# If a transformer consist of 770 turns in primary coil with 220 primary voltage produces 110 secondary voltage, then calculate the number of turns in secondary coil?

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Nam D. Share
Apr 24, 2018

$385$ turns in the secondary coil

#### Explanation:

Using the transformer equation,

${V}_{p} / {V}_{s} = {N}_{p} / {N}_{s}$

• ${V}_{p} , {V}_{s}$ are the voltages of the primary and secondary transformers

• ${N}_{p} , {N}_{s}$ are the numbers of coil turns in the primary and secondary transfomers

And so, we got:

$\frac{220 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{V")/(110color(red)cancelcolor(black)"V")=(770 \ "turns}}}}}{{N}_{s}}$

${N}_{s} = \frac{770 \setminus \text{turns}}{2}$

$= 385 \setminus \text{turns}$

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Apr 24, 2018

The number of turns is $= 385$

#### Explanation:

The equation of the transformer is

${V}_{p} / {V}_{s} = {n}_{p} / {n}_{s}$

The number of turns of the primary coil is ${n}_{p} = 770 \text{ turns}$

The number of turns of the secondary coil is ${n}_{s} = \text{? turns}$

The voltage in the primary coil is ${V}_{p} = 220 \text{ V}$

The voltage in the secondary coil is ${V}_{s} = 110 \text{ V}$

Therefore,

The number of turns of the secondary coil is

${n}_{s} = {V}_{s} / {V}_{p} \cdot {n}_{p} = \frac{110}{220} \cdot 770 = 385 \text{ turns}$

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