Introduction to SAT II Physics - FreeExamPapers

we drew. If we cross a resistor against the direction of the arrows, the voltage rises by IR.
Further, if our loop crosses a battery in the wrong direction—entering in the positive
terminal and coming out the negative terminal—the voltage drops by V. To summarize:
• Voltage drops by IR when the loop crosses a resistor in the direction of the
current arrows.
• Voltage rises by IR when the loop crosses a resistor against the direction of the
current arrows.
• Voltage rises by V when the loop crosses a battery from the negative terminal to
the positive terminal.
• Voltage drops by V when the loop crosses a battery from the positive terminal to
the negative terminal.
Let’s now put the loop rule to work in sorting out the current that passes through each of
the three resistors in the diagram we looked at earlier. When we looked at the junction
rule, we found that we could express the current from A to B—and hence the current from
E to D to A—as , the current from B to E as , and the current from B to C—and hence
the current from C to F to E—as – . We have two variables for describing the current,
so we need two equations in order to solve for these variables. By applying the loop rule
to two different loops in the circuit, we should be able to come up with two different
equations that include the variables we’re looking for. Let’s begin by examining the loop
described by ABED.
Remember that we’ve labeled the current between A and B as
and the current between
B and E as
. Because the current flowing from E to A is the same as that flowing from A
to B, we know this part of the circuit also has a current of .
Tracing the loop clockwise from A, the current first crosses
and the voltage drops by
. Next it crosses and the voltage drops by . Then the current crosses , and
the voltage rises by 12 V. The loop rule tells us that the net change in voltage is zero
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