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  • A circuit is as shown in the figure. Then, the current from is:

A circuit is as shown in the figure. Then, the current from \mathrm{A\: to \: B} is:

 

Option: 1

+500\: \mathrm{mA}


Option: 2

+250\: \mathrm{mA}


Option: 3

-250\: \mathrm{mA}


Option: 4

-500\: \mathrm{mA}


Answers (1)

best_answer

Applying Kirchhoff second law for a closed loop \mathrm{CABDC}, we get

\begin{aligned} & -10 \mathrm{I}_1-\left(\mathrm{I}_1-\mathrm{I}_2\right) 10+10=0 \\ \end{aligned}

\begin{aligned} & -20 \mathrm{I}_1+10 \mathrm{I}_2=-10 \\ \end{aligned}

\begin{aligned} & 2 \mathrm{I}_1-\mathrm{I}_2=1 \end{aligned}                                     (i)

Again, applying Kirchhoff second law for a closed loop \mathrm{ABFEA}, we get

\begin{aligned} & -\left(\mathrm{I}_1-\mathrm{I}_2\right) 10-5+15 \mathrm{I}_2=0 \\ \end{aligned}

\begin{aligned} & -10 \mathrm{I}_1+25 \mathrm{I}_2=5 \\ \end{aligned}

\begin{aligned} & 2 \mathrm{I}_1-5 \mathrm{I}_2=-1 \end{aligned}                           (ii)

Solving (i) and (ii), we get

\mathrm{I}_1=\frac{3}{4} \mathrm{~A}, \mathrm{I}_2=-\frac{1}{2} \mathrm{~A}

The current flows from \mathrm{A \: to\: B } is

\mathrm{=I_{1}-I_{2}=\frac{3}{4}A-\frac{1}{2}A=\frac{1}{4}A=0.25A=250\: mA}

Hence option 2 is correct.

Posted by

manish painkra

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