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:
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
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?