across the loop. We can express these changes in voltage as an equation, and then
substitute in the values we know for , , and :
Now let’s apply the loop rule to the loop described by BCFE.
Tracing the loop clockwise from B, the arrows cross , but in the wrong direction, from
positive to negative, meaning that the voltage drops by 8 V. Next, the current crosses ,
with an additional voltage drop of. Finally, it crosses , but in the opposite
direction of the arrows, so the current goes up by. Now we can construct a second
equation:
Plugging this solution for into the earlier equation of 4 + 3 = 12, we get:
So the current across is 28 / 13 A. With that in mind, we can determine the current
across and by plugging the value for into the equations we derived earlier: