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JEE Mains 6-Apr-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 56
Total: 90
An ideal transformer with purely resistive load operates at
12
kV
on the primary side. It supplies electrical energy to a number of nearby houses at
120
V
. The average rate of energy consumption in the houses served by the transformer is
60
kW
. The value of resistive load (Rs) required in the secondary circuit will be
m
Ω
.
[6-Apr-2023 shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
V
s
V
p
=
N
s
N
p
⇒
120
12000
=
N
s
N
p
⇒
N
s
N
p
=
1
100
−
−
−
(
i
)
For an ideal transformer, input power
=
Output power
And power is given by
P
=
i
V
i
p
V
p
=
i
s
V
s
=
60000
W
i
p
=
60000
12000
=
5
Now,
R
p
=
V
p
i
p
=
12000
5
=
2400
Ω
R
s
=
V
s
i
s
=
120
60000
∕
120
=
120
×
120
60000
=
120
500
=
0.240
Ω
=
240
m
Ω
© examsnet.com
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