Final answer:
The current that flows through a 3.200-V lithium cell with an internal resistance of 5.00Ω and a 1.00 kΩ voltmeter is approximately 3.20 mA. The voltmeter's resistance is added to the internal resistance to calculate the total resistance and apply Ohm's Law to find the current.
Step-by-step explanation:
The question involves determining the current flow through a lithium cell and understanding how internal resistance affects the measurement of a cell's terminal voltage using a voltmeter. To calculate the current, we will use Ohm's Law (V=IR), considering the total resistance of the circuit which is the sum of the internal resistance of the lithium cell and the resistance of the voltmeter.
The lithium cell has an electromotive force (emf) of 3.200 V and an internal resistance of 5.00Ω. The voltmeter, which measures terminal voltage, has a resistance of 1.00 kΩ (or 1000Ω). The total resistance in the circuit is therefore 1005Ω (5Ω internal + 1000Ω voltmeter).
Now, we apply Ohm's Law to find the current I that flows through the cell:
I = V / R
I = 3.200 V / 1005Ω
I = 0.003184 (~3.184 mA)
However, since we are looking for a value listed in the multiple-choice options, we round the current to the closest option, which is B) 3.20 mA.
Regarding the other parts of the question, the terminal voltage is simply the measured voltage across the terminals, which would be slightly less than the emf due to the drop across the cell's internal resistance when current is flowing. To find out how close the measured terminal voltage is to the emf, you would divide the measured terminal voltage by the emf and express it as a percentage or a ratio.