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.