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Topic
12th Grade
Physics
Circuit Networks
Question
In the circuit shown below
E
1
Â
=
Â
4.0
Â
V
,
R
1
=
2
Ω
Â
,
E
2
=
Â
6.0
Â
V
,
Â
R
2
=
4
Ω
and
R
3
Â
=
Â
2
Ω
. The current
I
1
is
Options:
A
.
 1.6Â A
B
.
 1.8Â A
C
.
 1.25Â A
D
.
 1.0Â A
Answer: Option B
:
B
Applying Kirchhoff's law for the loops (1) and (2) as shown infigure For loop(1)
−
2
i
1
−
2
(
i
1
−
i
2
)
+
4
=
0
⇒
2
i
1
−
i
2
=
2
.
.
.
.
(
i
)
For loop (2)
−
2
(
i
1
−
i
2
)
+
4
i
2
−
6
=
0
⇒
−
i
1
+
3
i
2
=
3
.
.
.
(
i
i
)
On solving equation (i) and (ii)
i
1
=
1.8
A
.
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