The ratio of a number of turns in the primary to secondary coil of a transformer is $3 : 2$ and the total number of turns in both the coils put together is $300$ . The output voltage is $200 V$ . Find the number of turns in each coil and the input voltage.

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Solution

Step 1: Given data
The ratio of the number of turns in the primary to the secondary coil $N_{p} : N_{s} = 3 : 2$
The total number of turns in both the coils put together is $300$
The output voltage $V_{s} = 200 V$
where $N_{s}$ and $N_{p}$ is number of turns in the secondary coil and primary coil respectively and $V_{s}$ and $V_{p}$ is voltage in the secondary and primary coil of the transformer.
Step 2: Finding $N_{p}$ and $N_{s}$
It is given that $\frac{N_{p}}{N_{s}} = \frac{3}{2}$ and $N_{p} + N_{s} = 300$
$\frac{N_{p}}{N_{s}} = \frac{3}{2}$ can be written as $N_{p} = N_{s} \times \frac{3}{2}$
Substituting the value of $N_{p}$ in $N_{p} + N_{s} = 300$ , we get
$N_{s} \times \frac{3}{2} + N_{s} = 300 \frac{5}{2} \times N_{s} = 300$
Cross multiply
$N_{s} = \frac{300 \times 2}{5} N_{s} = 120$
Substituting the value of $N_{s}$ in $N_{p} + N_{s} = 300$
$N_{p} + 120 = 300 N_{p} = 300 - 120 N_{p} = 180$
Step 3: Finding input voltage
We know that $\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{p}}$
Substituting values on the above formula
$\frac{120}{180} = \frac{200}{V_{p}}$
Cross multiplying
$V_{p} = \frac{180}{120} \times 200 V_{p} = 300 V$
Hence, the number of turns in the secondary coil is $120$ , the number of turns in the primary coil is $180$ and the input voltage is $300 V$ .

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