Question

A power (set up) transformer with an $1:8$ turn ratio has $60$ $H_z$, $120$ $V$ across the primary, the load in the secondary is ${10}^4$ $\Omega$. The current in the secondary is

• A. $96$ $A$
• B. $0.96$ $A$
• C. $9.6$ $A$
• D. $96$ $mA$

Answer 1

The correct option is D $96$ $mA$

$\frac{V_1}{V_2}=\frac{1}{8}$

$\Rightarrow \frac{I_1}{I_2}=8$

$\Rightarrow I_2=\frac{V_2}{R}$

$\Rightarrow I_2=\frac{8\times 120}{104}$

$\Rightarrow I_2=96$ $mA$