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Introduction to SAT II Physics - FreeExamPapers

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across the loop. We can express these changes in voltage as an equation, and then<br />

substitute in the values we know for , , and :<br />

Now let’s apply the loop rule <strong>to</strong> the loop described by BCFE.<br />

Tracing the loop clockwise from B, the arrows cross<br />

, but in the wrong direction, from<br />

positive <strong>to</strong> negative, meaning that the voltage drops by 8 V. Next, the current crosses ,<br />

with an additional voltage drop of . Finally, it crosses , but in the opposite<br />

direction of the arrows, so the current goes up by<br />

. Now we can construct a second<br />

equation:<br />

Plugging this solution for in<strong>to</strong> the earlier equation of 4 + 3 = 12, we get:<br />

So the current across<br />

is 28/13 A. With that in mind, we can determine the current<br />

across and by plugging the value for in<strong>to</strong> the equations we derived earlier:<br />

235

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