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JEE Main Physics Class 12 Current Electricity Part 4 Questions
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© examsnet.com
Question : 7
Total: 31
A galvanometer of resistance
100
Ω
when connected in series with
400
Ω
measures a voltage of upto
10
V
. The value of resistance required to convert the galvanometer into ammeter to read upto
10
A
is
x
×
10
−
2
Ω
. The value of
x
is :
[5 Apr 2024 Shift 2]
2
800
20
200
Validate
Solution:
👈: Video Solution
i
g
=
10
400
+
100
=
20
×
10
−
3
A
For ammeter
Let shunt resistance
=
S
i
g
R
=
(
i
−
i
g
)
S
20
×
10
−
3
×
100
=
10
S
S
=
20
×
10
−
2
Ω
© examsnet.com
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