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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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1 year ago
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

The shear stress will be 80,000,000 Pa or 80 MPa.

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