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Class 12 NEET Physics Moving Charges and Magnetism
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© examsnet.com
Question : 77
Total: 96
A galvanometer having a resistance of 9 ohm is shunted by a wire of resistance 2 ohm. If the total current is 1 amp, the part of it passing through the shunt will be
(1998)
0.2
amp
0.8 amp
0.25 amp
0.5 amp
Validate
Solution:
The shunt and galvanometer are in parallel.
Therefore,
1
R
e
q
=
1
9
+
1
2
or
R
e
q
=
18
11
Ω
Using Ohm's law,
V
=
I
R
e
q
=
1
×
18
11
=
18
11
V
∴
Current through shunt
=
V
R
s
=
18
11
2
=
9
11
≃
0.8
amp
© examsnet.com
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