In the circuit shown in figure E1 = 7V, E2=7VR1=R2=1Ω and R3=3 respectively. The current through the resistance R3 is:
2A
3.5A
1.75A
none of these
A
none of these
B
1.75A
C
2A
D
3.5A
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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=7
Now, applying Kirchoff's loop law in loop BCDEFAB-
i1R1−i2R2+E2=E1
⟹i1−i2+7=7
⟹i1=i2
Hence, in equation-
4i1+3i2=7
⟹4i1+3i1=7
⟹i1=1A
and i2=i1=1A
Current through R3 is i1+i2=2A
Answer-(A)
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