ch27-p034

# ch27-p034 - 34. (a) The voltage across R3 = 6.0 Ω is V3 =...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. 26-27) 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. ...
View Full Document

## 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.

Ask a homework question - tutors are online