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Questions  >>  CBSE XII  >>  Physics  >>  Alternating current
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Q)

The calculate the current drawn by the prime of a transformer which steps down $200\;v$ to $20\;v$ to operate a device of resistance $20 \Omega$ the efficiency of the transformer to be $80\%$

$\begin{array}{1 1} 0.125\;A \\ 1.25\;A \\ 12.50\;A \\ 125\;'A \end{array} $

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A)
Solution :
given $E_p= 200\;v$
$E_s= 20\;v$
$R_s= 20\; \Omega$
$I_s= \large\frac{E_s}{R_s}$
$\qquad= \large\frac{20}{10}$
$\qquad= 1\;A$
$n= \large\frac{E_sI_s}{E_pI_p}$
$\large\frac{80}{100}=\frac{20 \times 1}{200 \times I_p}$
$I_p = \large\frac{20 \times 1 \times 100}{80 \times 200}$
$I_P =\large\frac{1}{8}$$= 0.125\;A$
Answer : 0.125A
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