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GarryVolchara [31]
3 years ago
10

Find the size of the resistor which would limit the current to a value of 2 mA when connected to an 8 V supply

Engineering
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

R = 4000 ohms

Explanation:

Given that,

Current, I = 2 mA

The voltage of the supply, V = 8 V

We need to find the resistance of the resistor. We know that, Ohm's law is as follows :

V = IR

Where

R is resistance

So,

R=\dfrac{V}{I}\\\\R=\dfrac{8}{2\times 10^{-3}}\\\\\\R=4000\ \Omega

So, the resistance of the resistor is 4000 ohms.

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Identify the correct statements in the context of friction factors of laminar and turbulent flows
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Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 ·K, deter- mine the temperature of the
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Answer:

Hello your question is incomplete attached below is the complete question

answer :

a) 95.80°C

b) 8.23 MW

Explanation:

Convection heat transfer coefficient = 860 W/m^2 . k

<u>a) Calculate for the temp of sheet metal when it leaves the oil bath </u>

<em>first step : find the Biot number </em>

Bi = hLc / K  ------- ( 1 )

where : h = 860 W/m^2 , Lc = 0.0025 m ,  K = 60.5 W/m°C

Input values into equation 1 above

Bi = 0.036 which is < 1  ( hence lumped parameter analysis can be applied )

<em>next : find the time constant </em>

t ( time constant ) = h / P*Cp *Lc  --------- ( 2 )

where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C

Input values into equation 2 above

t ( time constant ) = 0.10092 s^-1

<em>Determine the elapsed time </em>

T = L / V = 9/20 = 0.45 min

∴<u>   temp of sheet metal when it leaves the oil bath </u>

= (T(t) - 45 ) / (820 - 45)  = e^-(0.10092 * 27 )

T∞ =  45°C

Ti = 820°C

hence : T(t) = 95.80°C

<u>b) Calculate the required rate of heat removal form the oil </u>

Q = mCp ( Ti - T(t) ) ------------ ( 3 )

m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s

Cp = 434 J/kg°C

Ti =  820°C

T(t) = 95.80°C

Input values into equation 3 above

Q = 8.23 MW

6 0
3 years ago
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