In the electrical circuit shown in the figure the current through the 4Ω resistor is:
0.5A
1A
0.25A
0.1A
A
0.25A
B
1A
C
0.5A
D
0.1A
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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+2I1−8(I−I1)=0 ⇒16I1−8I=0 ⇒I1=8I16⇒I1=12I....(i) In closed circuit ABEFA −9+3I+8(I−I1)+2l=0 ⇒13l−8I1=9 ⇒13l−8(12I)=9 ⇒I=99=1A...(ii) So current through 4Ω resistor I1=12×1=0.5A
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