Practicing Success
Current through wire XY of circuit shown is |
1 A 4 A 2 A 3 A |
2 A |
Let current through XY is i3. Applying Kirchhoff’s law to loop (1) and (2), $i_1+0 \times i_3-3 i_2=0$ ∴ $i_1=3 i_2$ (i) and $-2\left(i_1-i_3\right)+4\left(i_2+i_3\right)=0$ So, $2 i_1-4 i_2=6 i_3$ (ii) Also, $50=1i_1-2\left(i_1-i_3\right)$ $50=i_1-2\left(i_1-i_3\right)$ ∴ $3 i_1-2 i_3=50$ (iii) From eqs. (i), (ii) and (iii), we get $i_3=2 A$ |