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How much current is drawn by the primary coil of a transformer which steps down 220 V to 22 V to operate device with an impedance of 220 ohm.
1. 10 A
2. 1 A
3. 0.1 A
4. 0.01 A

1 Answer

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Best answer
Correct Answer - Option 4 : 0.01 A

\({I_s} = \frac{{{E_s}}}{{{R_s}}} = \frac{{22}}{{220}} = 0.1\;A\)

For an ideal transformer

\(\frac{{{I_p}}}{{{I_s}}} = \frac{{{E_s}}}{{{E_p}}} = \frac{{{N_s}}}{{{N_p}}} \Rightarrow {I_p} = \frac{{{E_s}}}{{{E_p}}} \times {I_s} = \frac{{22}}{{220}} \times 0.1 = {10^{ - 2}}A\)

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