
Applying Kirchhoffs law in mash $(I)$
$2 i_1+6 i_1-6 i_2+8 i_1+0.5 i_1-15=0$
$16.5 i_1-6 i_2=15 \ldots . \text { (I) }$
Applying Kirchhoff's law in mash $(II)$
$7 i _2+ i _2+10 i _2+6 i _2-6 i _1=0$
$24 i _2-6 i _1=0$
$4 i _2- i _1=0$
$i _1=4 i _2 \ldots \ldots \text { (II) }$
After solving of equation $(I)$ and $(II)$
Then,
$16.5 \times 4 i _2-6 i _2=15$
$66 i _2-6 i _2=15$
$60 i _2=15$
$i _2=\frac{1}{4} A$
Now put the value of $i_2$ in equation $(I)$
$i_1=4 i_2$
$i_1=4 \times \frac{1}{4}$
$i_1=1\,\,A$
Now, the current from the battery in the circuit is $1\,\,A$.

$1.$ Consider two different metallic strips ($1$ and $2$) of the same material. Their lengths are the same, widths are $w_1$ and $w_2$ and thicknesses are $d_1$ and $d_2$, respectively. Two points $K$ and $M$ are symmetrically located on the opposite faces parallel to the $x$ - $y$ plane (see figure). $V _1$ and $V _2$ are the potential differences between $K$ and $M$ in strips $1$ and $2$ , respectively. Then, for a given current $I$ flowing through them in a given magnetic field strength $B$, the correct statement$(s)$ is(are)
$(A)$ If $w _1= w _2$ and $d _1=2 d _2$, then $V _2=2 V _1$
$(B)$ If $w_1=w_2$ and $d_1=2 d_2$, then $V_2=V_1$
$(C)$ If $w _1=2 w _2$ and $d _1= d _2$, then $V _2=2 V _1$
$(D)$ If $w _1=2 w _2$ and $d _1= d _2$, then $V _2= V _1$
$2.$ Consider two different metallic strips ($1$ and $2$) of same dimensions (lengths $\ell$, width w and thickness $d$ ) with carrier densities $n_1$ and $n_2$, respectively. Strip $1$ is placed in magnetic field $B_1$ and strip $2$ is placed in magnetic field $B_2$, both along positive $y$-directions. Then $V_1$ and $V_2$ are the potential differences developed between $K$ and $M$ in strips $1$ and $2$, respectively. Assuming that the current $I$ is the same for both the strips, the correct option$(s)$ is(are)
$(A)$ If $B_1=B_2$ and $n_1=2 n_2$, then $V_2=2 V_1$
$(B)$ If $B_1=B_2$ and $n_1=2 n_2$, then $V_2=V_1$
$(C)$ If $B _1=2 B _2$ and $n _1= n _2$, then $V _2=0.5 V _1$
$(D)$ If $B_1=2 B_2$ and $n_1=n_2$, then $V_2=V_1$
Give the answer question $1$ and $2.$ 
What is the value of current at $t = 4\,s$ ? ............... $\mu A$

