Working out EMF and internal resistance

[PhysStudent81]

Homework Statement

Power supply of EMF = E and internal resistance of R. Need E and R.
This is connected to resistor of 10'000 Ohm. The current drawn from the power supply is 0.91A.
A voltmeter of internal resistance 10'000 Ohm is connected in parallel to the resistor. The voltmeter reads 8.3 Volts.

Original question (number 2) can be found here:

https://www.dropbox.com/s/n842i3ojnanhtxc/2013-01-13 14.37.43.jpg

Homework Equations

V = IR and Terminal PD = E - IR, presumably.

The Attempt at a Solution

Overall plan: Use the two situations to set up two simultaneous equations, then solve for E and R.
I'm assuming the questions is saying that the drawing of the current was independent of connecting the voltmeter.

From 1st part:

V = IR,
R = 10'000 Ohm
I = 0.91A
So V = 9'100 V
V = terminal PD so:
9'100 = E - 0.91×R (Call this 1)

From 2nd part:

Total resistance of Voltmeter and resistor is:

1/10000 + 1/10000 = 1/5000 so R_tot = 5000 Ohm.

Voltage measured is terminal voltage (8.3V)

Current is I = V/R = 8.3/5000 = (C 0.00166 A

So:

8.3 = E - 0.00166×R (Call this 2).

HOWEVER: combining 1 and 2 gives a negative internal resistance!

Any ideas what I've done wrong?
Thanks guys!

 

[gneill]

Homework Statement

Power supply of EMF = E and internal resistance of R. Need E and R.
This is connected to resistor of 10'000 Ohm. The current drawn from the power supply is 0.91A.
A voltmeter of internal resistance 10'000 Ohm is connected in parallel to the resistor. The voltmeter reads 8.3 Volts.

Original question (number 2) can be found here:

https://www.dropbox.com/s/n842i3ojnanhtxc/2013-01-13 14.37.43.jpg

Homework Equations

V = IR and Terminal PD = E - IR, presumably.

The Attempt at a Solution

Overall plan: Use the two situations to set up two simultaneous equations, then solve for E and R.
I'm assuming the questions is saying that the drawing of the current was independent of connecting the voltmeter.

From 1st part:

V = IR,
R = 10'000 Ohm
I = 0.91A
So V = 9'100 V
V = terminal PD so:
9'100 = E - 0.91×R (Call this 1)

From 2nd part:

Total resistance of Voltmeter and resistor is:

1/10000 + 1/10000 = 1/5000 so R_tot = 5000 Ohm.

Voltage measured is terminal voltage (8.3V)

Current is I = V/R = 8.3/5000 = (C 0.00166 A

So:

8.3 = E - 0.00166×R (Call this 2).

HOWEVER: combining 1 and 2 gives a negative internal resistance!

Any ideas what I've done wrong?



Thanks guys!

I note that they refer to it as "A power supply" and not a battery. Depending upon the complexity of the power supply and its ultimate purpose, it is possible that it will exhibit a negative internal resistance. This sort of thing is achievable with active components (amplifiers).

 

[TSny]

The potential difference across a 10 kΩ resistor carrying 0.91A would be 9100 V. I wonder if there's a typo in the book. Maybe the current was meant to be stated as .91 mA. Hard to say.

 

[PhysStudent81]

Ok thanks both of you.
I wanted to check whether I was missing something but I wasn't. It was a typo - setting the first current to 0.91mA gives the answer in the back of the book.

 

[Nickie]

It seems like you have made a mistake in your calculation for the current in the second part. The current should be 8.3/5000 = 0.00166 A, not 8.3/5000 = 0.00166 C. This would result in a positive internal resistance when combining the two equations.

Additionally, it may be helpful to label the equations as "equation 1" and "equation 2" rather than "call this 1" and "call this 2" to avoid confusion.

It is also important to note that the internal resistance of the power supply and the internal resistance of the voltmeter are not necessarily the same. So, it may be helpful to label them as "R1" and "R2" respectively in your equations.

Overall, your approach seems correct and it is good that you are setting up two simultaneous equations to solve for E and R. Just make sure to double check your calculations and labels to avoid any errors.