The two voltmeters in Figure 7-54 indicate the voltages shown. Apply logical thought and your knowledge of circuit operation to determine if there are any opens or shorts in the circuit and, if so, where they are located.

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Step1 : Determine if the voltmeter readings are correct. [latex]R_{1}[/latex],[latex]R_{2}[/latex] and [latex]R_{3}[/latex] act as a voltage divider. Calculate the voltage ( [latex] V_{A}[/latex]) across[latex]R_{3}[/latex] as follows;

[latex]V_{A} = \left(\frac{R_{3}}{R_{1} + R_{2} + R_{3}} \right)V_{S} = \left(\frac{3.3 \ k\Omega }{21.6 \ k\Omega } \right)24 \ V = 3.67 \ V[/latex]

The voltmeter A reading is correct. This indicates that [latex]R_{1}[/latex],[latex]R_{2}[/latex] and [latex]R_{3}[/latex] are connected and are not faulty.
Step 2 : See if the voltmeter B reading is correct.[latex] R_{6}[/latex] + [latex]R_{7}[/latex] is in parallel with [latex]R_{5}[/latex]. The series-parallel combination of [latex] R_{5}[/latex], [latex]R_{6}[/latex], and [latex] R_{7}[/latex], is in series with [latex] R_{4}[/latex]. Calculate the resistance of the [latex]R_{5}[/latex], [latex]R_{6}[/latex],and [latex]R_{7}[/latex] combination as follows:

[latex]R_{5\parallel (6+7)}= \frac{R_{5}(R_{6}+ R_{7})}{R_{5}+ R_{6}+ R_{7}}= \frac{(10 \ k\Omega )(17.2 \ k\Omega )}{27.2 \ k\Omega} = 6.32 \ k\Omega[/latex]

[latex]R_{5\parallel (6+7)}[/latex] and [latex]R_{4}[/latex] form a voltage divider, and voltmeter B measures the voltage across [latex]R_{5\parallel (6+7)}[/latex]. Is it correct? Check as follows:

[latex]V_{B}= \left(\frac{R_{5\parallel (6+7)}}{R_{4} + R_{5\parallel (6+7)}} \right) V_{S} = \left(\frac{6.32 \ k\Omega}{11 \ k\Omega} \right)24 \ V= 13.8 \ V [/latex]

Thus, the actual measured voltage (6.65 V) at this point is incorrect. Some logical thinking will help to isolate the problem.
Step 3 : [latex]R_{4}[/latex] is not open, because if it were, the meter would read 0 V. If there were a short across it, the meter would read 24 V. Since the actual voltage is much less than it should be,[latex]R_{5\parallel (6+7)}[/latex] must be less than the calculated value of 6.32 kΩ. The most likely problem is a short across [latex]R_{7}[/latex]. If there is a short from the top of [latex]R_{7}[/latex] to ground, [latex]R_{6}[/latex] is effectively in parallel with [latex]R_{5}[/latex]. In this case,

[latex]R_{5}\parallel R_{6}= \frac{R_{5}R_{6}}{R_{5} + R_{6}}= \frac{(10 \ k\Omega)(2.2 \ k\Omega)}{12.2 \ k\Omega} = 1.80 \ k\Omega[/latex]

Then [latex]V_{B}[/latex] is

[latex]V_{B}= \left(\frac{1.80 \ k\Omega}{6.5 \ k\Omega} \right) 24 \ V = 6.65 \ V[/latex]

This value for [latex]V_{B}[/latex] agrees with the voltmeter B reading. So there is a short across [latex]R_{7}[/latex]. If this were an actual circuit, you would try to find the physical cause of the short.

Question 7.21: The two voltmeters in Figure 7-54 indicate the voltages show...

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