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<p> </p>
<p style="margin-bottom: 0in; text-align: center;"><b><u>Simple series circuits</u></b></p>
<p style="margin-bottom: 0in"> </p>
<p style="margin-bottom: 0in">Match each symbol to the correct component:</p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="cell.jpeg" style="width: 181px; height: 137px; float: right;"></mdo:image></p>
<p> </p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="variable%20resistor.jpeg" style="width: 181px; height: 137px; float: left;"></mdo:image><mdo:image alt="" src="battery.jpeg" style="width: 222px; height: 137px; float: left;"></mdo:image></p>
<p style="margin-bottom: 0in;"> </p>
<p> </p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="inductor.jpeg" style="width: 181px; height: 97px; float: right;"></mdo:image><mdo:image alt="" src="capacitor.jpeg" style="width: 168px; height: 149px; float: left;"></mdo:image></p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="voltmeter.jpeg" style="width: 181px; height: 137px; float: left;"></mdo:image></p>
<p> </p>
<p style="margin-bottom: 0in;"> </p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="bulb.jpeg" style="width: 181px; height: 137px; float: right;"></mdo:image></p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="resistor.jpeg" style="width: 232px; height: 86px; float: right;"></mdo:image></p>
<p> </p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="ampmeter.jpeg" style="width: 181px; height: 137px;"></mdo:image></p>
<p style="margin-bottom: 0in;">Resistor, Cell, Ampmeter, Battery, Voltmeter, Variable resistor, Bulb, Capacitor, Inductor.</p>
<p style="margin-bottom: 0in;"> </p>
<p style="margin-bottom: 0in;"> </p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="circuit%201.jpeg" style="width: 305px; height: 283px; float: right;"></mdo:image></p>
<p style="margin-bottom: 0in;"> </p>
<p style="margin-bottom: 0in">For the following circuit, find</p>
<ol type="i">
<li>
<p style="margin-bottom: 0in;">the current supplied by the electric cell</p>
</li>
<li>
<p style="margin-bottom: 0in;">the voltage through the resistor</p>
</li>
<li>
<p style="margin-bottom: 0in;">the voltage through the bulb</p>
</li>
<li>
<p style="margin-bottom: 0in;">the power used by the bulb</p>
</li>
<li>
<p style="margin-bottom: 0in;">the total power supplied by the cell</p>
</li>
</ol>
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<p style="margin-bottom: 0in">Fast solutions:</p>
<p style="margin-bottom: 0in;"> Variable resistor
Ammeter Voltmeter</p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="variable%20resistor.jpeg" style="width: 181px; height: 137px; float: left;"></mdo:image><mdo:image alt="" src="ampmeter.jpeg" style="width: 181px; height: 137px;"></mdo:image><mdo:image alt="" src="voltmeter.jpeg" style="width: 181px; height: 137px;"></mdo:image></p>
<p style="margin-bottom: 0in;">
Battery
Resistor Bulb</p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="battery.jpeg" style="width: 222px; height: 137px;"></mdo:image><mdo:image alt="" src="resistor.jpeg" style="width: 218px; height: 68px;"></mdo:image><mdo:image alt="" src="bulb.jpeg" style="width: 181px; height: 137px;"></mdo:image></p>
<p style="margin-bottom: 0in;"> </p>
<p style="margin-bottom: 0in;"> Inductor
Cell Capacitor</p>
<p style="margin-bottom: 0in;"><mdo:image alt="" src="inductor.jpeg" style="width: 181px; height: 97px;"></mdo:image><mdo:image alt="" src="cell.jpeg" style="width: 181px; height: 137px;"></mdo:image><mdo:image alt="" src="capacitor.jpeg" style="width: 168px; height: 149px;"></mdo:image></p>
<p style="margin-bottom: 0in;"> </p>
<p style="margin-bottom: 0in;"> </p>
<ol type="i">
<li>
<p style="margin-bottom: 0in">1/3 A</p>
</li>
<li>
<p style="margin-bottom: 0in">1V</p>
</li>
<li>
<p style="margin-bottom: 0in">2V</p>
</li>
<li>
<p style="margin-bottom: 0in">2/3 W</p>
</li>
<li>
<p style="margin-bottom: 0in">1W</p>
</li>
</ol>
<p style="margin-bottom: 0in"> </p>
<p style="margin-bottom: 0in"> </p>
<p style="margin-bottom: 0in">Detailed solutions:</p>
<ol type="i">
<li>
<p style="margin-bottom: 0in;">the total resistance of the circuit is</p>
<p style="margin-bottom: 0in;">$3\Omega + 6\Omega = 9\Omega$ because this is a series circuit and therefore resistances add up</p>
<p style="margin-bottom: 0in;">V=IR so 3 = I x 9 so I = 3/9 = 1/3 A</p>
</li>
<li>
<p style="margin-bottom: 0in;">The voltage across the resistor:V=IR so V=1/3 x 3 = 1V because the current is the same through the bulb and the resistor</p>
</li>
<li>
<p style="margin-bottom: 0in;">the voltage through the bulb is:<br></br>
V=IR so V=1/3 x 6 = 2V or because the total voltage across the whole circuit is 3V and the resistor is taking up 1V, we know that the bulb is taking 3V - 1V = 2V.</p>
</li>
<li>
<p style="margin-bottom: 0in;">The power used by the bulb:<br></br>
Power=IV so P=1/3 x 2 = 2/3 Watts</p>
</li>
<li>
<p style="margin-bottom: 0in;">The total power supplied by the cell:<br></br>
P = IV so P = 1/3 x 3 = 1 Watt<br></br>
This means that the resistor must be dissipating 1 - 2/3 = 1/3 Watts</p>
</li>
</ol>
<p> </p>
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Electrical circuits I
Simple electronic circuit problem to revise series circuits.
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