In the figure shown, what is the current (in Ampere) drawn from the battery ? you are given 

$R_1=15\Omega ,R_2=10\Omega ,R_3=20\Omega ,R_4=5\Omega ,R_5=25\Omega ,R_6=30\Omega ,E=25V$

$$\begin{gathered} Apply series and parallel combination of resistors \\ 20\Omega , 5\Omega and 25\Omega are in series \\ \Rightarrow R_1=20+5+25=50\Omega \\ \Rightarrow R_1=50\Omega is parallel with 10\Omega \\ \Rightarrow R_2=\frac{50\left(10\right)}{50+10}=\frac{500}{60}=\frac{25}{3}\Omega \\ \Rightarrow R_2=\frac{25}{3}\Omega is in series with 15\Omega and 30\Omega \\ {R}_{eff}=\frac{25}{3}+15+ 30=\frac{160}{3}\Omega \\ i=\frac{\text{Voltage}}{\text{Net\hspace{0.17em}\hspace{0.17em}resistance\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}\hspace{0.17em}circuit}}=\frac{25}{\frac{160}{3}}=\frac{15}{32} \end{gathered}$$