91Ó°ÊÓ

Current is the term used to describe the speed at which charge moves.

a)

\(\begin{array}{c}{V_p} = 120V\\{N_p} = 500\\{N_s} = 4\end{array}\)

Transformer equation,

\(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}\)

b)

\(\begin{array}{c}{V_p} = 120\;V\\{V_s} = 0.96\;V\\{I_s} = 4A\end{array}\)

Transformer equation,

\(\frac{{{V_s}}}{{{V_p}}} = \frac{{{I_p}}}{{{I_s}}}\)

c)

\(\begin{array}{c}{V_p} = 12V\\\;{I_p} = 0.032\;A\end{array}\)

Power equation,

\(P{\rm{ }} = {\rm{ }}V{\rm{ }}I\)

a)

Consider the transformer equation,

\(\begin{array}{c}\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}\\\frac{{{V_s}}}{{120}} = \frac{4}{{500}}\\{V_s} = 0.96\;V\end{array}\)

Therefore, Voltage output is \(0.96\;V\).

b)

Consider the transformer equation,

\(\begin{array}{c}\frac{{{V_s}}}{{{V_p}}} = \frac{{{I_p}}}{{{I_s}}}\\\frac{{0.96}}{{120}} = \frac{{{I_p}}}{4}\\{I_p} = 0.032\;A\end{array}\)

Therefore, input current value is\(0.032\;A\).

c)

Consider the power equation,

\(\begin{array}{c}{P_p} = {V_p}{I_p}\\{P_p} = 120 \times 0.032\\ = 3.84W\end{array}\)

Therefore, input power is obtained as \(3.84W\).