Let's find the current in the 10-Ω resistor:
[tex]R_{eq1}=\frac{30\cdot60}{30+60^}[/tex][tex]R_{eq1}=20\Omega[/tex]
[tex]R_{eq2}=20+10=30\Omega[/tex]
For the 10Ω resistor, the voltage is:
[tex]\begin{gathered} I_{10}=\frac{V}{R_{eq2}}=\frac{6}{30}=0.2A \\ so: \\ V_{10}=0.2(10)=2V \end{gathered}[/tex]
For the 30-Ω resistor
[tex]\begin{gathered} V_{30}=0.04V \\ \end{gathered}[/tex]
For the 60-Ω resistor:
[tex]V_{60}=4V[/tex]
And for the 20-Ω resistor:
[tex]V_{20}=6V[/tex]
Therefore, we can conclude:
d- the 20-Ω resistor
