n the circuit shown below, the voltage across the battery terminals is measured to be 12 V when the switch, S, is open. When S is closed, this voltage falls to 8 V and the resistor, R, is then dissipating a power of 100 W. The internal resistance of the battery in Ohms to two decimal places is:
Question
n the circuit shown below, the voltage across the battery terminals is measured to be 12 V when the switch, S, is open. When S is closed, this voltage falls to 8 V and the resistor, R, is then dissipating a power of 100 W. The internal resistance of the battery in Ohms to two decimal places is:
Solution
The power dissipated by the resistor (P) when the switch is closed is given by:
P = I^2 * R
where: I is the current through the resistor, and R is the resistance of the resistor.
We can rearrange this equation to solve for I:
I = sqrt(P / R)
We know that P = 100 W, but we don't know R. However, we can find R using Ohm's law (V = I * R) and the fact that the voltage across the resistor when the switch is closed is 8 V (since this is the voltage across the battery terminals when the switch is closed):
R = V / I R = 8 V / I
Substituting this into the equation for I gives:
I = sqrt(P / (V / I)) I = sqrt(I^2 * P / V) I^2 = I^2 * P / V I^2 * V = I^2 * P V = P
So, the current through the resistor when the switch is closed is sqrt(100 W / 8 V) = 3.54 A.
The internal resistance of the battery (r) can be found using the formula:
r = (V_open - V_closed) / I
where: V_open is the voltage across the battery terminals when the switch is open, and V_closed is the voltage across the battery terminals when the switch is closed.
Substituting the given values gives:
r = (12 V - 8 V) / 3.54 A = 1.13 Ω
So, the internal resistance of the battery is approximately 1.13 Ω to two decimal places.
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