664
14 Electromagnetism II
Rise in temperature
T = Heat / thermal capacity = 2.914/0.9 = 3.24
◦ C
R = R 0 (1 + ααT ) = 10
4
(1 + 0.004 × 3.24)
= 1.01296 × 10
4
R = R − R 0 = 129.6
∴
t
t
=
R
R 0
=
129.6
10 4 = 0.01296
Percentage error =
t
t
× 100 = 1.3%
14.32 The amplitude i m of the current oscillations is given by
i m =
E m
(ω L − 1/ω c) 2 + R 2
At resonance, ω = ω and i m =
E m
R
Set i m =
E m
(ωL − 1/ωc) 2 + R 2
=
1
2
E m
R
Squaring and simplifying
ωL −
1
ωc
2
= 3 R
2
or ω
2 LC ±
√
3ω RC − 1 = 0
The only acceptable solutions are
ω 1 =
√
3
2
R
L
+
3R 2
4L 2 +
1
LC
ω 2 = −
√
3
2
R
L
+
3R 2
4L 2 +
1
LC
Subtracting the last equation from the previous one
ω = ω 1 − ω 2 =
√
3
R
L
∴
ω
ω
=
√
3
ω
R
L
14 Electromagnetism II
Rise in temperature
T = Heat / thermal capacity = 2.914/0.9 = 3.24
◦ C
R = R 0 (1 + ααT ) = 10
4
(1 + 0.004 × 3.24)
= 1.01296 × 10
4
R = R − R 0 = 129.6
∴
t
t
=
R
R 0
=
129.6
10 4 = 0.01296
Percentage error =
t
t
× 100 = 1.3%
14.32 The amplitude i m of the current oscillations is given by
i m =
E m
(ω L − 1/ω c) 2 + R 2
At resonance, ω = ω and i m =
E m
R
Set i m =
E m
(ωL − 1/ωc) 2 + R 2
=
1
2
E m
R
Squaring and simplifying
ωL −
1
ωc
2
= 3 R
2
or ω
2 LC ±
√
3ω RC − 1 = 0
The only acceptable solutions are
ω 1 =
√
3
2
R
L
+
3R 2
4L 2 +
1
LC
ω 2 = −
√
3
2
R
L
+
3R 2
4L 2 +
1
LC
Subtracting the last equation from the previous one
ω = ω 1 − ω 2 =
√
3
R
L
∴
ω
ω
=
√
3
ω
R
L