High School

A real battery is modeled as an ideal EMF in series with an internal resistance, both unknown quantities. When a load resistance of 6 ohms is placed across the battery terminals, the terminal voltage is 40 V. When a load resistance of 12 ohms is placed across the battery terminals, the terminal voltage is 48 V.

What is the battery’s EMF?

What is the internal resistance of the battery?

Answer :

This implies that -8V = 0, which is not possible

To determine the battery's EMF and internal resistance, we can use the concept of a voltage divider. The voltage divider equation is given by:

V = EMF - (I * R_internal)

where V is the terminal voltage, EMF is the electromotive force (battery's voltage), I is the current flowing through the circuit, and R_internal is the internal resistance of the battery

We have two equations based on the given information:

Equation 1: 40V = EMF - (I * R_internal) (for the load resistance of 6 ohms)

Equation 2: 48V = EMF - (I * R_internal) (for the load resistance of 12 ohms)

To find EMF and R_internal, we can solve these two equations simultaneously. Let's begin:

Subtracting Equation 2 from Equation 1, we eliminate the EMF:

40V - 48V = EMF - (I * R_internal) - (EMF - (I * R_internal))

-8V = 0

This implies that -8V = 0, which is not possible. Therefore, there is no solution to this system of equations.

The given information seems to be contradictory or inconsistent. Please double-check the values or conditions provided for the terminal voltage with different load resistances.

learn more about implies from this link.

https://brainly.com/question/34063676

#SPJ11

Other Questions