<h2>Current and Voltage</h2>
<p>In an ideal DC circuit, current and voltage are constant. In an AC circuit, current and voltage vary with time. The value of the current or voltage at any specific time is called the <em>instantaneous current or voltage</em> and is calculated as follows:</p>
<p>\begin{align*} i &amp; = {I}_{\max}\sin\left(\text{2}\pi ft \right)\\ v &amp; = {V}_{\max}\sin\left(\text{2}\pi ft\right) \\ i &amp; = {I}_{\max}\sin\left(\omega t \right)\\ v &amp; = {V}_{\max}\sin\left(\omega t\right) \end{align*}</p>
<p>\(i\) is the instantaneous current. \({I}_{\max}\) is the maximum current. \(v\) is the instantaneous voltage. \({V}_{\max}\) is the maximum voltage. \(f\) is the frequency of the AC and t is the time at which the instantaneous current or voltage is being calculated.</p>
<p>The value we use for AC is known as the root mean square (rms) average. This is the same as what the DC voltage would be for the same source and is defined as:</p>
<p>\begin{align*} {I}_{rms}&amp; = \frac{{I}_{\max}}{\sqrt{\text{2}}}\\ {V}_{rms}&amp; = \frac{{V}_{\max}}{\sqrt{\text{2}}} \end{align*}</p>
<p>Since AC varies sinusoidally, with as much positive as negative, doing a straight average would get you zero for the average voltage. The rms value by-passes this problem.</p>
<div class="ns-school-emp">
<h2>Example: Laptop Transformer</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>The transformer for the laptop on which this book was written has the following information:</p>
<ul>
<li>INPUT: \(\text{100}\)-\(\text{240}\) \(\text{V}\); \(\text{1.5}\) \(\text{A}\); \(\text{50}\)/\(\text{60}\) \(\text{Hz}\)</li>
<li>OUTPUT:\(\text{20}\) \(\text{V}\); \(\text{3.25}\) \(\text{A}\)</li>
</ul>
<p>What changes from input to output, apart from the voltage and current values, and what does that imply? In addition, calculate the rms (root mean square) current and voltage values for the input and/or output as appropriate.</p>
</div>
<div>
<h3>Step 1: Comparing input and output</h3>
<p>The input description includes a frequency because it is designed for regular household use where we use alternating current. The output doesn't include a frequency. This implies that the output is not alternating current. This means that the output voltage and current will be constant with time.</p>
</div>
<div>
<h3>Step 2: RMS values</h3>
<p>Root mean square values are only applicable when dealing with alternating current. The transformer takes alternating current input and produces direct current output, this means that we only need to determine the rms values for the input.</p>

\begin{align*} V_{\text{rms}} &amp; = \frac{V_{\text{max}}}{\sqrt{\text{2}}} \\ V_{\text{rms}} &amp; = \frac{\text{240}\text{ V}}{\sqrt{\text{2}}} \\ &amp; = \text{169.71}\text{ V} \end{align*}

<p>Therefore \(V_{\text{rms}}=\text{169.71}\text{ V}\)</p>

\begin{align*} I_{\text{rms}} &amp; = \frac{I_{\text{max}}}{\sqrt{\text{2}}} \\ I_{\text{rms}} &amp; = \frac{\text{1.5}\text{ A}}{\sqrt{\text{2}}} \\ &amp; = \text{1.06}\text{ A} \end{align*}

<p>Therefore \(I_{\text{rms}}=\text{1.06}\text{ A}\)</p>
</div>
</div>
</div>
<div class="ns-school-emp">
<h2>Example: Camera Battery Charger</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>A camera charger has the following information:</p>
<ul>
<li>INPUT: \(\text{100}\)-\(\text{240}\) \(\text{V}\); \(\text{0.085}\) \(\text{A}\) (\(\text{100}\) \(\text{V}\)) - \(\text{0.05}\) \(\text{A}\) (\(\text{240}\) \(\text{V}\)) ; \(\text{50}\)/\(\text{60}\) \(\text{Hz}\)</li>
<li>OUTPUT:\(\text{4.2}\) \(\text{V}\); \(\text{0.7}\) \(\text{A}\)</li>
</ul>
<p>Calculate the rms (root mean square) current and voltage values for both \(\text{100}\) \(\text{V}\) and \(\text{240}\) \(\text{V}\) input.</p>
</div>
<div>
<h3>Step 1: Understanding the two cases</h3>
<p>The reason the transformer has the different input voltages listed is because it may be used internationally and not all countries use the same household voltage. The transformers purpose is to ensure that the output is consistent regardless of the input voltage. The different input voltages of \(\text{100}\) \(\text{V}\) and \(\text{240}\) \(\text{V}\) result in different input current values. This is why two different current values are listed under input but the voltage in parentheses tells you which case they are applicable to.</p>
<p>The cases are:</p>
<ul>
<li><strong>\(\text{100}\) \(\text{V}\):</strong> \(\text{0.085}\) \(\text{A}\)</li>
<li><strong>\(\text{240}\) \(\text{V}\):</strong> \(\text{0.05}\) \(\text{A}\)</li>
</ul>
</div>
<div>
<h3>Step 2: Input voltage of \(\text{100}\) \(\text{V}\)</h3>

\begin{align*} V_{\text{rms}} &amp; = \frac{V_{\text{max}}}{\sqrt{\text{2}}} \\ V_{\text{rms}} &amp; = \frac{\text{100}\text{ V}}{\sqrt{\text{2}}} \\ &amp; = \text{70.71}\text{ V} \end{align*}

<p>Therefore \(V_{\text{rms}}=\text{70.71}\text{ V}\)</p>

\begin{align*} I_{\text{rms}} &amp; = \frac{I_{\text{max}}}{\sqrt{\text{2}}} \\ I_{\text{rms}} &amp; = \frac{\text{0.085}\text{ A}}{\sqrt{\text{2}}} \\ &amp; = \text{0.06}\text{ A} \end{align*}

<p>Therefore \(I_{\text{rms}}=\text{0.06}\text{ A}\)</p>
</div>
<div>
<h3>Step 3: Input voltage of \(\text{240}\) \(\text{V}\)</h3>

\begin{align*} V_{\text{rms}} &amp; = \frac{V_{\text{max}}}{\sqrt{\text{2}}} \\ V_{\text{rms}} &amp; = \frac{\text{240}\text{ V}}{\sqrt{\text{2}}} \\ &amp; = \text{169.71}\text{ V} \end{align*}

<p>Therefore \(V_{\text{rms}}=\text{169.71}\text{ V}\)</p>

\begin{align*} I_{\text{rms}} &amp; = \frac{I_{\text{max}}}{\sqrt{\text{2}}} \\ I_{\text{rms}} &amp; = \frac{\text{0.5}\text{ A}}{\sqrt{\text{2}}} \\ &amp; = \text{0.35}\text{ A} \end{align*}

<p>Therefore \(I_{\text{rms}}=\text{0.35}\text{ A}\)</p>
</div>
</div>
</div>
<section></section>

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Current and Voltage

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Current and Voltage

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