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Byju's Answer
Standard XII
Physics
Heat Engines
For 100
Question
For
100
% modulation (AM) the useful part of the total power radiated is
A
1
2
of the total power
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B
1
3
of the total power
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C
1
4
of the total power
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D
2
3
of the total power
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Solution
The correct option is
B
1
3
of the total power
100
% modulation
⇒
m
a
=
1
U
s
e
f
u
l
p
o
w
e
r
T
o
t
a
l
p
o
w
e
r
r
a
d
i
a
t
e
d
=
m
2
a
2
+
m
a
2
=
1
1
+
2
=
1
3
⇒
Useful power
=
1
3
(total power radiated)
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