. (a) E = 10V in each case
 i1 = i3 = × i = 0.5 A (b) (A) as source (B) as load
4 (C) (A) E – ir
2 = 10 –1
i2 = i = 1A = 9V
4
(B) E + ir
2
E2
10  E  = 10 + 1
19. =   R = (when in series) = 11 volt
3  3R  9R (D) (A) output (B) Output
P = VI P = VI
E2 = 9 × 1 = 9 watt = 11 × 1
 = 30
R = 11 watt
(E) (A) i 2R (B) i2R
E2 = 1 Watt = 1Watt
Parallel = 30
R (F) EI = 10 × 1 = 10 (in ech case)
So total 3 × 30 = 90 watt (G) (A) 10 – ir – V = 0
25 25 25 V = 10 – 1
25 = 9 volt
10V 5V 20V 30V (B) 10 + ir – V = 0
25V
15 30 5V 55 V = 11 volt
(H) (A) P = – [Vbox I] (B) Vbox I
20. 5 10 5 11 = – [9 × 1] = 11 × 1
3A 3A 1 5A = – 9 Watt = 11 Watt
O O O i3 5 i4 5
2 2
      6
21.   . R . t =   . R2 . t
 R1  r   R2  r 
1
4
i2
i5
  R2 i1 2
R1  r = R2  r . R1 25.
i
 r= R1R 2
V

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 Solutions Slot – 2 (Physics) Page # 17

1 1 1 5
i1 : i2 : i3 = : :  = 15 : 5 : 6 5+ × 1 = 6 Volt
2 6 5 1 R
5 = 1 + R  R = 4
15 5 6 4
i1 = × u, i2 = , i3 = × i, i4 = i
26 26 26 9 30. R1 × 60 R2 × 40 ....(1)
5 R 2  10
i5 = check i2R for all R1 × 50 = R  10 × 50 ....(2)
9 2

5A 24/5 2V

26.
10m
96 480 O
O
2volt
96 1 24 31. 30
i= Ampere  + R = 96  R = 20
480 5 5 G
480
1 1.5 volt
i2
10m  2volt
G 1 volt  5 meter
i 20 i1
1.5 volt  7.5 meter
27.
(a)
20V
2V 5
20 20  500 50 A
i= = , i=
10  20  480 5000  9600 73
500 2
 0  10V 35
10 10m
O 30
2
 30
35
12 G
2V  volt
5V 7
1 O
3V 3
28. V G
+ – B 12 70
2V volt  10m, 1 volt  m
1 7 12
A
3 70
4V 5 1.5 V  × 1.5 m
10V 12
10 O (b)

40 cm
10
O
29. G
5V
– +
10 G
5 1
5
5  1  6 Volt
R 1 R 5

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