Answer:
(a) -55.6%, -31.25 V
(b) -11.1%, -6.25 V
(c) -0.93%, -0.52 V
Explanation:
The open-circuit voltage is (75V)(750/(250+750)) = 56.25 V.
The circuit to which the meter is connected has a Thevenin equivalent impedance of 1/(1/250 +1/750) = 187.5 kΩ. So, the relative error for a given meter impedance of R (in kΩ) is ...
(R/(R +187.5) -1) × 100% = -187.5/(R +187.5) × 100%
(a) For a meter impedance of (1 kΩ/V)×150V = 150 kΩ, the error is ...
-187.5/(150 +187.5) × 100% ≈ -55.6%
The voltage error is (56.25 V)(-55.6%) = -31.25 V.
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(b) For a meter impedance of (20 kΩ/V)×75V = 1500 kΩ, the error is ...
-187.5/(1500+187.5) × 100% ≈ -11.1%
The voltage error is (56.25 V)(-5.88%) = -6.25 V.
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(c) For a meter impedance of 20,000 kΩ, the error is ...
-187.5/(20000+187.5) × 100% ≈ -0.93%
The voltage error is (56.25 V)(-0.93%) = -0.52 V.
