A) 0.125 A
B) 0.225 A
C) 0.325 A
D) 0.425 A
Correct Answer: A
Solution :
(a) 0.125 A Given: \[{{V}_{P}}=200\,\,V,\,R=20\,\Omega \,{{V}_{S}}=20\,V,\,\eta =80%\] Current through the secondary coil is \[{{l}_{S}}=\frac{{{V}_{S}}}{R}=\frac{20\,V}{20\,\Omega }=1\,A\] Efficiency of transformer, \[\eta =\frac{Output\,\,power}{Input\,\,power}=\frac{{{V}_{S}}{{l}_{S}}}{{{V}_{P}}{{l}_{P}}}\] Or \[{{l}_{P}}=\frac{{{V}_{S}}{{l}_{S}}}{{{V}_{P}}\eta }=\frac{20\times 1\times 100}{200\times 80}=0.125\,\,A\]
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