Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book
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570 12 Electric Circuits
∴ 10 = i
1
+ 30i
or 31i
1
+ 30i
2
= 10 (3)
where we have used (1). Solving (2) and (3)
i
1
=−3.04 A, i
2
= 3.48 A
Power dissipated through E
1
is i
2
1
r
1
= (3.04)
2
× 1 = 9.24 W
Power dissipated through E
2
is i
2
2
r
2
= (3.48)
2
× 2 = 24.2W
Power dissipated through R is i
2
R = (0.44)
2
× 30 = 5.8W
(∵ i = i
1
+i
2
=−3.04 + 3.48 = 0.44 A)
12.44 Referring to Fig. 12.44
i
1
+i
2
= i (junction theorem) (1)
Traversing the loop BE
2
AE
1
B counterclockwise the loop theorem gives
ξ
2
−r
2
i
2
− (ξ
1
−r
1
i
1
) = 0
∴ 1 − 2i
1
− (2 − 1i
1
) = 0
or i
1
− 2i
2
= 1(2)
For the loop BE
1
ARB
ξ
1
−i
1
r
1
−iR = 0
∴ 2 − 1i
1
− 10i = 0
or 11i
1
+ 10i
2
= 2(3)
Fig. 12.44
12.3 Solutions 571
where we have used (1). Application of Kirchhoff’s laws to the loop
BE
2
ARB does not yield anything extra. From (1), (2) and (3) we find
i
1
= 0.44 A, i
2
=−0.28 A and i = 0.16 A. The negative sign of i
2
shows
that its direction is opposite to that has been assumed.
12.45
i
1
+i
2
+i
3
= i = 5A (1)
As ρ = const and R = π d
2
/4 (Fig. 12.45)
R
1
: R
2
: R
3
=
l
1
d
2
1
:
l
2
d
2
2
:
l
3
d
2
3
∴ R
1
: R
2
: R
3
=
2
3
2
:
3
4
2
:
4
5
2
=
2
9
:
3
16
:
4
25
(2)
Since P.D across all the resistors is identical,
i
1
R
1
= i
2
R
2
= i
3
R
3
(3)
∴ i
2
= i
1
R
1
R
2
=
2
9
×
16
3
i
1
=
32
27
i
1
(4)
i
3
= i
1
R
1
R
3
=
2
9
×
25
4
=
25
18
i
1
(5)
∴ i
1
+
32
27
i
1
+
25
18
i
1
= i = 5(6)
∴ i
1
= 1.4A, i
2
= 1.66 A, i
3
= 1.94 A
where we have used (1), (4) and (5).
Fig. 12.45
12.46 The P.D across 8 and 2 resistors are equal, Fig. 12.46:
2i
2
= 8i
3
(1)
As P.D across 3 and 6 resistors are equal
3i
4
= 6i
1
(2)
As P.D across 4 and 6 resistors are equal
6i
6
= 4i
5
(3)
572 12 Electric Circuits
Fig. 12.46
Applying junction theorem
i = i
1
+i
2
+i
3
+i
4
(4)
i
2
+i
3
= i
5
+i
6
(5)
V
AB
= V
AD
+ V
DB
∴ 3i
4
= 8i
3
+ 4i
5
(6)
Applying the loop theorem to CAFBC
ξ − ir −3i
4
= 0
2
3
i +3i
4
= 1.8(7)
Solving (1), (2), (3), (4), (5), (6) and (7), i
4
= 0.4 A and i = 0.9 A. Applying
the loop theorem to the entire circuit
ξ − ir −iR = 0(8)
where R is the equivalent resistance of the circuit
R =
ξ
i
=
1.8
0.9
−
2
3
=
4
3
Power dissipated in the entire circuit is
P = i
2
(R +r) = (0.9)
2
4
3
+
2
3
= 1.62 W
12.47 Let A and B be the midpoints of the coils. As no current flows through the
galvanometer, P.D across AB i s zero. Applying the loop theorem to the main
circuit, Fig. 12.47
12.3 Solutions 573
Fig. 12.47
ξ
1
+ ξ
2
−i(R + R + r
1
+r
2
) = 0
∴ i =
ξ
1
+ ξ
2
2R +r
1
+r
2
=
1.5 + 3
2 × 10 + 5 + r
2
=
4.5
r
2
+ 25
(1)
The P.D of the point B with respect to the negative terminal of the first cell is
V
B
= ξ
1
−
10
2
+r
1
i = 1.5 −10 i (2)
The P.D of the point A with respect to the negative terminal of the first cell is
V
A
= ξ
1
+ ξ
2
−
10
2
+r
2
+ 10 +r
1
i
= 1.5 +3 −(5 +r
2
+ 10 + 5)i = 4.5 −(20 +r
2
)
But V
B
= V
A
(∵ no current flows through the galvanometer)
∴ 1.5 − 10i = 4.5 −(20 +r
2
)i
∴ i =
3
10 +r
2
(3)
From (1) and (2), r
2
= 20
12.48
(a) (i)
R
eq
=
R
1
R
2
R
1
+ R
2
+
R
3
R
4
R
3
+ R
4
=
8 × 2
8 + 2
+
3.2 × 3.2
3.2 + 3.2
= 3.2k
(ii)
V
p
1
= 320 ×
1.6
3.2
= 160 V
(iii)
V
p
2
= 320 ×
1.6
3.2
= 160 V
574 12 Electric Circuits
(iv)
I
T
=
ξ
R
eq
=
320
3.2 × 10
3
= 0.1A
I
1
= I
T
×
R
2
R
1
+ R
2
= 0.1 ×
2
8 + 2
= 0.02 A
I
3
= I
T
×
R
4
R
3
+ R
4
= 0.1 ×
3.2
3.2 + 3.2
= 0.05 A
(v)
W = I
2
T
R
eq
= (0.1)
2
× 3.2 × 10
3
= 32 J
W
3
= I
2
3
R
3
= (0.05)
2
× 8 × 10
3
= 20 J
(b)
P =
ξ
2
r
=
(24)
2
0.01
= 5.76 ×10
4
J
12.49
(i)
i =
ξ
R +r
=
24
140 + 0.02
= 0.1714 A
(ii)
V = ξ −ir = 24 −0.1714 × 0.02 = 23.9966 V
(iii)
P
R
= i
2
R = (0.1714)
2
× 140 = 4.113 W
P
r
= i
2
r = (0.1714)
2
× 0.02 = 5.87 × 10
−4
W
(iv)
V =ξ =24 V. Full voltage is available in the absence of load resistance.
12.50
(i)
R
eq
=
R
1
R
2
R
1
+ R
2
+ R
5
+
R
3
R
4
R
3
+ R
4
=
80 × 80
80 + 80
+ 20 +
40 × 40
40 + 40
= 80 k
(ii)
I
T
=
V
R
eq
=
300
80 × 10
3
= 3.75 ×10
−3
A
I
1
= I
T
×
R
2
R
2
+ R
1
= 3.75 ×10
−3
×
80
80 + 80
= 1.875 ×10
−3
A
I
3
= I
T
×
R
4
R
4
+ R
1
= 3.75 ×10
−3
×
40
40 + 40
= 1.875 ×10
−3
A
(iii)
V
1
=
V
R
eq
×
R
1
R
2
(R
1
+ R
2
)
=
300
80
×
80 × 80
(80 + 80)
= 150 V
V
2
=
VR
5
R
eq
= 300 ×
20
80
= 75 V
V
3
=
V
R
eq
×
R
3
R
4
(R
3
+ R
4
)
=
300
80
×
40 × 40
(40 + 40)
= 75 V
(iv)
P
5
= I
2
T
R
5
= (3.75 ×10
−3
)
2
× 20 × 10
3
= 0.281 W
P = I
2
T
R
eq
= (3.75 ×10
−3
)
2
× 80 × 10
3
= 1.125 W
12.3 Solutions 575
Fig. 12.48
12.51
i
1
= i
2
+i
3
(junction theorem) (1)
Traversing the top loop counterclockwise (Fig. 12.48)
ξ
1
−i
2
R
3
− ξ
3
−i
2
R
2
− ξ
2
−i
1
R
1
= 0 (loop theorem)
∴ 13 − 300 i
2
− 3 − 200 i
2
− 30 − 100 i
1
= 0
or 100 i
1
+ 500i
2
+ 20 = 0(2)
Traversing the bottom loop counterclockwise
ξ
4
−i
3
R
4
− ξ
5
+i
2
R
2
− ξ
3
+i
2
R
3
= 0
∴ 5 − 400 i
3
− 8 + 200 i
2
− 3 + 300 i
2
= 0
or 500 i
2
− 400 i
3
− 6 = 0(3)
Required equations for the unknown currents are (1), (2) and (3).
12.52
(i) Kirchhoff’s rule 1 (junction theorem) At any junction of an electric
network (branched circuit) the algebraic sum of the currents flowing
towards that junction is zero (Fig. 12.49).
Kirchhoff’s rule 2 (loop theorem)
Sum of the changes in the potential encountered in traversing a loop
(closed circuit) in a particular direction (clockwise or counterclockwise)
is zero.
If a resistor is traversed in the direction of the current, the change in the
potential is −iR, while in the opposite direction it is +iR.
If a seat of emf is traversed in the direction of emf, the change in poten-
tial is +ξ, while in the opposite direction it is −ξ.
576 12 Electric Circuits
(ii)
i
1
= i
2
+i
3
(junction theorem) (1)
Traversing the top loop counterclockwise
− ξ
1
−i
2
R
2
+ ξ
3
+ ξ
2
−i
1
R
1
= 0 (loop theorem)
∴ −11 − 200 i
2
+ 33 + 22 − 100 i
1
= 0
or 100 i
1
+ 200i
2
− 44 = 0(2)
Traversing the bottom loop counterclockwise
− ξ
4
−i
3
R
3
+ ξ
5
− ξ
3
+i
2
R
2
= 0
∴ −44 − 300 i
3
+ 55 − 33 + 200 i
2
= 0
or 200 i
2
− 300 i
3
− 22 = 0(3)
Equations (1), (2) and (3) are the required equations in the three
unknown currents i
1
, i
2
and i
3
.
Fig. 12.49
12.53
(i) The equivalent resistance of the circuit is
R
eq
= R
1
+ R
2
+
R
3
R
4
R
3
+ R
4
= 5 +10 +
20 × 60
20 + 60
= 30 k
V
1
= V ×
R
1
R
eq
= 300 ×
5
30
= 50 V
V
2
= V ×
R
2
R
eq
= 300 ×
10
30
= 100 V
12.3 Solutions 577
(ii)
V
p
= V ×
R
3
R
4
R
eq
(R
3
+ R
4
)
= 300 ×
20 × 60
30(20 + 60)
= 150 V
(iii)
I
T
=
V
R
eq
=
300
30 × 10
3
= 0.01A
I
2
= I
T
R
4
R
4
+ R
3
= 0.01 ×
60
60 + 20
= 0.0075 A
I
4
= I
T
R
3
R
4
+ R
3
= 0.01 ×
20
60 + 20
= 0.0025 A
12.54
i
1
−i
2
−i
3
= 0 (junction theorem) (1)
Traversing clockwise the top loop (Fig. 12.50)
Fig. 12.50
ξ
1
+i
1
R
1
− ξ
2
− ξ
4
+i
2
R
2
+ ξ
3
= 0 (loop theorem)
∴ 5 + 10i
1
− 10 − 20 + 20i
2
+ 15 = 0
or i
1
+ 2i
2
− 1 = 0(2)
Traversing clockwise the bottom loop
− ξ
3
−i
2
R
2
+ ξ
4
− ξ
6
+i
3
R
4
+ ξ
5
+i
3
R
3
= 0 (loop theorem)
∴ −15 − 20i
2
+ 20 − 30 + 40i
3
+ 25 + 30i
3
= 0
or 7i
3
− 2i
2
= 0(3)
The required equations are (1), (2) and (3) in three unknown currents.
578 12 Electric Circuits
12.55
i
1
−i
2
−i
3
= 0 (junction theorem) (1)
Traversing the top loop clockwise (Fig. 12.51)
ξ
4
−i
1
R
4
− ξ
5
−i
2
R
3
− ξ
2
−i
2
R
2
= 0
∴ 50 − 200i
1
− 30 − 80i
2
− 10 − 120i
2
= 0
or 20i
1
+ 20i
2
− 1 = 0(2)
Traversing the outer loop clockwise
ξ
4
−i
1
R
4
− ξ
5
−i
3
R
5
− ξ
3
−i
3
R
1
+ ξ
1
−i
3
R
6
= 0
50 − 200i
1
− 30 − 100i
3
− 20 − 60i
3
+ 40 − 40i
3
= 0
or 5i
1
+ 5i
2
− 1 = 0(3)
Fig. 12.51
The required equations are (1), (2) and (3) in three unknown currents i
1
,
i
2
and i
3
. Note that by considering the bottom loop no new information is
provided as it is already contained in the other two loops. In general it is
sufficient to consider any two loops out of three.
12.56 Traversing the loop ABCDA clockwise
−iR−i
1
R
1
+ ξ
2
−ir
2
− ξ
1
−ir
1
= 0
i
1
= 2i/3, i
2
= i/3
∴ −4.5 i −3 ×
2 i
3
+ 8 − i −4 −0.5i = 0
∴ i = 0.5A, i
1
= 0.33 A, i
2
= 0.165 A
P.D over E
1
V
1
= ξ
1
+ir
1
= 4 +0.5 ×0.5 = 4.25 V
P.D over E
2
V
2
= ξ
2
−ir
2
= 8 −0.5 ×1 = 7.5V
Chapter 13
Electromagnetism I
Abstract Chapters 13 and 14 are devoted to electromagnetism concerned with
motion of charged particles in electric and magnetic fields, Lorentz force, cyclotron
and betatron, magnetic induction, magnetic energy and torque, magnetic dipole
moment, Faraday’s law, Hall Effect, RLC circuits, r esonance frequency, Maxwell’s
equations, Electromagnetic waves, Poynting vector, phase velocity and group veloc-
ity, dispersion relations, waveguides and cut-off frequency.
13.1 Basic Concepts and Formulae
Motion of Charged Particles in Electric and Magnetic Fields
Assuming that the magnetic field (B) acts perpendicular to the plane of orbit of a
particle of charge q and mass m moving with velocity v, the radius of curvature (r)
is given by
r =
mv
qB
(13.1)
the angular velocity by
ω =
v
r
=
qB
m
(13.2)
the frequency by
f =
ω
2π
=
qB
2πm
(cyclotron frequency) (13.3)
the kinetic energy (K )by
K =
1
2
q
2
r
2
B
2
m
(13.4)
579