Current Electricity Part 2

According to given circuit diagram, As we know that, parallel equivalent resistance, $\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$ and series equivalent resistance, $R_{eq}=R_{1}+R_{2}+R_{3}+...$ Let the net resistance across a and b be R’ $\therefore \frac{1}{R'}=\frac{1}{2+2+6}+\frac{1}{4+6}$ $a=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}$ R' = 5 Ω Hence, total resistance, $R=R'+1=5+1=6 \Omega$ According to current division rule, current in upper branch, $I_{1}=I \cdot \frac{4+6}{(2+2+6)+(4+6)}$ $=1 \cdot \frac{10}{20}=\frac{1}{2}=\frac{1}{2} \cdot \frac{V}{R}=\frac{1}{2} \times \frac{12}{6}=1 \text{A}$ Again, according to current division rule, current in 15Ω resistor $I_{15}=I_{1} \cdot \frac{10}{10+15}=1 \times \frac{2}{5}=0.4 \text{A}$ ∴ Voltage drop across15Ω resistor, $V_{15}=I_{15} \times 15=0.4 \times 15=6 \text{V}$