In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: $$R_1 = 15 \; \Omega$$, $$R_2 = 10 \; \Omega$$, $$R_3 = 20 \; \Omega$$, $$R_4 = 5 \; \Omega$$, $$R_5 = 25 \; \Omega$$, $$R_6 = 30 \; \Omega$$, $$E = 15$$ V

 Step 1: Identify series on the right side

$$R_3​,R_4​,R_5​are\ in\ series$$

$$R_{345}=20+5+25=50Ω$$

 Step 2: Parallel combination

$$R_{345}​\ is\ in\ parallel\ with\ R_2​$$

$$R_{p\ =\ \ \frac{\ 10\times\ 50}{10\ +\ 50}=\ \frac{\ 500}{60}=\ \frac{\ 25}{3}Ω}$$

 Step 3: Series with remaining resistors

$$R_1​,R_P​,R_{6\ }​are\ in\ series$$

$$Req​=15+\ \frac{\ 25}{3}+30=\ \frac{\ 160}{3}​Ω$$

 Step 4: Current from battery

$$I=\ \frac{\ E}{Req​}=\ \frac{\ 15}{\ \frac{\ 160}{3}}=\ \frac{\ 45}{160}​=\ \frac{\ 9}{32}​A$$