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stira [4]
2 years ago
11

A transformer has input voltage and current of 12 V and 4 A

Physics
1 answer:
Advocard [28]2 years ago
4 0

Explanation:

It is given that,

Input voltage, V_i=12\ V

Input current, I_i=4\ A

Output current, I_o=0.8\ A

Number of turns in the secondary side of transformer, N_s=1177

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}

Substituting all the above values

So,

N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885

So, the number of turns in primary side of the transformer is 5885.

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Answer:

e)

Explanation:

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3 years ago
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Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational
vampirchik [111]

Answer:

I) c=1385.667\frac{J}{kg K}

II)The difference from the value obtained on part I is: 2000-1385.67 =614.33 \frac{J}{Kg K}

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

Explanation:

From the problem we have the molar mass given M=18\frac{gr}{mol} of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.

Part I

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Let C_v (\frac{J}{Kg K}) the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.

Let C (\frac{J}{Kg K}) the specific heat capcity this value represent the heat capacity aboserbed by mass.

For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by C_v =\frac{R}{2} so the total for the 6 degrees of freedom would be:

C_v =6*\frac{R}{2}=3R=3x8.314\frac{J}{mol K}=24.942\frac{J}{mol K}

And by definition we know that the specific heat capacity is defined:

c=\frac{C_V}{M}

If we replace all the values we have:

c=\frac{24.942\frac{J}{mol K}}{0.018\frac{kg}{mol}}=1385.667\frac{J}{kg K}

So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is c=1385.667\frac{J}{kg K}

Part II

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3 years ago
Help fast please!!!!!
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