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BITSAT 2018 Slot 2 Solved Paper
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© examsnet.com
Question : 9
Total: 150
In the circuit given below, the value of resistance X, when the potential difference between the points B and D is zero, will be
9 Ω
8 Ω
4 Ω
6 Ω
Validate
Solution:
The given circuit is equivalent to the balanced Wheatstone's network.
Let, resistance in
a
r
m
A
B
is
P
,
P
=
15
+
6
=
21
Ω
Resistance in
a
r
m
B
C
is
Q
,
Q
=
(
8
X
8
+
X
)
+
(
3
)
Ω
Resistance in
a
r
m
D
A
is
R
,
R
=
(
(
6
)
(
6
)
6
+
6
)
+
(
15
)
=
3
+
15
=
18
Ω
Resistance in
a
r
m
C
D
is
S
,
S
=
(
(
4
)
(
4
)
4
+
4
)
+
(
4
)
=
2
+
4
=
6
Ω
Using balancing condition for Wheatstone's network.
P
Q
=
R
S
21
(
8
X
8
+
X
)
+
(
3
)
=
18
6
21
(
8
X
8
+
X
)
+
(
3
)
=
3
(
8
X
8
+
X
)
+
(
3
)
=
21
3
(
8
X
8
+
X
)
+
(
3
)
=
7
8
X
8
+
X
=
4
8
X
=
32
+
4
X
4
X
=
32
X
=
8
Ω
© examsnet.com
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