Question

1. the probable fault in this circuit is

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

Explanation:

Step1: Calculate inductive reactance

The formula for inductive reactance is \(X_{L} = 2\pi fL\). Given \(f = 10\times10^{3}\text{ Hz}\) and \(L=2\times 10^{- 3}\text{ H}\), then \(X_{L}=2\pi\times10\times10^{3}\times2\times10^{-3}\).

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Since \(X_{L}\ll R = 2\times10^{3}\Omega\), in a series \(RL\) circuit, the voltage across the resistor \(V_{R}=\frac{R}{R + X_{L}}V_{s}\approx V_{s}\) and voltage across inductor \(V_{L}=\frac{X_{L}}{R + X_{L}}V_{s}\approx0\) when \(X_{L}\ll R\). But here \(V_{L} = 15V\) and \(V_{R}=0V\), which is not the case for a normal \(RL\) circuit.

Step2: Analyze inductor fault

If the inductor is open - circuit, the current in the circuit \(I = 0\) (because an open - circuit has infinite resistance). Using Ohm's law \(V = IR\), for the resistor \(V_{R}=IR = 0\) (since \(I = 0\)). The source voltage will appear across the open - circuit inductor.

Answer:

B. an open inductor