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Class 12 NEET Physics Moving Charges and Magnetism
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© examsnet.com
Question : 56
Total: 96
A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
(2008)
6050
4450 Ω
5050 Ω
5550 Ω
Validate
Solution:
Total initial resistance
=
R
G
+
R
1
=
(
50
+
2950
)
Ω
=
3000
Ω
ε
=
3
V
∴
C
u
r
r
e
n
t
=
3
V
3000
Ω
=
1
×
10
−
3
m
A
If the deflection has to be reduced to 20 divisions, current
i
=
1
m
A
×
2
3
as the full deflection scale for 1 mA = 30 divisions.
3
V
=
3000
Ω
×
1
m
A
=
x
Ω
×
2
3
m
A
⇒
x
=
3000
×
1
×
3
2
=
4500
Ω
But the galvanometer resistance
=
50
Ω
Therefore, the resistance to be added
=
(
4500
−
50
)
Ω
=
4450
Ω
© examsnet.com
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