<p><img class=\"mceBlock\" src=\"https://s3.amazonaws.com/answer-board-image/596

Question

<p><img class="mceBlock" src="https://s3.amazonaws.com/answer-board-image/59619db6-f573-471b-a8e3-8c556fac5ae2.jpeg" border="0" alt="uploaded image" /><br />In the lab, the coils will be connected in series in a circuit like the one shown below. A small resistance r called shunt is included in the circuit to allow us to easily measure the current running through the coil with a voltmeter. (Resistances used in parallel with a meter in this way often are called shunts; in effect, the resistor &#8220;shunts&#8221; most of the current so that it bypasses the meter. Making the current measurement in this way provides a degree of protection for the meter.)</p>
<p>&#160;</p>
<p><span>If the potential difference across the shunt is measured to be 0.04 mV, and </span><span><em>r</em></span><span> = 0.03 Ohms, what is the current in each coil, in mA?</span></p>

Explanation / Answer

i = V/R = 0.04e-3/0.03 = 1.333 mA = 1.333*10-3 A

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