Question

# The primary coil of an idea step up transformer has $$100\ turns$$ and transformation ratio is also $$100$$. The input voltage and power are respectively $$220\ V$$ and $$1100\ W$$ calculate $$(2M)$$1. Number of turns in secondary2. Current in primary3. Voltage across secondary4. Current in secondary5. Power in secondary

Solution

## Given that, Input voltage $${{V}_{input}}=220V$$ Power $${{P}_{input}}=1100\,W$$ Number of turns $${{N}_{P}}=100$$ $$r=100$$ (I). Number of turns in secondary We know that,   $$r=\dfrac{{{N}_{s}}}{{{N}_{p}}}$$  $${{N}_{_{S}}}=r\times {{N}_{p}}$$  $${{N}_{s}}=100\times 100$$  $${{N}_{s}}=10000$$   (II).The current in primary We know that   $${{I}_{p}}=\dfrac{{{P}_{input}}}{{{V}_{p}}}$$  $${{I}_{P}}=\dfrac{1100}{220}$$  $${{I}_{p}}=5\,A$$  (III). The Voltage across secondary We know that   $$r=\dfrac{{{V}_{s}}}{{{V}_{p}}}$$  $${{V}_{s}}=r\times {{V}_{p}}$$  $${{V}_{s}}=100\times 220$$  $${{V}_{s}}=22000\,V$$    (IV). The Current in secondary We know that,   $$r=\dfrac{{{I}_{p}}}{{{I}_{S}}}$$  $${{I}_{s}}=\dfrac{{{I}_{p}}}{r}$$  $${{I}_{s}}=\dfrac{5}{100}$$  $${{I}_{s}}=0.05\,A$$   (V). The power in secondary We know that, For ideal transformer Power in secondary = power in primary $${{P}_{s}}=1100\,W$$ This is the solution  Physics

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