A 240 V ac supply is connected to the primary coil and a 6 W resistor is corrected to the secondary coil. What is the current in the primary coil?
A) 0.10 A
B) 0.14 A
C) 2 A
D) 40 A
Correct Answer: A
Solution :
[a] The equivalent primary load is \[{{R}_{1}}={{\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right)}^{2}}\,{{R}_{2}}={{\left( \frac{20}{1} \right)}^{2}}\,(6.0)\,=2400\,\Omega \] Current in the primary coil \[=\frac{240}{{{R}_{1}}}=\frac{240}{2400}=0.1A\]
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