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JEE Main Physics Class 12 Alternating Currents Part 1 Questions
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© examsnet.com
Question : 43
Total: 100
An alternating voltage v(t) = 220 sin 100Ât volt is applied to a purely resistive load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is :
[8 April 2019 I]
5 ms
2.2 ms
7.2 ms
3.3 ms
Validate
Solution:
As
V
(
t
)
=
220
sin
100
π
t
so,
I
(
t
)
=
220
50
sin
100
π
t
i.e.,
I
=
I
m
=
sin
(
100
π
t
)
For
I
=
I
m
t
1
=
π
2
×
1
100
π
=
1
200
sec
.
and for
I
=
I
m
2
⇒
I
m
2
=
I
m
sin
(
100
π
t
2
)
⇒
π
6
=
100
π
t
2
⇒
t
2
=
1
600
s
∴
t
req
=
1
200
−
1
600
=
2
600
=
1
300
s
=
3.3
m
s
© examsnet.com
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