Question

1. the probable fault in this circuit is

a a changed value component

b an open inductor

c a shorted inductor

d there is no fault

Explanation:

Step1: Analyze the voltage across the resistor

If the inductor is normal, there should be a voltage drop across the resistor according to Ohm's law \(V = IR\). But here the voltage across the \(2k\Omega\) resistor is \(0V\).

Step2: Consider the inductor's state

If the inductor is open - circuit, the current in the circuit would be \(0A\) (since an open - circuit breaks the current path), and both the inductor and resistor would have \(0V\) (by \(V = IR\)). But the inductor has \(15V\). If the inductor is shorted (\(R_{L}=0\Omega\)), the total resistance in the circuit is mainly from the resistor. Using Ohm's law \(I=\frac{V}{R}\) (where \(V = 15V\) and \(R = 2k\Omega\)), the current \(I=\frac{15V}{2000\Omega}=7.5mA\). But because the inductor is shorted (\(V_{L}=IR_{L}=0V\) as \(R_{L} = 0\)), and the source voltage appears across the inductor (a shorted inductor has \(0\) resistance, so all the source voltage is across it in this simple series - like situation considering the measurement points). A changed - value component would not lead to \(0V\) across the resistor in this way. So the inductor is shorted.

Answer:

C. a shorted inductor