Question

1. the probable fault in this circuit is

a r1 open
b c1 open
c there is no fault
d a changed value

Explanation:

1. Analyze R1 branch: The voltage across R1 is 0V, but predicted current is 5.39 mA. For a resistor \( R = 1\ k\Omega \), if there were no fault, \( V = IR = 5.39\ mA\times1\ k\Omega = 5.39\ V \), but it's 0V.
2. Analyze C1's role: Capacitor C1 is in series with R1. If C1 is open, the circuit in the R1 branch is open, so no current flows through R1 (hence 0V across R1), matching the observation.
3. Eliminate other options:

• Option a (R1 open): If R1 were open, voltage across it would be equal to source voltage (or related), but here 0V, so incorrect.
• Option c (no fault): Discrepancy in R1 voltage/current shows fault, so incorrect.
• Option d (changed value): No indication of component value change; open circuit explains 0V across R1 better.

Answer:

b. C1 open