For the circuit shown in figure, choose then correct options:
 

(CD) From the loop,$badb 2i_1-2+i_1+1+2(i_1-i_2)=0$

$5i_1-2i_2=1$   …………..$\left(1\right)$

From the loop $dcbd,$

3 - 3$i_2$- 1$\times i_2$-1+2($i_1-i_2)=0$

or, 3$i_2-i_1=1 ...........\left(2\right)$

From eqs. (1)and( 2).

$i_1=\frac{5}{13}amp, and i_2=\frac{6}{13} amp$

So,  $i_1-i_2=-\frac{1}{13}amp$

$V_b-V_d=\frac{2}{13}volt$

The $PD$ across the ceil $\left(C\right)$ is 

$V_c-V_d=-3i_2+3=\frac{21}{13}volt$

The $PD$ across the ceil $\left(B\right)$ is 

$V_c-V_b=i_2+1=\frac{19}{13}volt$