Question

A transformer is based on the principle of mutual induction. Input is supplied to primary coil and output is taken across the secondary coil of transformers. It is found that $E_s/E_p=i_p/i_s$ when there is no energy loss, the efficiency of a transformer is given by
$\eta =\frac{P_{\text{output }}}{P_{\text{input }}}=\frac{E_s{i}_s}{E_p{i}_p}$
How much current is drawn by the primary coil of a transformer which steps down $220V$ to $44V$ to operate a device with an impedance of $880\Omega$ ?

• A. 1 A
• B. 0.1 A
• C. 0.01 A
• D. 0.02 A

Answer 1

Answer: (C)

Here, $i_p=?,E_p=220V,E_s=44V,R_s=880\Omega$
Current in secondary coil, i.e., $i_s=\frac{E_s}{R_s}=\frac{44}{880}=\frac{1}{20}A$
As, $E_p{i}_p=E_s{i}_s$
Current drawn by primary coil i.e.,
$\therefore i_p=\frac{E_s{i}_s}{E_p}=\frac{44}{220}\times \frac{1}{20}$
$=0.01A$