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 is \(X_{L}=2\pi fL\). 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: Calculate the impedance of the circuit

In an \(RL\) series circuit, \(Z=\sqrt{R^{2}+X_{L}^{2}}\). Here \(R = 2\times10^{3}\Omega\) and \(X_{L}\approx1256.6\Omega\). But if the inductor is open, the current \(I = 0\). The voltage across an open - circuit component is equal to the source voltage (by Kirchhoff's voltage law, \(V = V_{R}+V_{L}\), when \(I = 0\), \(V_{R}=IR = 0\) and \(V_{L}=V\)). If the resistor is open, the voltage across the resistor would be the source voltage. If there is no fault, \(V_{R}=\frac{R}{R + X_{L}}V\) and \(V_{L}=\frac{X_{L}}{R + X_{L}}V\). Since the voltage across the inductor is \(0V\) (which is like an ideal wire for voltage measurement when open - circuit inductor), and the voltage across the resistor is \(10V\) (source voltage)

Step3: Analyze each option

• Option a: If a component has a changed value, we would expect some non - zero voltage across both components (not \(0V\) across one and source voltage across the other in a simple series circuit analysis).
• Option b: When the inductor is open, the current in the circuit is \(0\). Using \(V = IR\), the voltage across the resistor \(V_{R}=0\) (if \(I = 0\)), but in reality, when the inductor is open, the full source voltage appears across the open (inductor is open, so the voltmeter across the inductor is measuring the source voltage through the resistor. But if we consider ideal voltmeters with infinite impedance, for a series \(RL\) circuit with open inductor, the current \(I=\frac{V}{R+\infty}=0\). The voltage across the resistor \(V_{R}=IR = 0\) and the voltage across the inductor \(V_{L}=V\) (by \(V = V_{R}+V_{L}\)). But in the given circuit, the voltage across the inductor is \(0V\) (which is wrong for a non - open inductor in a normal \(RL\) circuit). Wait, no, if the inductor is open, the current \(I = 0\). The voltage across the resistor \(V_{R}=IR=0\) (but in the figure \(V_{R} = 10V\)). Wait, no, reverse - if the inductor is open, the current path is broken. The voltmeter across the resistor is connected in a way that if the inductor is open, the voltmeter across the resistor is actually measuring the source voltage (because there is no current, so no voltage drop across the inductor in terms of \(V = IR\) for the inductor. But the inductor as an open - circuit: \(V_{L}=V\) (source) and \(V_{R}=0\) (if we consider \(V_{R}=IR\)). But in the figure \(V_{R} = 10V\) (source voltage) and \(V_{L}=0V\). If the resistor is open, \(V_{R}=V\) (source) and \(V_{L}=0\) (since \(I = 0\)).
• Option c: If the resistor is open, the current \(I = 0\). Using \(V = IR\), \(V_{L}=0\) (because \(I = 0\)) and \(V_{R}=V\) (source voltage) which matches the given voltage measurements (\(V_{L}=0V\) and \(V_{R}=10V\))
• Option d: If there is no fault, \(V_{R}=\frac{R}{R + X_{L}}V=\frac{2000}{2000 + 1256.6}\times10\approx6.1V\) and \(V_{L}=\frac{X_{L}}{R + X_{L}}V=\frac{1256.6}{2000+1256.6}\times10\approx3.9V\)

Answer:

C. an open resistor