(d) The direction of i2is clearly downward. (e) Using our conclusion from part (a) in Eq. 27-17, we have P = i1ε1= (4.0 A)(16 V) = 64 W. (f) Using results from part (a) in Eq. 27-17, we obtain P = i'2= (2.0 A)(8.0 V) = 16 W. (g) Energy is being supplied in battery 1. (h) Energy is being absorbed in battery 2. 105. (a) The six resistors to the left of 1= 16 V battery can be reduced to a single resistor R= 8.0 Ω, through which the current must be iR= 1/R= 2.0 A. Now, by the loop rule, the current through the 3.0 Ωand 1.0 Ωresistors at the upper right corner is
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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.