Question

if the turns ratio is 5, the secondary voltage is greater than the primary
voltage by a factor of
​

Answer 1

The secondary voltage is greater than the primary  voltage by a factor of 1/5.

Explanation:

The turn ratio of a transformer is given by :

$a=\dfrac{n_p}{n_s}=\dfrac{V_p}{V_s}$

Here,

$n_p\ and\ n_s$ are number of turns in primary and second coil

$V_p\ and\ V_s$ are voltage in primary and second coil

$\dfrac{n_p}{n_s}=\dfrac{5}{1}=\dfrac{V_p}{V_s}$

$\dfrac{V_p}{V_s}=\dfrac{5}{1}$

So, the secondary voltage is greater than the primary  voltage by a factor of 1/5. Hence, this is the required solution.

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Answer 2

The secondary voltage is greater than the primary voltage by a factor of 1/5.

Explanation:

The turn ratio of a transformer is given by :

$a=\frac{n_p}{n_s}=\frac{V_p}{V_s}$

Here,

$n_p$ are $n_s$ number of turns in a primary and second coil

$V_p$ are $V_s$ voltage in the primary and second coil

$\frac{n_p}{n_s}=\frac{5}{1}= \frac{V_p}{V_s}$

$\frac{V_p}{V_s} =\frac{5}{1}$

So, the secondary voltage is greater than the primary  voltage by a factor of $1/5$. Hence, this is the required solution.

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