Answer:
[tex]44\Omega[/tex]
Explanation:
When N resistors are placed in series, they are placed along the same branch of the circuit, and the current flowing through them is the same.
The equivalent resistance of a series of N resistors is given by
[tex]R=R_1+R_2+...+R_N[/tex]
So, it is equal to the sum of the individual resistances.
In this problem, the three resistor's resistances are:
[tex]R_1=6.00\Omega\\R_2=15.0\Omega\\R_3=23.0\Omega[/tex]
Therefore, the equivalent resistance of the circuit is:
[tex]R=6+15+23=44\Omega[/tex]
So, the circuit has an equivalent resistance of 44 Ω.
