Question

1. the probable fault in this circuit is

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

Explanation:

Step1: Calculate the inductive reactance

The formula for inductive reactance \(X_{L}=2\pi fL\). Given \(f = 10\times10^{3}\text{ Hz}\) and \(L=20\times10^{- 3}\text{ H}\), then \(X_{L}=2\pi\times10\times10^{3}\times20\times10^{-3}=400\pi\approx1256.6\Omega\).

Step2: Analyze the voltage distribution in a normal circuit

In a series \(RL\) circuit, \(V = V_{R}+V_{L}\). If there is no fault, \(V_{L}
eq0\). Here \(V_{L} = 0V\) and \(V_{R}=10V\) (equal to the source voltage).

Step3: Determine the fault

If the inductor is open, the current in the circuit is \(I = 0\). Using \(V_{R}=IR\), with \(I = 0\), \(V_{R}\) should be \(0\). But if the resistor is open, the current \(I = 0\), and the voltage across the inductor \(V_{L}=IX_{L}=0\), and the voltage across the open - resistor will be equal to the source voltage.

Answer:

C. an open resistor