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?
r = Vr/I = = (9-7.8)/1.5*10^-3=800 Ohms
To find the internal resistance of the battery, we can use Ohm's Law. Ohm's Law states that the voltage across a resistor (in this case, the internal resistance) is equal to the current passing through the resistor multiplied by the resistance.
The given emf (E) of the battery is 9.0 V, the terminal voltage (Vt) is 7.8 V, and the current passing through the battery (I) is 1.5 mA.
We can determine the voltage across the internal resistance (Vr) by subtracting the terminal voltage from the emf:
Vr = E - Vt
Vr = 9.0 V - 7.8 V
Vr = 1.2 V
Now, we can use Ohm's Law to find the internal resistance (R):
Vr = I * R
R = Vr / I
R = 1.2 V / 0.0015 A
R = 800 Ω
Therefore, the internal resistance of the battery is 800 Ω.
To find the internal resistance of the battery, you can use Ohm's law. Ohm's law states that the voltage across a component is equal to the current flowing through it multiplied by its resistance.
In this case, the voltage across the battery terminals is the emf minus the voltage drop across the internal resistance. Let's call the internal resistance "r". So, the voltage across the battery terminals is:
V = Emf - (current x r)
Given that the emf is 9.0 V, the terminal voltage is 7.8 V, and the current is 1.5 mA (which is 0.0015 A), we can calculate the internal resistance by rearranging the equation and solving for "r".
7.8 V = 9.0 V - (0.0015 A x r)
Rearranging the equation gives:
0.0015 A x r = 9.0 V - 7.8 V
0.0015 A x r = 1.2 V
Now, solve for "r" by dividing both sides of the equation by 0.0015 A:
r = (1.2 V) / (0.0015 A)
r ≈ 800 Ω
Therefore, the internal resistance of the battery is approximately 800 Ω.