From the indicated voltmeter reading in Figure 7-51. determine if there is a fault by applying the APM approach. If there is a fault, identify it as either a short or an open.

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Step 1: Analysis
Determine what the voltmeter should be indicating as follows. Since [latex]R_{2}[/latex] and [latex]R_{3}[/latex] are in parallel, their combined resistance is

[latex]R_{2 \parallel 3}= \frac{R_{2}R_{3}}{R_{2}+ R_{3}} = \frac{(4.7 \ k\Omega )(10 \ k\Omega )}{14.7 \ k\Omega } = 3.20 \ k\Omega [/latex]

Determine the voltage across the parallel combination by the voltage-divider formula,

[latex]V_{2 \parallel 3}= \left(\frac{R_{2 \parallel 3}}{R_{1}+ R_{2 \parallel 3}} \right) V_{S} = \left(\frac{3.2 \ k\Omega}{18.2 \ k\Omega} \right)24 \ V= 4.22 \ V [/latex]

This calculation shows that 4.22 V is the voltage reading that you should get on the meter. However, the meter reads 9.6 V across [latex]R_{2 \parallel 3}[/latex]. This value is incorrect, and, because it is higher than it should be. either [latex]R_{2}[/latex] or [latex]R_{3}[/latex] is probably open. Why? Because if either of these two resistors is open, the resistance across which the meter is connected is larger than expected. A higher resistance will drop a higher voltage in this circuit.

Step 2: Planning
Start trying to find the open resistor by assuming that is open. If it is, the voltage across [latex]R_{3}[/latex] is

[latex]V_{3}= \left(\frac{R_{3}}{R_{1}+ R_{3}} \right) V_{S} = \left(\frac{10 \ k\Omega}{25 \ k\Omega} \right)24 \ V= 9.6 \ V [/latex]

Since the measured voltage is also 9.6 V, this calculation shows that [latex]R_{2 }[/latex] is probably open.

Step 3: Measurement
Disconnect power and remove [latex]R_{2 }[/latex]. Measure its resistance to verify it is open. If it is not, inspect the wiring, solder, or connections around [latex]R_{2 }[/latex], looking for the open.

Question 7.19: From the indicated voltmeter reading in Figure 7-51. determi...

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