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Current through wire \mathrm{XY} of circuit shown is

 

Option: 1

1\mathrm{A}


Option: 2

4\mathrm{A}


Option: 3

2\mathrm{A}


Option: 4

3\mathrm{A}


Answers (1)

best_answer

Let current through \mathrm{XY} is \mathrm{i}_3. Applying Kirchhoff's law to loop (1) and (2),

\begin{gathered} \mathrm{i}_1+0 \times \mathrm{i}_3-3 \mathrm{i}_2=0 \\ \end{gathered}

\begin{gathered} \therefore \quad \mathrm{i}_1=3 \mathrm{i}_2 \\ \end{gathered}                      .................(i)

\begin{gathered} \text { and }-2\left(\mathrm{i}_1-\mathrm{i}_3\right)+4\left(\mathrm{i}_2+\mathrm{i}_3\right)=0 \end{gathered}

So, \mathrm{\quad 2 i_1-4 i_2=6 i_3}                 .............(ii)

Also,

50=\mathrm{i}_1-2\left(\mathrm{i}_1-\mathrm{i}_3\right)

\mathrm{ 50=i_1-2\left(i_1-i_3\right) }

\mathrm{ \therefore \quad 3 i_1-2 i_3=50 }      ..............(iii)

From eqs. (i), (ii) and (iii), we get

\mathrm{i}_3=2 \mathrm{~A}

Hence option 3 is correct.
 

Posted by

chirag

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