Question 15.27: The circuit represented by the schematic in Figure 15-76 has...

Apply the APM method to this troubleshooting problem.

Analysis: First think of the possible causes for the circuit to have no output voltage.
1. There is no source voltage or the frequency is so high that the capacitive reactance is almost zero.

2. There is a short between the output terminals. Either the capacitor could be internally shorted, or there could be some physical short in the circuit.
3. There is an open between the source and the output. This would prevent current and thus cause the output voltage to be zero. The resistor could be open, or the conductive path could be open due to a broken or loose connecting wire or a bad protoboard contact.
4. There is an incorrect component value. The resistor could be so large that the current and, therefore, the output voltage are negligible. The capacitor could be so large that its reactance at the input frequency is near zero

Planning : You decide to make some visual checks for problems such as the function generator power cord not plugged in or the frequency set at an incorrect value. Also,broken leads, shorted leads, as well as an incorrect resistor color code or capacitor label often can be found visually. If nothing is discovered after a visual check, then you will make voltage measurements to track down the cause of the problem. You decide to use a digital oscilloscope and a DMM to make the measurements.
Measurement: Assume that you find that the function generator is plugged in and the
frequency setting appears to be correct. Also, you find no visible opens or shorts during your visual check, and the component values are correct.
The first step in the measurement process is to check the voltage from the source with the scope. Assume a 10 V rms  sine wave with a frequency of 5 kHz is observed at the circuit input as shown in Figure 15-77(a). The correct voltage is present, so the first possible cause has been eliminated.
Next, check for a shorted capacitor by disconnecting the source and placing a DMM (set on the ohmmeter function) across the capacitor. If the capacitor is good, an open will be indicated by an OL (overload) in the meter display after a short charging time. Assume the capacitor checks okay, as shown in Figure 15-77(b). The second possible cause has been eliminated. Since the voltage has been “lost” somewhere between the input and the output, you must now look for the voltage. Reconnect the source and measure the voltage across the resistor with the DMM (set on the voltmeter function) from one resistor lead to the
other. The voltage across the resistor is zero. This means there is no current, which indicates an open somewhere in the circuit.
Now, begin tracing the circuit back toward the source looking for the voltage (you could also start from the source and work forward). You can use either the scope or the DMM but decide to use the multimeter with one lead connected to ground and the other used to probe the circuit. As shown in Figure 15-77(c), the voltage on the right lead of the resistor, point (1), reads zero. Since you already have measured zero voltage across the resistor, the voltage on the left resistor lead at point (2) must be zero as the meter indicates. Next, moving the meter probe to point (3). you read 10 V. You have found the voltage! Since there is zero volts on (he left resistor lead, and there is 10 V at point (3), one of the two contacts in the protoboard hole into which the wire leads are inserted is bad. It could be that the small contacts were pushed in too far and were bent or broken so that the circuit lead does not make contact.Move either or both the resistor lead and the wire to another hole in the same row.Assume that when the resistor lead is moved to the hole just above, you have voltage at the output of the circuit (across the capacitor).