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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pychu [463]

Answer:

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