(a) The plug-in transformer for a laptop computer puts out \(7.50V\) and can supply a maximum current of \(2.00A\). What is the maximum input current if the input voltage is \(240V\)? Assume \(100\% \) efficiency. (b) If the actual efficiency is less than \(100\% \), would the input current need to be greater or smaller? Explain.

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

a)

\(\begin{array}{c}{V_p} = 240\;V\\{V_s} = 7.5\;V\\{I_s} = 2\;A\end{array}\)

Transformer equation,

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

b)

\(\begin{array}{c}{V_p} = 240\;V\\{V_s} = 7.5\;V\\{I_s} = 2\;A\end{array}\)

Transformer equation,

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

Where \(\eta \) is the efficiency

a)

Consider the transformer equation,

\(\begin{array}{c}\frac{{{V_s}}}{{{V_p}}} = \frac{{{I_p}}}{{{I_s}}}\\\frac{{7.5}}{{240}} = \frac{{{I_p}}}{2}\\{I_p} = 0.0625\;A\end{array}\)

Therefore, input current value is \(0.0625\;A\).

b)

Consider the transformer equation.

\(\begin{array}{c}\eta {V_p}{I_p} = {V_s}{I_s}\\\eta \times 240 \times {I_p} = \eta \times 7.5 \times 2\\{I_p} = \frac{{0.0625}}{\eta }\end{array}\)

We know that\(\eta \)is less than\(1\), which means\({I_p}\) is more than\(0.0625\;A\). So, input current need to be more.

Therefore, input current need to be more.