When switch S in the figure is open, the voltmeter V of the battery reads 3.13V . When the switch is closed, the voltmeter reading drops to 2.90V , and the ammeter A reads 1.70A. Assume that the two meters are ideal, so they do not affect the circuit.
emf= V
sorry I left out the V and the A earlier.
Both posts are statements. Nowhere do you ask a question.
To find the emf (electromotive force) of the battery, we can use the formula:
emf = V + IR
Where:
emf is the electromotive force (or the voltage of the battery)
V is the voltmeter reading (with the switch open)
I is the current flowing through the circuit (measured by the ammeter)
R is the resistance of the circuit
Let's calculate the resistance of the circuit first:
Since the voltmeter is connected in parallel to the battery, it measures the same potential difference as the battery (emf).
So, when the switch is open, the voltmeter reading V is equal to the emf of the battery, which is 3.13V.
Now, when the switch is closed, the voltmeter reading drops to 2.90V, and the ammeter reads a current of 1.70A.
We can use Ohm's law, V = IR, to calculate the resistance (R).
R = V / I = 2.90V / 1.70A = 1.7059 Ω (rounded to four decimal places).
Now, we can find the emf (V) of the battery:
emf = V + IR = 2.90V + (1.70A * 1.7059 Ω) ≈ 5.79V.
Therefore, the electromotive force (emf) of the battery is approximately 5.79V.