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algol13
3 years ago
7

For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples

of 5 units has been 3.4 volts, and the range has been 1.3 volts.
Required:
What would the upper and lower control limits be for the resulting control charts (average and range)?
Physics
1 answer:
astra-53 [7]3 years ago
8 0

Answer:

Average :

UCL = 4.15

LCL = 2.65

Range :

UCL = 2.75

LCL = 0

Explanation:

Given :

Sample size, n = 5

Average, X = 3.4

Range, R = 1.3

A2 for n = 5 ; equals 0.577 ( X chart table)

For the average :

Upper Control Limit (UCL) :

X + A2*R

3.4 + 0.577(1.3) = 4.1501

Lower Control Limit (LCL) :

X - A2*R

3.4 - 0.577(1.3) = 2.6499

FOR the range :

Upper Control Limit (UCL) :

UCL = D4*R

D4 for n = 5 ; equals = 2.114

UCL = 2.114*1.3 = 2.7482

Lower Control Limit (LCL) :

LCL = D3*R

D3 for n = 5 ; equals = 0

LCL = 0 * 1.3 = 0

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The height of a circular tower is 179 meters and its diameter is 48 meters. (Assume that the building is a perfect cylinder). Wh
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3 years ago
An l-c cirucit with a 70 mh inductor and a ,54 F capacitor oscillates. The maximum charge on the capacitor is 11.5 C. What are t
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This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

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