Alternating Current Ques 37
37. A step-up transformer operates on a $230$ $ V$ line and supplies a load of $2$ ampere. The ratio of the primary and secondary windings is $1: 25$. The current in the primary is
[1998]
(a) $25 $ $A$
(b) $50$ $ A$
(c) $15$ $ A$
(d) $12.5 $ $A$
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Answer:
Correct Answer: 37.(b)
Solution:
- (b) $\frac{N_p}{N_s}=\frac{E_p}{E_s}=\frac{1}{25}$
$\therefore E_s=25$ $ E_p$
But $E_s I_s=E_p I_p$
$25$ $ E_p \times 2=E_p \times I_p \Rightarrow I_p=50 $ $A$
If $n_p$ is number of turns in primary coil and $n_s$ is number of turns in secondary coil, then
$\frac{E_s}{E_p}=\frac{I_p}{I_s}=\frac{n_s}{n_p}=k$
Here, $I_p=$ current in primary
$k=$ Transformation ratio
$I_s=$ current is secondary
For a step up transformer, $k>1$
For step down transformer, $k<1$
BETA
