Question

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

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

Answer 1

The correct option is D $96mA$

Given,

Number of turns in primacy coil, $N_p$

Number of turns in secondary coil, $N_s$

Number of turn ratio $\frac{N_p}{N_s}=\frac{1}{8}$

Resistance in secondary, $R={10}^4\Omega$

Let,

Voltage in primacy coil, $V_p$

Voltage in secondary coil, $V_s$

Current in secondary, $I_s$

Transformer formula

$\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{1}{8}$

$V_s=8V_p=8\times 120=960V$

$I_s=\frac{V_s}{R}=\frac{960}{{10}^4}=0.096A$

$I_s=96mA$

Current in secondary, $I_s=96mA$