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laila [671]
3 years ago
10

A 250 kilo ohms and a 750 kilo ohms resistor are connected in series across a 75-volt source. Determine the error in measuring t

he voltage across each of these resistances when the voltage is read with (a)0-150 volt, 1000 ohms/volt meter
(b)0-75 volt, 20000 ohms/volt meter
(c)voltmeter with an input impedance of 20 megaohms
​
Engineering
1 answer:
Tom [10]3 years ago
3 0

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.

__

(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.

__

(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.

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