Question

1. make inferences how does the current from a 1.5 v flashlight battery compare to the current from a 12 v car battery? explain.

Explanation:

Step1: Recall Ohm's Law

Ohm's Law is \(I=\frac{V}{R}\), where \(I\) is current, \(V\) is voltage, and \(R\) is resistance.

Step2: Analyze the relationship

Assume the resistance \(R\) (of the circuit connected to the battery) is the same for both cases. For the flashlight battery \(V_1 = 1.5\space V\) and for the car battery \(V_2=12\space V\). Then \(I_1=\frac{V_1}{R}=\frac{1.5}{R}\) and \(I_2=\frac{V_2}{R}=\frac{12}{R}\).

Step3: Compare the currents

Dividing \(I_1\) by \(I_2\) gives \(\frac{I_1}{I_2}=\frac{\frac{1.5}{R}}{\frac{12}{R}}=\frac{1.5}{12}=\frac{1}{8}\). So \(I_1=\frac{1}{8}I_2\)

Answer:

The current from the \(1.5\space V\) flashlight battery is \(\frac{1}{8}\) of the current from the \(12\space V\) car battery (assuming the resistance of the circuits they are connected to is the same).