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Question

In the electrical circuit shown in the figure the current through the 4Ω resistor is:
470379.PNG
  1. 0.5A
  2. 1A
  3. 0.25A
  4. 0.1A

A
0.25A
B
1A
C
0.5A
D
0.1A
Solution
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The current through the various branches of the circuit will be shown as given
According to Kirchhoff's second law in closed circuit BCDEB
2I1+4I1+2I18(II1)=0
16I18I=0
I1=8I16I1=12I....(i)
In closed circuit ABEFA
9+3I+8(II1)+2l=0
13l8I1=9
13l8(12I)=9
I=99=1A...(ii)
So current through 4 Ω resistor I1=12×1=0.5A
500991_470379_ans.png

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