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AP EAMCET Engineering 2013 Solved Paper
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© examsnet.com
Question : 30
Total: 160
In the circuit shown in the figure, the current ‘I’ is
6 A
2 A
4 A
7 A
Validate
Solution:
Applying junction law
We have
I
=
I
1
+
I
2
24
−
V
3
=
10
−
V
2
+
9
−
V
1
⇒
24
−
V
3
=
28
−
3
V
2
⇒
2
(
24
−
V
)
=
3
(
28
−
3
V
)
⇒
48
−
2
V
=
84
−
9
V
⇒
7
V
=
36
⇒
V
=
5.14
V
From Ohm's law
Δ
V
=
I
R
Δ
V
=
24
−
5.14
=
18.86
,
R
=
3
Ω
∴
I
=
18.86
3
≈
6
A
© examsnet.com
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