When switch S is open, the voltmeter V of the battery reads 3.13 . When the switch is closed, the voltmeter reading drops to 2.90 , and the ammeter A reads 1.70 . Assume that the two meters are ideal, so they do not affect the circuit.
emf= V
I don't see a question here.
To find the electromotive force (emf) of the battery, we can use the equation:
emf = V + I * r
where:
emf is the electromotive force of the battery,
V is the voltage reading on the voltmeter when the switch is open,
I is the current flowing through the circuit, and
r is the internal resistance of the battery.
In this case, we have the following information:
V (voltmeter reading when the switch is open) = 3.13 V
V (voltmeter reading when the switch is closed) = 2.90 V
I (current reading on the ammeter) = 1.70 A
When the switch is closed, the voltmeter reading drops to 2.90 V, and the ammeter reads 1.70 A. This means that the voltmeter is measuring the voltage across the external resistance (r_ext) and the ammeter is measuring the current flowing through the circuit.
To find the internal resistance (r) of the battery, we can use Ohm's Law:
r = (V_open - V_closed) / I
where:
V_open is the voltmeter reading when the switch is open,
V_closed is the voltmeter reading when the switch is closed, and
I is the current flowing through the circuit.
Substituting the given values into the equation, we get:
r = (3.13 V - 2.90 V) / 1.70 A
= 0.23 V / 1.70 A
≈ 0.1353 Ω
Now that we have the internal resistance (r) of the battery, we can find the electromotive force (emf) using the equation:
emf = V_open + I * r
Substituting the given values into the equation, we get:
emf = 3.13 V + 1.70 A * 0.1353 Ω
= 3.13 V + 0.2301 V
≈ 3.36 V
Therefore, the electromotive force (emf) of the battery is approximately 3.36 V.