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3 years ago

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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A composite wall is to be used to insulate a freezer chamber at -350C. Two insulating materials are to be used with conductiviti

choli [55]

Answer:

thickness1=1.4m

thickness2=2.2m

convection coefficient=0.33W/m^2K

Explanation:

you must use this equation to calculate the thickness:

L=K(T2-T1)/Q

L=thickness

T=temperature

Q=heat

L1=0.04*(0--350)/10=1.4m

L2=0.1(220-0)/10=2.2m

Then use this equation to calculate the convective coefficient

H=Q/(T2-T1)

H=10/(250-220)=0.33W/m^2K

7 0

4 years ago

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

$\epsilon_{car}=1-\frac{T_c}{T_H}$

Where $T_c$ is the atmospheric temperature and $T_H$ is the maximum burn temperature.

For the case (B), the efficiency we will use is:

$\epsilon_{b}=0.4\epsilon_{car}$

The work done by the engine can be calculated as:

$W=\epsilon Q=\epsilon H_v\cdot m_{fuel}$ where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

$m=\frac{P\cdot t}{\epsilon H_v}$