An old battery with an emf of 9.0 V has a terminal voltage of 7.8 V when it is supplying a current of 1.5 mA. What is the internal resistance of the battery?
800 Ω
900 Ω
400 Ω
600 Ω
See 12:35 PM post.
Henry, that doesn't help the time travelers from the future....
To find the internal resistance of the battery, we can use Ohm's Law and the equation for terminal voltage.
Ohm's Law states that V = IR, where V is the voltage across a component, I is the current passing through the component, and R is the resistance.
In this case, the terminal voltage (V) is 7.8 V and the current (I) is 1.5 mA (or 0.0015 A). The equation becomes:
7.8 V = 0.0015 A * R
To solve for R, divide both sides of the equation by 0.0015 A:
7.8 V / 0.0015 A = R
R = 5200 Ω
However, the resistance we found here is the total resistance, which is the sum of the internal resistance and any external resistance in the circuit. Since we are interested in the internal resistance, we need to subtract the external resistance.
In this case, the external resistance is 0 because it is not given. Therefore, the internal resistance of the battery is equal to the total resistance, which is 5200 Ω.
Since none of the given options match the calculated value of 5200 Ω, it's likely that there is a mistake in the options or the calculation.