Current through \(R _{ S }= I =\frac{7}{35}=\frac{1}{5} A\)
Current through \(90 \Omega= I _{2}=\frac{15}{90}=\frac{1}{6} A\)
Current through zener \(=\frac{1}{5}-\frac{1}{6}=\frac{1}{30} A\)
Power through zener diode
\(P = VI\)
\(P =15 \times \frac{1}{30}=0.5 watt\)
\(P =5 \times 10^{-1} watt\)