# Ideal Voltage in Series w/ resistance

1. Jun 16, 2010

### shane317

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

KVL, KCL, Ohm's Law

3. 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 eachother or parallel with respect to eachother.

I assume the resistances given in the table are Req 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

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2. Jun 16, 2010

### CEL

I believe that each resistor replaces the precedent one.

3. Jun 16, 2010

### shane317

OK, say the ideal voltage source and the internal resistance Rsource 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(Rsource) and from there I can figure out the rest.

Shane

4. Jun 16, 2010

### CEL

You're welcome!

5. Jun 16, 2010

### 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.