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# In the circuit shown in figure E1 = 7V, E2=7V R1=R2=1Ω and R3=3 respectively. The current through the resistance R3 is:2A3.5A1.75Anone of these

A
2A
B
1.75A
C
none of these
D
3.5A
Solution
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#### Using Kirchoff's loop law in loop BCFAB-i1R1+(i1+i2)R3=E1⟹i1+3(i1+i2)=7⟹4i1+3i2=7Now, applying Kirchoff's loop law in loop BCDEFAB-i1R1−i2R2+E2=E1⟹i1−i2+7=7⟹i1=i2Hence, in equation-4i1+3i2=7⟹4i1+3i1=7⟹i1=1Aand i2=i1=1ACurrent through R3 is i1+i2=2AAnswer-(A)

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