# ass15 - Solutions Assignment 15 25.33 IDENTIFY V = E Ir SET...

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Solutions, Assignment 15 25.33. I DENTIFY : V Ir = - E . S ET U P : The graph gives 9.0 V V = when 0 I = and 2.0 A I = when 0. V = E XECUTE : (a) E is equal to the terminal voltage when the current is zero. From the graph, this is 9.0 V. (b) When the terminal voltage is zero, the potential drop across the internal resistance is just equal in magnitude to the internal emf, so rI = E , which gives r = E /I = (9.0 V)/(2.0 A) = 4.5 Ω. E VALUATE : The terminal voltage decreases as the current through the battery increases. 25.36. I DENTIFY : The sum of the potential changes around the circuit loop is zero. Potential decreases by IR when going through a resistor in the direction of the current and increases by E when passing through an emf in the direction from the - to + terminal. S ET U P : The current is counterclockwise, because the 16 V battery determines the direction of current flow. E XECUTE : 16.0 V 8.0 V (1.6 5.0 1.4 9.0 ) 0 I + - - Ω+ Ω+ Ω+ Ω = 16.0 V 8.0 V 0.47 A 1.6 5.0 1.4 9.0 I - = = Ω+ Ω+ Ω+ (b) 16.0 V (1.6 ) b a V I V + - Ω = , so 16.0 V (1.6 )(0.47 A) 15.2 V. a b ab V V V - = = - = (c) 8.0 V (1.4 5.0 ) c a V I V + + Ω+ Ω = so (5.0 )(0.47 A) (1.4 )(0.47 A) 8.0 V 11.0 V. ac V = + + = (d) The graph is sketched in Figure 25.36.

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