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Question

- 2A
- 3.5A
- 1.75A
- none of these

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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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