Question

1. the probable fault in this circuit is

a a changed value component
b a shorted inductor
c an open inductor
d there is no fault

Explanation:

Step1: Analyze the voltage across the inductor

In a normal RL - circuit, the inductor would have a voltage drop. Here, the voltage across the inductor is \(0V\).

Step2: Consider the properties of a shorted component

A shorted component has a very low resistance (ideally \(0\Omega\)). According to Ohm's law \(V = IR\), if \(R = 0\) (for a shorted inductor), then \(V=0\) (assuming a non - zero but finite current \(I\)).

Step3: Rule out other options

• If it were an open inductor (option C), the current in the circuit would be \(0\) (since an open circuit has infinite resistance), and the voltage across the resistor would be \(0V\) (by \(V = IR\), \(I = 0\) implies \(V = 0\) for the resistor), which is not the case here (voltage across resistor is \(120V\)).
• If there were a changed - value component (option A), it is not indicated by the given voltage measurements in a straightforward way as compared to the shorted inductor case.
• Since there is a \(0V\) across the inductor (which is not normal in a non - DC steady - state RL - circuit where \(V_L

eq0\) in the transient and \(V_L = 0\) only in DC steady - state, but here we have an AC source), option D (no fault) is incorrect.

Answer:

B. a shorted inductor