Question

1. the probable fault in this circuit is

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

Explanation:

Step1: Analyze the voltage across components

In a series circuit (inductor and resistor in series here), if the resistor is open (has infinite resistance), the current in the circuit is zero. According to \(V = IR\), for the inductor \(V_L=I X_L\) (where \(X_L = 2\pi fL\)) and for the resistor \(V_R=IR\). When \(I = 0\), \(V_L=0\) (as measured in the circuit) and all the source voltage (\(120 V\)) appears across the open - resistor (since there is no current to cause a voltage drop across the inductor).

Step2: Rule out other options

• If it were a changed - value component (a), we would expect some non - zero voltage across both components (depending on the new values).
• If the resistor were shorted (b), the voltage across the resistor would be \(V_R = 0\) and the inductor would have a non - zero voltage (\(V_L=120 V\) as \(V = V_L+V_R\) in series and \(V = 120 V\) source).
• Since the measured voltages (\(V_L = 0 V\) and \(V_R=120 V\)) are consistent with an open resistor, option (d) (no fault) is incorrect.

Answer:

C. an open resistor