Since the energy of the charge decreases, point A is at a higher potential than point B; that is, VA– VB= 50 V. (b) The end-to-end potential difference is given by VA– VB= +iR+ ε, where is the emf of element C and is taken to be positive if it is to the left in the diagram. Thus, = VA– VB– iR= 50 V – (1.0 A)(2.0
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.