ch27-p036 - enough V 2 will result in a negative value for...

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(c) Looking at the point where the upward-sloping i 2 line crosses the axis (at V 2 = 4 V), we note that i 1 = 0.1 A there and that the loop rule around the right-hand loop should give V 1 i 1 R 1 = i 1 R 2 when i 1 = 0.1 A and i 2 = 0. This leads directly to R 2 = 40 . 36. (a) For typing convenience, we denote the emf of battery 2 as V 2 and the emf of battery 1 as V 1 . The loop rule (examining the left-hand loop) gives V 2 + i 1 R 1 V 1 = 0. Since V 1 is held constant while V 2 and i 1 vary, we see that this expression (for large
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Unformatted text preview: enough V 2 ) will result in a negative value for i 1 – so the downward sloping line (the line that is dashed in Fig. 27-47(b)) must represent i 1 . It appears to be zero when V 2 = 6 V. With i 1 = 0, our loop rule gives V 1 = V 2 which implies that V 1 = 6.0 V. (b) At V 2 = 2 V (in the graph) it appears that i 1 = 0.2 A. Now our loop rule equation (with the conclusion about V 1 found in part (a)) gives R 1 = 20 Ω ....
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