Question

The current drawn by the primary of a transformer, which steps down 200 V to 20 V to operate a device of

resistance 20$\Omega$is

(Assume the efficiency of the transformer to be 80%)

• A. 0.125 A
• B. 0.225 A
• C. 0.325 A
• D. 0.425 A

Answer 1

Answer: (A)

0.125 A

Answer 2

:given:

$V_{P} = 200V,R = 20\Omega,V_{s} = 20V,\eta = 80\%$

Current through the secondary coil is

$I_{s} = \frac{V_{s}}{R} = \frac{20V}{20\Omega} = A$

Efficiency of transformer $\eta = \frac{outputpower}{Inputpower} = \frac{V_{s}I_{s}}{V_{P}I_{P}}$

or , $I_{P} = \frac{V_{s}I_{s}}{V_{P}\eta} = \frac{20 \times 1 \times 100}{200 \times 80} = 0.125A.$