Question

How much current is drawn by the primary coil of a transformer which steps down $220\, V$ to $22\, V$ to operate a device with an impedance of $220\, \Omega$ ?

• A. 1 A
• B. 0.25 A
• C. 0.01 A
• D. 1.50 A

Answer 1

Answer: (C)

Here, $\varepsilon_{p}=220\, V , \varepsilon_{s}=22\, V$\
and $Z=220\, \Omega$ If $i_{s}$ is the current through the secondary, then
$i_{s}=\frac{\varepsilon_{s}}{Z}=\frac{22\, V }{220\, \Omega}=0.1\, A$
We know that, $\varepsilon_{p} i_{p}=\varepsilon_{s} i_{s}$
where, $i_{p}$ is the current drawn by the primary coil
Thus, $220 \times i_{p}=22 \times 0.1\, A$
$\Rightarrow i_{p}=0.01\, A$