Question 348: AC fundamentals

Question

The number of turns in the primary and the secondary of a transformer are 1000 and 3000 , respectively. If $80 \mathrm{~V} \mathrm{~AC}$ is applied to the primary coil of the transformer, then the potential difference per turn of the secondary coil would be

Accepted Answer

$\because$ We know that, Transformation ratio is
$k=\frac{V_2}{V_1}$ .... (i)
and Turn ratio is
$\alpha=\frac{1}{k}=\frac{V_1}{V_2}=\frac{N_1}{N_2}$ ..... (ii)
Here, $N_1=$ Primary winding turns
$N_2=$ Secondary winding turns
$V_1=$ Primary winding voltage
$V_2=$ Secondary winding voltage
$\begin{array}{rlrl}
	ext { Given : } N_1 & =1000 \
N_2 & =3000 	ext { and } V_1=80 \mathrm{~V} \
V_2 & =? \
	herefore \quad \frac{V_2}{V_1} & =\frac{N_2}{N_1} \
\Rightarrow \quad & V_2 =\frac{N_2}{N_1} \cdot V_1 \quad 	ext{..... (iii)}
\end{array}$
put the values of $N_1, N_2$ and $V_1$ in above equation, we get
$V_2=\frac{3000}{1000} 	imes 80=240 \mathrm{~V}$
Now, potential difference across each turn
$=\frac{240}{3000}=0.08 \mathrm{~V}$

Answer

240 V

Answer

2400 V

Answer

0.24 V

Answer

0.08 V

AC fundamentals

The number of turns in the primary and the secondary of a transformer are 1000 and 3000 , respectively. If 80 \mathrm{~V} \mathrm{~AC} is applied to the primary coil of the transformer, then the potential difference per turn of the secondary coil would be

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