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Unformatted text preview: 34. (a) The voltage across R3 = 6.0 Ω is V3 = iR3= (6.0 A)(6.0 Ω) = 36 V. Now, the
voltage across R1 = 2.0 Ω is
(VA – VB) – V3 = 78 − 36 = 42 V,
which implies the current is i1 = (42 V)/(2.0 Ω) = 21 A. By the junction rule, then, the
current in R2 = 4.0 Ω is
i2 = i1− i = 21 A − 6.0 A = 15 A.
The total power dissipated by the resistors is (using Eq. 2627)
2
i12 (2.0 Ω) + i2 (4.0 Ω) + i 2 (6.0 Ω) = 1998 W ≈ 2.0 kW . By contrast, the power supplied (externally) to this section is PA = iA (VA  VB) where iA =
i1 = 21 A. Thus, PA = 1638 W. Therefore, the "Box" must be providing energy.
(b) The rate of supplying energy is (1998 − 1638 )W = 3.6×102 W. ...
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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.
 Spring '08
 Any
 Physics, Current

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