In the circuit shown in figure $E_{1}$ = 7V, $E_{2} = 7 V$ $R_{1} = R_{2} = 1 Ω$ and $R_{3} = 3$ respectively. The current through the resistance $R_{3}$ is:

Question

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Answer 1

Using Kirchoff's loop law in loop BCFAB-

$i_{1} R_{1} + (i_{1} + i_{2}) R_{3} = E_{1}$

$⟹ i_{1} + 3 (i_{1} + i_{2}) = 7$

$⟹ 4 i_{1} + 3 i_{2} = 7$

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

$i_{1} R_{1} - i_{2} R_{2} + E_{2} = E_{1}$

$⟹ i_{1} - i_{2} + 7 = 7$

$⟹ i_{1} = i_{2}$

Hence, in equation-

$4 i_{1} + 3 i_{2} = 7$

$⟹ 4 i_{1} + 3 i_{1} = 7$

$⟹ i_{1} = 1 A$

and $i_{2} = i_{1} = 1 A$

Current through $R_{3}$ is $i_{1} + i_{2} = 2 A$

Answer-(A)

Answer 2

Using Kirchoff's loop law in loop BCFAB-

i_{1} R_{1} + (i_{1} + i_{2}) R_{3} = E_{1}

⟹ i_{1} + 3 (i_{1} + i_{2}) = 7

⟹ 4 i_{1} + 3 i_{2} = 7

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

i_{1} R_{1} - i_{2} R_{2} + E_{2} = E_{1}

⟹ i_{1} - i_{2} + 7 = 7

⟹ i_{1} = i_{2}

Hence, in equation-

4 i_{1} + 3 i_{2} = 7

⟹ 4 i_{1} + 3 i_{1} = 7

⟹ i_{1} = 1 A

and

i_{2} = i_{1} = 1 A

Current through

R_{3}

is

i_{1} + i_{2} = 2 A

Answer-(A)