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Apply the loop rule to loop abcdefgha in Figure 21.25.
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<b>Figure 21.25</b>. A circuit schematic.
Figure 21.25. A circuit schematic.
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Solution Video

OpenStax College Physics Solution, Chapter 21, Problem 31 (Problems & Exercises) (1:33)

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Video Transcript

This is College Physics Answers with Shaun Dychko. This question asks us to traverse a loop beginning at point <i>a</i> and going in this direction to point <i>b, c, d, e, f, g</i> and then back to <i>a</i>. So we’re going in this direction to the loop here and we have to create our equation from the loop rule. So, one step at a time. We’ll start at point <i>a</i> and we’re going to traverse in the direction of the current across the resistors that means we have a negative. Negative <i>I2</i> multiplied by <i>R2</i> is that term. And then going from the negative terminal to the positive terminal is a positive change in potential. And so we have plus EMF one and then minus <i>I2</i> times little <i>r1</i>. And then we get to this point it gets a little bit strange because we’re traversing the loop in the opposite direction to the current through this resistor number three. And so this is going to be a positive current three multiplied <i>R3</i>. And then likewise for this resistor we are traversing it from right to left, despite the current going from left to right. So we’re traversing the opposite direction from the current. So this resistor also is a plus <i>I3</i> times little <i>r2</i>. And then across this EMF, we end at the negative terminal and we begin at the positive terminal so this is a reduction in a potential. And so this is minus EMF two and we end up at point <i>a</i>again. And so all of this equals zero.