66.7 Watts
Explanation:
Let [tex]R_{1}=1.0[/tex] ohms, [tex]R_{2}=3.0[/tex]ohms and [tex]R_{3}=6.0\:[/tex]ohms. Since [tex]R_{2}[/tex] and [tex]R_{3}[/tex] are in parallel, their combined resistance [tex]R_{23}[/tex] is given by
[tex]\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}[/tex]
or
[tex]R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms[/tex]
The total current flowing through the circuit I is given
[tex]I=\dfrac{V_{s}}{R_{Total}}[/tex]
where
[tex]R_{Total}=R_{1}+R_{23}= 3.0\:ohms[/tex]
Therefore, the total current through the circuit is
[tex]I=\dfrac{30\:V}{3.0\:ohms}=10\:A[/tex]
In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor [tex]V_{1}[/tex]:
[tex]V_{1}=(10\:A)(1.0\:ohms)=10\:V[/tex]
This means that voltage drop across the 6-ohm resistor [tex]V_{3}[/tex] is 20 V. The power dissipated P by the 6-ohm resitor is given by
[tex]P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W[/tex]
