Answer: Less than 4 ohms
Explanation:
We have three resistors with the following resistance:
[tex]R_{1}=4\Omega[/tex]
[tex]R_{2}=6\Omega[/tex]
[tex]R_{3}=8\Omega[/tex]
Now, when the resistors are connected in parallel, the total resistance [tex]R[/tex] is calculated as follows:
[tex]\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}[/tex]
Isolating [tex]R[/tex]:
[tex]R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}[/tex]
Rewriting with th known values:
[tex]R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}[/tex]
Finally:
[tex]R=1.84 \Omega <4 \Omega[/tex]
Hence, the correct option is less than 4 ohms.
