Question

The primary coil of an ideal step up transformer has 100 turn and transformation

ratio is also 100. The input voltage and power are 220V and 1100W

respectively. Calculate

(a) The number of turns in secondary

(b) The voltage across the secondary

(c) The current in the secondary

(d) The current in the primary

(e) The power in the secondary

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{N}}_{\mathrm{p}}=100 \\ \mathrm{r}=100 \\ {\mathrm{V}}_{\mathrm{p}}=220\mathrm{V} \\ {\mathrm{P}}_{\mathrm{input}}=1100\mathrm{W} \\ \left(1\right)\mathrm{r}=\frac{{\mathrm{N}}_{\mathrm{s}}}{{\mathrm{N}}_{\mathrm{p}}} \\ {\mathrm{N}}_{\mathrm{s}}=\mathrm{r}\times {\mathrm{N}}_{\mathrm{p}} \\ {\mathrm{N}}_{\mathrm{s}}=100\times 100=10000 \\ \left(2\right)\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}}}=\mathrm{r} \\ {\mathrm{V}}_{\mathrm{s}}=\mathrm{r}\times {\mathrm{V}}_{\mathrm{p}} \\ {\mathrm{V}}_{\mathrm{s}}=100\times 220=22000\mathrm{volt} \\ \left(3\right){\mathrm{P}}_{\mathrm{input}}={\mathrm{V}}_{\mathrm{p}}{\mathrm{I}}_{\mathrm{p}} \\ {\mathrm{I}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{\mathrm{input}}}{{\mathrm{V}}_{\mathrm{p}}} \\ {\mathrm{I}}_{\mathrm{p}}=\frac{1100}{220}=5\mathrm{A} \\ \frac{{\mathrm{I}}_{\mathrm{p}}}{{\mathrm{I}}_{\mathrm{s}}}=\mathrm{r} \\ {\mathrm{I}}_{\mathrm{s}}=\frac{{\mathrm{I}}_{\mathrm{p}}}{\mathrm{r}} \\ {\mathrm{I}}_{\mathrm{s}}=\frac{5}{100}=0.05\mathrm{A} \\ \left(4\right){\mathrm{I}}_{\mathrm{p}}=5\mathrm{A} \\ \left(5\right)\mathrm{For}\mathrm{ideal}\mathrm{transformer}, \\ \mathrm{Power}\mathrm{in}\mathrm{secondary}=\mathrm{Power}\mathrm{in}\mathrm{primary}=1100\mathrm{W} \end{gathered}$$