Interface circuit for a device with internal resistance

[jegues]

Homework Statement

You are given a device with which to operate with an input resistance of 50 W. The device requires 9 V to operate. However, the only available power supply is a 12 V source.

Part A:Using only resistors, design an interface circuit to lower the 12 V to 9 V for operating the device.

Part B:If the power supply has a 10 W internal resistance, what will be the voltage of the load?

Part C:How would you adjust it so that you get a 9 V voltage again?

Homework Equations

$$V_{out} = \frac{R_{2}}{R_{1} + R_{2}} \cdot V_{in}$$

The Attempt at a Solution

For Part A, can't I simply use a voltage divider to drop the 12V to 9V? If I set ,

$$R_{1} = 1 \Omega$$

and,

$$R_{2} = 3 \Omega$$

Then I should obtain a $$V_{out} = 9V$$.

Part B & C is where I start to get a little confused.

For Part B, if the power supply has 10 W internal resistance and it says in the question that the device has an input resistance of 50 W does that mean the other 40 W will be on the load?

Part C is hooked into Part B so until I fully understand that, I don't think I have a shot at answering Part C.

Any ideas/comments/suggestions?

Thanks again!

 

[CEL]

Homework Statement

You are given a device with which to operate with an input resistance of 50 W. The device requires 9 V to operate. However, the only available power supply is a 12 V source.

Part A:Using only resistors, design an interface circuit to lower the 12 V to 9 V for operating the device.

Part B:If the power supply has a 10 W internal resistance, what will be the voltage of the load?

Part C:How would you adjust it so that you get a 9 V voltage again?

Homework Equations

$$V_{out} = \frac{R_{2}}{R_{1} + R_{2}} \cdot V_{in}$$

The Attempt at a Solution

For Part A, can't I simply use a voltage divider to drop the 12V to 9V? If I set ,

$$R_{1} = 1 \Omega$$

and,

$$R_{2} = 3 \Omega$$

Then I should obtain a $$V_{out} = 9V$$.

Part B & C is where I start to get a little confused.

For Part B, if the power supply has 10 W internal resistance and it says in the question that the device has an input resistance of 50 W does that mean the other 40 W will be on the load?

Part C is hooked into Part B so until I fully understand that, I don't think I have a shot at answering Part C.

Any ideas/comments/suggestions?

Thanks again!

I believe that the resistances are 50 ohms and 10 ohms, instead of watts.
For part A you are right, only R2 should be 50 ohms, instead of 3.
For part B, you use the voltage divider again, to calculate the voltage at the load.
For part C, you add a resistance that is the difference between the value calculated in part A and the internal resistance of the source.

 

[jegues]

I believe that the resistances are 50 ohms and 10 ohms, instead of watts.
For part A you are right, only R2 should be 50 ohms, instead of 3.
For part B, you use the voltage divider again, to calculate the voltage at the load.
For part C, you add a resistance that is the difference between the value calculated in part A and the internal resistance of the source.

Okay so for Part A:

$$R_{2} = 50 \Omega$$

and

$$R_{1} = 16.67 \Omega$$

I'm still a little confused about part b. When I'm using my voltage divider equation,

$$V_{out} = \frac{R_{2}}{R_{1} + R_{2}} \cdot V_{in}$$

this time I'm computing Vout right? Isn't that the voltage that will remain at the load?

If that's the case I don't know which resistors are what. I'm assuming R2 = 10 ohm, how do I find R1 in order to solve for Vout?

Once I figure all this stuff out I'll move on to Part C.

Thanks again!

 

[CEL]

Okay so for Part A:

$$R_{2} = 50 \Omega$$

and

$$R_{1} = 16.67 \Omega$$

I'm still a little confused about part b. When I'm using my voltage divider equation,

$$V_{out} = \frac{R_{2}}{R_{1} + R_{2}} \cdot V_{in}$$

this time I'm computing Vout right? Isn't that the voltage that will remain at the load?

If that's the case I don't know which resistors are what. I'm assuming R2 = 10 ohm, how do I find R1 in order to solve for Vout?

Once I figure all this stuff out I'll move on to Part C.

Thanks again!

For part B, R1 is the source resistance (10 ohm) and R2 the load resistance (50 ohm).

 

[jegues]

For part B, R1 is the source resistance (10 ohm) and R2 the load resistance (50 ohm).

What told you that the source resistance was R1, and not R2? That's the piece of understanding I'm missing. Can you explain?

Aside from that little gap in my understanding, from what you've told I think I can finish off the problem.

So for Part B:

$$V_{out} = \frac{50}{60} \cdot 12$$

$$V_{out} = 10 V$$

So then for Part C: I want to drop Vout by 1V so I would simply add another 6.6 ohm resistor in series with R1.

Does everything look okay?

 

[CEL]

What told you that the source resistance was R1, and not R2? That's the piece of understanding I'm missing. Can you explain?

Aside from that little gap in my understanding, from what you've told I think I can finish off the problem.

So for Part B:

$$V_{out} = \frac{50}{60} \cdot 12$$

$$V_{out} = 10 V$$

So then for Part C: I want to drop Vout by 1V so I would simply add another 6.6 ohm resistor in series with R1.

Does everything look okay?

R2 is the resistance across which the voltage is measured.
Your calculations are OK.