In the given circuit, the battery has an emf E = 60 V and an internal resistance $r=1.5\mathrm{\Omega }$ Find the current at a,b,c and d when all the switches are closed.

$15\mathrm{\Omega }, 25\mathrm{\Omega }$ are in series. Their effective $40\mathrm{\Omega }$ is in parallel to $60\mathrm{\Omega }$

$\therefore {\mathrm{R}}_{\mathrm{PQ}}=\frac{40\times 60}{40+60}=24\mathrm{\Omega }.$ 

This is in series with $6\mathrm{\Omega }.$ Their effective $30\mathrm{\Omega }$ is parallel to $15\mathrm{\Omega }$

$\therefore {\mathrm{R}}_{\mathrm{QS}}=\frac{30\times 15}{30+15}=10\mathrm{\Omega }$

this $10\mathrm{\Omega }, 8.5\mathrm{\Omega }, \mathrm{r}=1.5\mathrm{\Omega }$ are in series.

$\therefore R_{\text{net }}=10+8.5+1.5=20\mathrm{\Omega }$

$I=\frac{E}{R}=\frac{60}{20}=3A=I_d$

For loop (1), $-\mathrm{E}+8.5\mathrm{I}+15{\mathrm{I}}_1+\mathrm{r}\left(\mathrm{I}\right)=0$

$-60+8.5\left(3\right)+15{\mathrm{I}}_1+1.5\left(3\right)=0$

$\Rightarrow 15{\mathrm{I}}_1=30\Rightarrow {\mathrm{I}}_1=2\mathrm{A}={\mathrm{I}}_{\mathrm{C}}$

$\text{ For loop (2), }60{\mathrm{I}}_2-15{\mathrm{I}}_1+6\left(\mathrm{I}-{\mathrm{I}}_1\right)=0$

$60{\mathrm{I}}_2-15\left(2\right)+6(3-2)=0$

$\Rightarrow 60{\mathrm{I}}_2=24\Rightarrow {\mathrm{I}}_2=0.4\mathrm{A}={\mathrm{I}}_{\mathrm{b}}$

$\Rightarrow {\mathrm{I}}_{\mathrm{a}}=\mathrm{I}-{\mathrm{I}}_1-{\mathrm{I}}_2=3-2-0.4=0.6\mathrm{A}$