# 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?

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

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

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:

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

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

$= 385 \setminus \text{turns}$