The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit

Respuesta :

Answer:

The resistance of the original circuit is [tex]32\,\,\Omega[/tex]

Explanation:

In the original circuit, we have an unknown resistor that we call R, an unknown power supply that we call V, and the current is 15 Amps. in the second circuit with an added 8 Ω resistor in series, which gives an equivalent resistance of R+8 Ω, using the same power supply V, the current is 12 Amps. SO, we can write a system of two equations with two unknowns as follows:

[tex]V=R\,(15)\\V=(R+8)\,(12)\\then\\15\,R=12\,R+92\\3\,R=96\\R=\frac{96}{3} \,\Omega\\R=32\,\,\Omega[/tex]