I = V/Rl = 1.453/2.50 = 0.5812A.
r = (1.492-1.453)/0.5812 = 0.0671 Ohms = Internal resistance.
r = (1.492-1.453)/0.5812 = 0.0671 Ohms = Internal resistance.
First, let's understand the given information:
- When the battery is not used in a circuit, the voltmeter measures a potential difference of 1.492V across its terminals. This potential difference is known as the EMF of the battery.
- When a 2.50Ω light bulb is connected to the battery, the potential difference across the battery's terminals is 1.453V.
According to Ohm's Law, the potential difference (V) across a resistor is given by the equation V = I * R, where I is the current flowing through the resistor and R is the resistance of the resistor.
In this case, when the circuit is closed and the light bulb is connected, the potential difference across the battery's terminals (V1) can be calculated as the difference between the EMF of the battery and the potential difference across the resistor:
V1 = EMF - V
Given:
EMF = 1.492V
V = 1.453V
Substituting the values, we get:
V1 = 1.492V - 1.453V
V1 = 0.039V
Now, we need to find the current flowing through the circuit (I) using Ohm's Law:
V1 = I * R
Given:
V1 = 0.039V
R = 2.50Ω
Substituting the values, we can solve for I:
0.039V = I * 2.50Ω
Rearranging the equation:
I = 0.039V / 2.50Ω
I = 0.0156 A (Ampere)
Now, we can use the formula for calculating the internal resistance (r) of the battery:
r = (EMF - V1) / I
Given:
EMF = 1.492V
V1 = 0.039V
I = 0.0156 A
Substituting the values, we can solve for r:
r = (1.492V - 0.039V) / 0.0156 A
Calculating:
r = 1.453V / 0.0156 A
r = 93.15 Ω (Ohm)
Therefore, the internal resistance of the battery is 93.15 Ω.