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Ohm’s Law is a fundamental principle for understanding electric circuits. It states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. This relationship is mathematically represented as:\[ V = IR \]where:

• \(V\) is the voltage across the conductor in volts (V).
• \(I\) is the current flowing through the conductor in amperes (A).
• \(R\) is the resistance of the conductor in ohms (\(\Omega\)).

This law is particularly useful when analyzing circuits involving motors and generators, as it allows us to calculate unknown values like voltage, current, or resistance. In the motor problem we're discussing, Ohm’s Law is used to find the back electromotive force (EMF) by considering the voltage drop across the motor's internal resistance.

Motor resistance, particularly armature resistance, is a critical factor when dealing with electric motors. It refers to the electrical resistance encountered by current passing through the wire coils inside the motor. For any given motor, this resistance can influence the current flow significantly.
When a motor runs, its armature resistance often acts in series with the back EMF and supply voltage. In our exercise, the motor resistance was provided as \(0.720 \, \Omega\). This resistance is crucial for calculating both the back EMF—when the motor is in motion—and the initial current, which occurs when the motor is just turned on. Managing and calculating the correct resistance is vital to ensure proper motor function and to avoid excessive current that can potentially damage the motor at start-up.

When a motor is initially turned on, it hasn't yet developed the rotational motion required to generate back EMF. At this point, the current drawn by the motor is only determined by the applied voltage and the armature resistance. The initial current is very high because there’s no opposing force from back EMF.
We calculate it using Ohm's Law rearranged to \[ I_0 = \frac{V}{R} \]. For our problem, this means applying the full voltage across the armature, leading to:\[ I_0 = \frac{120.0}{0.720} = 166.67 \, \text{A} \]This calculation highlights why additional series resistance might be needed—without it, the enormous initial current could harm the motor. To mitigate such risks, engineers often add extra resistance to limit the starting current to a safer level, as demonstrated by finding a series resistance needed to reduce the initial current to 15.0 A in this specific exercise.