Problem 4-63

A voltmeter with a resistance of 85.5 KW is used to measure the voltage Vab in the circuit shown.
a) What is the voltmeter reading?
b) What is the percent error in the voltmeter reading? [(measured - actual) / actual] X 100%]

Note that the "actual" voltage is what we might call Voc.

Although we could do a brute force analysis to determine the "actual" voltage (Voc) and another to determine the "measured" voltage, it seems sensible to first determine the Thevenin equivalent, unless it appears that doing so would require MORE effort.

Since there are no dependent sources, one's initial thought is to use source transforms and circuit reduction. In this case, this is a viable option. (See problem 4-62 for a situation where this is not true.) Since source transforms are ususally rather simple, finding the Thevenin equivalent by this method looks easier than two full-blown analyses of the original circuit.

First transform the 5K and 50V in series.

Now combine the two current sources in parallel and the 5K and 20K in parallel. Note that since the current sources point in opposite directions, I will keep the orientation of the larger and subtract the value of the smaller.

Transform the current source and the 4K resistor.

Combine the 4K and 1K in series to give 5K, then transorm the 5 K and the 60 V source. (Hopefully you can follow those two steps without my including the intermediate step explicitly.)

Combine the 5K and 45 K in parallel to give 4.5K, then one final source transform. (Again, I am doing two steps in one.)

It should now be apparent that the "actual" voltage is 54 V.

To find the "measured" voltage, connect the 85.5K voltmeter to terminals a and b.

Voltage division seems the obvious method to find the "measured" voltage, VM.

VM = 54 [85.5K/(4.5K+85.5K)]
VM = 51.3 V (answer to part a)

The percent error is

E = [(54 - 51.3) / 51.3] X 100%
E = - 5.26% (answer to part b)

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