654
14 Electromagnetism II
14.12 For the parallel RLC circuit, Fig. 14.1
Q = C V, I 2 = −
dQ
dt
= −C
dV
dt
V = R p (I 2 + I 1 ) = −L
dI 1
dt
dV
dt
= R p
dI 2
dt
+
dI 1
dt
= −C R p
d 2 V
dt 2 − R p
V
L
∴
d 2 V
dt 2 +
1
R p C
dV
dt
+
V
LC
= 0
(parallel arrangement)
Compare the above equation with the given equation
d 2 V
dt 2 +
R
L
dV
dt
+
V
LC
= 0
∴ R p =
L
CR
.
14.13 We can find the dimensions of μ 0 and ε 0 from the following set of formulae:
F = ilB (force on the current-carrying wire)
(1)
B =
μ 0 i
2πr
(magnetic field due to a current-carrying wire)
(2)
F =
Q 2
4πε 0 r 2 (electrostatic force between charges)
(3)
i = Q/t
(4)
Combining (1)–(4), [μ 0 ε 0 ] = [T 2 /L 2 ]
or
1
√
μ 0 ε 0
=
L
T
= [v] = [c]
14.14 We first derive the dimensional formulae for R, C and L from the defining
equation Power = i 2 R
[Power] = [Ml
2 T
−3
] = [A
2 R] = [A
2
][R]
∴ [R] = [Ml
2 T
−3 A
−2
]
(1)
Energy of a capacitor E =
1
2
Q 2
C
and Q = it
∴ [C] = [M
−1 l
−2 T
4 A
2
]
(2)
14 Electromagnetism II
14.12 For the parallel RLC circuit, Fig. 14.1
Q = C V, I 2 = −
dQ
dt
= −C
dV
dt
V = R p (I 2 + I 1 ) = −L
dI 1
dt
dV
dt
= R p
dI 2
dt
+
dI 1
dt
= −C R p
d 2 V
dt 2 − R p
V
L
∴
d 2 V
dt 2 +
1
R p C
dV
dt
+
V
LC
= 0
(parallel arrangement)
Compare the above equation with the given equation
d 2 V
dt 2 +
R
L
dV
dt
+
V
LC
= 0
∴ R p =
L
CR
.
14.13 We can find the dimensions of μ 0 and ε 0 from the following set of formulae:
F = ilB (force on the current-carrying wire)
(1)
B =
μ 0 i
2πr
(magnetic field due to a current-carrying wire)
(2)
F =
Q 2
4πε 0 r 2 (electrostatic force between charges)
(3)
i = Q/t
(4)
Combining (1)–(4), [μ 0 ε 0 ] = [T 2 /L 2 ]
or
1
√
μ 0 ε 0
=
L
T
= [v] = [c]
14.14 We first derive the dimensional formulae for R, C and L from the defining
equation Power = i 2 R
[Power] = [Ml
2 T
−3
] = [A
2 R] = [A
2
][R]
∴ [R] = [Ml
2 T
−3 A
−2
]
(1)
Energy of a capacitor E =
1
2
Q 2
C
and Q = it
∴ [C] = [M
−1 l
−2 T
4 A
2
]
(2)