Question

Answer 1

\(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{15.10}{15+10}=6\left(\Omega\right)\)

\(R_{tđ}=R_1+R_{23}=9+6=15\left(\Omega\right)\)

\(U=I.R_{tđ}=0,3.15=4,5\left(V\right)\)

Answer 2

\(R=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=9+\left(\dfrac{15.10}{15+10}\right)=15\left(\Omega\right)\)

\(\Rightarrow U=IR=0,3.15=4,5V\)