Question

An iron transformer assumed to be $100\%$ efficient. The ration of the secondary truns to the primary truns is $1:20$. A $240V$ ac supply is connected to the primary coil. What is the current in the primary coil?

• A. $0.1A$
• B. $0.14A$
• C. $2A$
• D. $40A$

Answer 1

The correct option is A $0.1A$

As this is $100\%$ efficient this means the power in the secondary coil will be same as the power in the primary coil
i.e, $I_p{V}_p=I_s{V}_s$
$I_p=\frac{I_s{V}_s}{V_p}$
As we know that the ratio of emf is equal to the ratio of the number of turns.
So, $I_p=I_s.\frac{N_s}{N_p}=\frac{V_s}{R_s}.\frac{N_s}{N_p}=\left(\frac{N_S}{N_P}\right).\frac{V_p}{R_s}.\frac{N_s}{N_p}$
$I_p=\frac{N_s^2}{N_p^2}\frac{V_p}{R_s}$
$I_p={\left(\frac{1}{20}\right)}^2.\frac{240}{6}$
$I_p=\frac{1}{10}=0.1A$