Question

State Kirchoff's rules. Use these rules to find the values of current $I_1,I_2,I_3$ in the circuit diagram given

Answer 1

Given by $c$ in $1sec$

$v=\frac{2m}{1}=2m/s$

$\frac{\pi }{15}={12}^{\circ }$

$\cos {12}^{\circ }\cos$

$I_3=I_1+I_1......\left(3\right)$

In Loop $1$:

$(+4v)-\left(1v\right)-(I_2.3)-2I_3=0$

$3=3I_1+2I_3.....\left(1\right)$

In Loop $2$;

$(+5v)-\left(2v\right)-4I_1+3I_2=0$

$3I_2=4I_1=1v.....\left(2\right)$

$3_2-4(I_3-I_2)=1v$

$7I_2-4I_3=2v.....\left(4\right)$

From eqn $\left(1\right)$ and $\left(4\right)$,

$13I_2=7$

$I_2=\frac{7}{13}$ and

$I_3=\frac{3}{2}\times \frac{6}{13}=\frac{9}{13}$ and

$\therefore I_1=I_3-I_2=\frac{2}{13}A$.