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AMU Engineering 2020 solved paper Solved Paper
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© examsnet.com
Question : 41
Total: 150
Figure shows a simple potentiometer circuit for measuring a small emf produced by a thermocouple
The meter wire
P
Q
has a resistance of
5
Ω
and the driver cell has an emf of
2.00
V
. If a balance point is obtained at
0.600
m
along
P
Q
, when measuring an emf of
6.00
m
V
, then what is the value of resistance
R
?
95
Ω
995
Ω
195
Ω
1995
Validate
Solution:
The voltage per unit length on the meter wire
P
Q
is
6.00
m
V
0.60
m
=
10
m
V
∕
m
Hence potential across the meter wire
P
Q
is 10
m
V
.
Current drawn from the drives cell is.
I
=
10
m
V
5
Ω
=
2
m
A
Value of resistanceis,
R
=
2
V
−
10
m
V
2
m
A
=
1990
m
V
2
m
A
=
995
Ω
© examsnet.com
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