A voltmeter whose resistance is 1000 ohms measures the voltage of a worn out 1.5V battery as .9V. What is the internal Resistance
The Attempt at a Solution
In order to do this, I Solved for the Equiv Resistance of the circuit (leaving in the unknown Ri) by assuming they were two resistors in parallel. as:
Using ohms law I calculated the current through the circuit:
Plugging in my answer for Ri and simplifying I got:
Now assuming the resistors are in parallel, I would know that the current through each resistor is the same, meaning that I would get:
assuming the voltage drop across Ri as .6V (because 1.5-.9 is .6) I plugged that in and got:
However, when I solved for Ri, I got a negative number!
Help would be great!!
Next, thinking that they were in parallel, the current through both resistors was the same, so