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stira [4]
3 years ago
11

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

Physics
1 answer:
Advocard [28]3 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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The blackbody emission spectrum of object A peaks in the ultraviolet region of the electromagnetic spectrum at a wavelength of 2
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Answer:

A) Object A is 3.25 times hotter.

B) Object A radiates 111.6 times more energy per unit of area.

Explanation:

Wiens's law states that there is an inverse relationship between the wavelength in which there is a peak in the emission of a black body and its temperature, mathematically,

\lambda_{peak}= \dfrac{0.0028976}{T},

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T=\dfrac{0.0028976}{\lambda_{peak}}.

Now, we can easily compute the temperatures.

For object A:

T_{A}=\dfrac{0.0028976}{200*10^{-9}}

T_{A}=14488K.

For object B:

T_{B}=\dfrac{0.0028976}{650*10^{-9}}

T_{B}=4458K

From this, we get that

T_{A}/T_{B}=3.25,

which means that object A is 3.25 times hotter.

Stefan's Law states that a black body emits thermal radiation with power proportional to the fourth power of its temperature.

This is

E=\sigma T^{4},

where  \sigma=5.67*10^{-8}\ Wm^{-2}K^{-4} is call the Stefan-Boltzmann constant.

From this, power can be easily compute:

E_{A}=(5.67*10^{-8}*(14488)^{4})=2.5*10^{9}W\\E_{B}=(5.67*10^{-8}*(4458)^{4})=22.4*10^{6}}W,

and we can notice that

E_{A}/E_{B}=111.6,

which means that object A radiates 111.6 time more energy per unit of area.

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