Question

to initiate the self - excitation process, the rotor requires an initial excitation voltage from the battery, via the ignition switch and the charging system indicator light on the dashboard. once the rotor is excited and begins rotating within the stator, the alternator is able to generate its own current for continued self - excitation. during initial excitation, the charging system indicator lamp will be illuminated due to the voltage difference on either side of the lamp. after the alternator starts producing current, the voltage generated by the alternator is sent to d+. this causes the charging system indicator lamp to switch off, because the same charging system voltage is now present on either side of the lamp. the charging system indicator lamp is not illuminated when the engine is running, and the alternator is working properly. how is it that the indicator lamp is switched off? the voltage drops at test point 15 in the diagram. the voltage generated by the alternator comes to d+. it drops 12 volts across the charging system indicator lamp. incorrect. the voltage at test point 15 is 12 volts. the voltage generated by the alternator at d+ is 12 volts. how much does the voltage drop across the charging current indicator?

Explanation:

Step1: Analyze the voltage situation

When the alternator works properly, the voltage generated by the alternator is sent to D+. Before the alternator works, there is a voltage difference across the charging - system indicator lamp (so it lights up). After the alternator works, the voltage on both sides of the lamp is the same (because the voltage generated by the alternator is sent to D+).

Step2: Calculate the voltage drop

Since the voltage at test point 15 is 12 volts and the voltage generated by the alternator at D+ is 12 volts, according to the formula for voltage drop \(V = V_1 - V_2\) (where \(V_1\) and \(V_2\) are the voltages on both sides of the component). Here, \(V_1=V_2 = 12\) volts. So the voltage drop \(V=12 - 12=0\) volts.

Answer:

0