hen a battery is connected to a 100-Ω resistor, the current is 4.00 A. When the same battery is connected to a 449-Ω resistor, the current is 1.09 A. Find the emf supplied by the battery and the internal resistance of the battery.
I know that emf is equal to current times resistance but i must be missing something here because i always come up with the wrong answer.
See previous post: Thu, 2-19-15, 6:56 PM.
To find the emf supplied by the battery and the internal resistance, you need to set up a system of equations using Ohm's Law and the equation for electromotive force (emf).
First, let's establish the equations using Ohm's Law:
For the first scenario:
Ohm's Law: V = I1 * R1
where V is the voltage across the resistor, I1 is the current, and R1 is the resistance.
Given: I1 = 4.00 A, R1 = 100 Ω
For the second scenario:
Ohm's Law: V = I2 * R2
Given: I2 = 1.09 A, R2 = 449 Ω
Now, let's consider the equation for emf:
emf = V + Ir
where emf is the electromotive force, V is the voltage across the resistor, and Ir is the voltage drop due to the internal resistance of the battery.
Here's where the internal resistance comes into play. When a current flows through a battery, there is a voltage drop caused by the internal resistance of the battery itself. This voltage drop affects the total voltage output.
Assuming the internal resistance is denoted as r, we can rewrite the equations as follows:
For the first scenario:
emf = I1 * R1 + I1 * r
For the second scenario:
emf = I2 * R2 + I2 * r
Now, we have two equations with two unknowns (emf and r), which we can solve simultaneously to find the values.
1. Substitute the known values for the first scenario:
emf = (4.00 A) * (100 Ω) + (4.00 A) * r
2. Substitute the known values for the second scenario:
emf = (1.09 A) * (449 Ω) + (1.09 A) * r
Now, we have two equations:
emf = 400 Ω + 4r ...........(1)
emf = 488.41 Ω + 1.09r ...........(2)
3. Solve the system of equations simultaneously. Subtract equation (1) from equation (2):
0 = 88.41 Ω - 2.91r
4. Rearrange the equation:
2.91r = 88.41 Ω
5. Solve for r:
r ≈ 30.41 Ω
Now that we have the value of r, we can substitute it back into either of the original equations to find the value of emf.
Using equation (1):
emf = 400 Ω + 4r ≈ 400 Ω + (4 * 30.41 Ω) ≈ 527.64 Ω
Therefore, the emf supplied by the battery is approximately 527.64 Ω, and the internal resistance of the battery is approximately 30.41 Ω.