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Question

In the circuit shown in figure E1 = 7V, E2=7V R1=R2=1Ω and R3=3 respectively. The current through the resistance R3 is:
598037_66ddfaf010244cd697a674eab2a3214e.png
  1. 2A
  2. 3.5A
  3. 1.75A
  4. none of these

A
none of these
B
1.75A
C
2A
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=7

Now, applying Kirchoff's loop law in loop BCDEFAB-

i1R1i2R2+E2=E1

i1i2+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)

850592_598037_ans_625fae4f548f4ac188c05c766deb19fe.jpg

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