Ideal Voltage in Series w/ resistance

[shane317]

Homework Statement

A device can be modeled using an ideal voltage source in series with a resistance. That
device is shown in Figure P2.2. A set of resistors was connected to the terminals of the device, one at a time, and the voltage at the device terminals, vt, was measured in each case. The results are shown in Table P2.2. Then, a 10[Ω] resistor was connected to the terminals of the device. Find the power absorbed by this 10[Ω] resistor, when it is connected to the device terminals.

Please see image

Homework Equations

KVL, KCL, Ohm's Law

The Attempt at a Solution

I need help setting up the problem. The resistors are connected one at a time, but are they connected in series with respect to each other or parallel with respect to each other.

I assume the resistances given in the table are R_eq as opposed to the value of the individual resistors added to the circuit. I am also assuming the final 10 ohm resistor is connected after the three resistor set is removed.

Thank you, any help is greatly appreciated.

Shane

 

[CEL]

I believe that each resistor replaces the precedent one.

 

[shane317]

I believe that each resistor replaces the precedent one.

OK, say the ideal voltage source and the internal resistance R_source are in series to illustrate the device in question between terminals A and B. Each resistor in the table is connected to the terminals A and B one at a time and the resistances in the table are the resistances of the three resistors. So now to model the device I need to solve for the internal resistance(R_source) and from there I can figure out the rest.

Thanks for the reply,

Shane

 

[CEL]

You're welcome!

 

[vk6kro]

There might be a problem with this question.

The current into the load resistor is increasing as the resistor value increases. (0.6A 1.6A 5.6A)

If the device in the box was a voltage source in series with a resistor, you would expect the current into the load to decrease with increasing resistance.

This increased resistance and reduced current are what causes the internal voltage drop to be less.