Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 19 - DC Circuits - Search and Learn - Page 559: 4

Answer

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Work Step by Step

When the galvanometer has a null/zero reading, there is no current in it, or through the emf that is being measured. All of the current is flowing through the long uniform slide wire. Put the unknown emf in place, and apply Kirchhoff’s loop rule to the lower loop. $$\epsilon_x-IR_x=0$$ Note that there is no current through the emf, so any internal resistance in the emf or any resistance in the galvanometer won’t cause any voltage drops. Now put the standard emf in place, and again apply Kirchhoff’s loop rule to the lower loop. $$\epsilon_s-IR_s=0$$ Note that the current flowing through the slide wire, I, is the same in both situations. It is independent of the emf in the lower loop because there’s no current flowing in the lower loop. Solve the 2 loop rule equations for the current I, and set them equal. $$I=\frac{\epsilon_x }{R_x}=\frac{\epsilon_s }{R_s}$$ $$ \epsilon_x = (\frac{ R_x}{R_s})\epsilon_s $$ This was the relationship to be shown.
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