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JEE Main 17 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 13
Total: 90
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of
15
Ω
resistance is connected across
B
D
. Calculate the current through the galvanometer when a potential difference of
10
V
is maintained across
A
C
.
[17 Mar 2021 Shift 2]
2.44
µ
A
2.44
m
A
4.87
m
A
4.87
µ
A
Validate
Solution:
As, A is directly connected to the positive terminal of the battery,
V
A
=
10
V
and
V
C
=
0
By nodal analysis at
B
,
V
B
−
10
100
+
V
B
−
V
D
15
+
V
B
−
0
10
=
0
53
V
B
−
20
V
D
=
30
...(i)
By nodal analysis at
D
,
V
D
−
10
60
+
V
D
−
V
B
15
+
V
D
−
0
5
=
0
−
4
V
B
+
17
V
D
=
10
....(ii)
Solving Eqs. (i) and (ii) by substitution method, we get
V
D
=
0.792
V
⇒
V
B
=
0.865
V
The current through the galvanometer,
I
=
V
B
−
V
D
R
Substituting the values in the above equation, we get
I
=
0.865
−
0.792
15
⇒
I
=
4.87
m
A
© examsnet.com
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