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Tardigrade
Question
Physics
Three resistance of 10Ω , 5Ω , 2Ω respectively are connected in parallel. New resistance will be
Q. Three resistance of 10
Ω
, 5
Ω
, 2
Ω
respectively are connected in parallel. New resistance will be
1636
218
Current Electricity
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A
more than 10
Ω
19%
B
between 10
Ω
and 5
Ω
18%
C
between 5
Ω
and 2
Ω
20%
D
less than 2
Ω
.
43%
Solution:
R
p
1
=
R
1
1
+
R
1
1
+
R
2
1
+
R
3
1
=
10
1
+
5
1
+
2
1
=
10
8
Hence
R
p
=
1.25
Ω