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laila [671]
3 years ago
6

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

he capacitor is charged to 5.03 V, then discharged through the resistor. The time for the potential difference across the plates to decrease to 0.833 V is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from 11.5 μs to 6.33 ms, what should be the (a) lower value and (b) higher value of the resistance range of the resistor?
Engineering
1 answer:
anyanavicka [17]3 years ago
4 0

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

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Answer:

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given data

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to find out

rate of corrosion in (a) mpy and (b) mm/yr

solution

we get here the rate of corrosion that is express as

rate of corrosion = (k × W) ÷ (D × A × T)     ..................1

here k is constant  and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel

so put her value in equation 1 we get

Rate of corrosion = \frac{534*485*1000}{7.9*150*24hr*365day/yr}

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and

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All machines have three fundamental hazards: moving parts, point of operation, and?
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A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
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Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

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\dfrac{V_3}{V_2}=\dfrac{3200}{520}

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So expansion ratio\dfrac{V_4}{V_3}=3.25.

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\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

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4 years ago
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