The output of a step-down transformer is meas | vedclass

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(B) The output power of the transformer is given by $P_s = V_s I_s$,where $V_s$ is the $RMS$ voltage and $I_s$ is the $RMS$ current.Given $P_s = 12 \

Vedclass · Accepted Answer

$0.71$

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(B) The output power of the transformer is given by $P_s = V_s I_s$,where $V_s$ is the $RMS$ voltage and $I_s$ is the $RMS$ current.Given $P_s = 12 \ W$ and $V_s = 24 \ V$.Therefore,the $RMS$ current $I_s = \frac{P_s}{V_s} = \frac{12}{24} = 0.5 \ A$.The peak current $I_m$ is related to the $RMS$ current $I_s$ by the formula $I_m = \sqrt{2} I_s$.Substituting the value of $I_s$,we get $I_m = \sqrt{2} 	imes 0.5 \approx 1.414 	imes 0.5 = 0.707 \ A$.Rounding to two decimal places,the peak current is $0.71 \ A$.

The output of a step-down transformer is measured to be $24 \ V$,when connected to a $12 \ W$ light bulb. The value of peak current is $...... \ A$. (in $A$)

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