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Byju's Answer
Standard XI
Physics
Simple Circuit
Consider the ...
Question
Consider the circuit shown in the figure. The current i
3
is equal to
A
5 amp
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B
3 amp
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C
– 3 amp
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D
−
5
6
amp
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Solution
The correct option is
D
−
5
6
amp
Suppose current through different paths of the circuit is as follows.
After applying KVL for loop (1) and loop (2)
We get
28
i
1
=
−
6
−
8
⇒
i
1
=
−
1
2
A
and
54
i
2
=
−
6
−
12
⇒
i
2
=
−
1
3
A
Hence
i
3
=
i
1
+
i
2
=
−
5
6
A
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0
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