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JEE Mains 03-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 7
Total: 75
A
750
H
z
,
20
V
(rms) source is connected to a resistance of
100
a
,
an inductance of
0.1803
H
and a capacitance of 10
µ
F all in series. The time in which the resistance (heat capacity
2
J
∕
°
C
) will get heated by
10
°
C
. (assume no loss of heat to the surroundings) is close to :
418
s
245
s
348
s
365
s
Validate
Solution:
f
=
750
H
z
,
V
rms
=
20
V
R
=
100
a
,
L
=
0.1803
H
C
=
10
µ
F
,
S
=
2
J
∕
°
C
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
=
√
R
2
+
(
ω
L
−
1
∕
ω
C
)
2
=
√
R
2
+
(
2
π
f
L
−
1
2
π
f
C
)
2
Putting values
|
Z
|
=
834
Ω
In
A
C
power
P
=
V
rms
i
rms
cos
ϕ
cos
ϕ
=
R
|
Z
|
i
rms
=
V
rms
|
Z
|
=
V
rms
2
R
(
|
Z
|
)
2
=
(
20
834
)
2
×
100
=
0.0575
J
∕
s
H
=
P
t
=
S
Δ
θ
t
=
2
(
10
)
0.0575
=
348
s
e
c
© examsnet.com
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