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Class 12 NEET Physics Current Electricity Questions Part 1
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© examsnet.com
Question : 7
Total: 100
A potentiometer wire of length
L
and a resistance
r
are connected in series with a battery of e.m.f.
E
0
and a resistance
r
1
. An unknown e.m.f.
E
is balanced at a length
l
of the potentiometer wire. The e.m.f. E will be given by
[NEET 2015]
E
0
l
L
L
E
0
r
(
r
+
r
1
)
l
L
E
0
r
l
r
1
E
0
r
(
r
+
r
1
)
⋅
l
L
Validate
Solution:
👈: Video Solution
The current through the potentiometer wire is
I
=
E
0
(
r
+
r
1
)
and the potential difference across the wire is
V
=
I
r
=
E
0
r
(
r
+
r
1
)
The potential gradient along the potentiometer wire is
k
=
V
L
=
E
0
r
(
r
+
r
1
)
L
As the unknown e.m.f.
E
is balanced against length I of the potentiometer wire,
∴
E
=
k
l
=
E
0
r
(
r
+
r
1
)
l
L
© examsnet.com
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