The current flowing in the circuit is given by the Ohm's First Law.
\(i_{eq}= \frac{V}{R_{eq}} \\ i_{eq}= \frac{6}{4+6} \\ i_{eq}= 0.6A\)
The current flowing in the 6-ohm resistor is the total current of circuit, therefore the voltage drop across this resistor is given by:
\(V_{6}=R_{6}*I_{eq} \\ V_{6}=6*0.6 \\ \boxed {V_{6}=3.6V}\)
Obs: the drawing is attached
If you notice any mistake in my english, please let me know, because i am not native.
