# Lesson 2 Flashcards — Circuit Fundamentals and Ohm's Law

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<summary><strong>1. What is the difference between charge, current, and voltage?</strong></summary>
<div class="card-answer"><p>Charge is the physical quantity itself, measured in coulombs. Current is the <em>rate</em> of charge flow, <span class="math notranslate nohighlight">\(I = dq/dt\)</span>, in amperes. Voltage is the energy available per unit charge, <span class="math notranslate nohighlight">\(V = \Delta E/dq\)</span>, in volts.</p></div>
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<summary><strong>2. Is current measured through a point or across two points? What about voltage?</strong></summary>
<div class="card-answer"><p>Current is measured <strong>through</strong> a point — it is a flow. Voltage is measured <strong>between two points</strong> — it is a difference. This is why an ammeter goes in series and a voltmeter goes in parallel.</p></div>
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<summary><strong>3. Which direction does conventional current flow?</strong></summary>
<div class="card-answer"><p>From high potential to low potential, i.e. from the + terminal to the − terminal. Note this is opposite to the direction electrons actually move; the convention predates the discovery of the electron.</p></div>
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<summary><strong>4. What does a battery actually store?</strong></summary>
<div class="card-answer"><p><strong>Charge</strong> — not voltage and not current. A battery rated 24 A·h holds <span class="math notranslate nohighlight">\(24 \times 3600 = 86{,}400\)</span> coulombs.</p></div>
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<summary><strong>5. State Ohm's Law and the power equation.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(V = I R\)</span> and <span class="math notranslate nohighlight">\(P = I V\)</span>. Combining them gives the two other useful forms, <span class="math notranslate nohighlight">\(P = I^2 R\)</span> and <span class="math notranslate nohighlight">\(P = V^2/R\)</span> — use whichever matches what you already know.</p></div>
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<summary><strong>6. What determines a resistor's resistance?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(R = \rho L / A\)</span>: the material's resistivity <span class="math notranslate nohighlight">\(\rho\)</span>, the length <span class="math notranslate nohighlight">\(L\)</span>, and the cross-sectional area <span class="math notranslate nohighlight">\(A\)</span>. Longer means more resistance; fatter means less.</p></div>
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<summary><strong>7. Copper has lower resistivity than gold or aluminum. Why is either ever used?</strong></summary>
<div class="card-answer"><p>Gold does not corrode, so it is used on contacts and connectors. Aluminum is far lighter and cheaper per unit of conductance, which is why transmission lines and aircraft wiring use it. Cost, weight, and corrosion beat raw resistivity.</p></div>
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<summary><strong>8. What is engineering notation, and how does it differ from scientific notation?</strong></summary>
<div class="card-answer"><p>The exponent is always a <strong>multiple of 3</strong>, so every result maps onto a standard prefix. <span class="math notranslate nohighlight">\(0.000\,030\,56\)</span> W becomes <span class="math notranslate nohighlight">\(30.56\,\mu\text{W}\)</span> rather than <span class="math notranslate nohighlight">\(3.056\times10^{-5}\)</span> W.</p></div>
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<summary><strong>9. Should a voltmeter's internal resistance be large or small? Why?</strong></summary>
<div class="card-answer"><p><strong>Large</strong>, ideally infinite. The meter connects in parallel with the device, so a low internal resistance would draw current and change the very voltage it is trying to measure.</p></div>
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<summary><strong>10. Should an ammeter's internal resistance be large or small? Why?</strong></summary>
<div class="card-answer"><p><strong>Small</strong>, ideally zero. The meter is inserted in series, so any resistance it adds would reduce the current it is measuring. This is also why connecting an ammeter directly across a source creates a short circuit.</p></div>
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<summary><strong>11. Energy costs 10&cent; per kW·h. A 120 W bulb runs for a year. What does it cost?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(E = P\,\Delta t = 120\,\text{W} \times 8760\,\text{h} = 1051.2\ \text{kW}\cdot\text{h}\)</span>, so the cost is about <strong>$105</strong>. Note the utility bills for <em>energy</em>, not power.</p></div>
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