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JEE Main Physics Class 12 Moving Charges and Magnetism Part 1 Questions
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© examsnet.com
Question : 43
Total: 100
To know the resistance
G
of a galvanometer by half deflection method, a battery of
e
m
f
V
E
and resistance
R
is used to deflect the galvanometer by angle
theta
. If a shunt of resistance
S
is needed to get half deflection then
G
,
R
and S related by the equation:
[Online April 9, 2016]
S
(
R
+
G
)
=
R
G
2
S
(
R
+
G
)
=
R
G
2
G
=
S
2
S
=
G
Validate
Solution:
According to Ohm's Law,
I
=
V
R
I
g
=
V
R
+
G
where,
I
g
-Galvanometer current, G-Galvonometer resistance
When shunt of resistance
S
is connected parallel to the Galvanometer then
G
=
G
S
G
+
S
∴
I
=
V
R
+
G
S
G
+
S
Equal potential difference is given by
I
′
g
G
=
(
I
−
I
′
g
)
S
I
′
g
(
G
+
S
)
=
I
S
⇒
I
g
2
=
I
S
G
+
S
⇒
V
2
(
R
+
G
)
=
V
R
+
G
S
G
+
S
×
S
G
+
S
⇒
1
2
(
R
+
G
)
=
S
R
(
G
+
S
)
+
G
S
⇒
R
(
G
+
S
)
+
G
S
=
2
S
(
R
+
G
)
⇒
R
G
+
R
S
+
G
S
=
2
S
(
R
+
G
)
⇒
R
G
=
2
S
(
R
+
G
)
−
S
(
R
+
G
)
∴
R
G
=
S
(
R
+
G
)
© examsnet.com
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