This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 31. (a) We ﬁrst ﬁnd the currents. Let i1 be the current in R1 and take it to be positive if it is upward. Let i2 be the current in R2 and take it to be positive if it is to the left. Let i3 be the current in R3 and take it to be positive if it is to the right. The junction rule produces i1 + i2 + i3 = 0 . The loop rule applied to the lefthand loop produces E1 − i3 R3 + i1 R1 = 0 and applied to the righthand loop produces E2 − i2 R2 + i1 R1 = 0 . We substitute i1 = −i2 − i3 , from the ﬁrst equation, into the other two to obtain E1 − i3 R3 − i2 R1 − i3 R1 = 0 and E2 − i2 R2 − i2 R1 − i3 R1 = 0 . The ﬁrst of these yields i3 = E1 − i2 R1 . R1 + R3 Substituting this into the second equation and solving for i2 , we obtain i2 = = E2 (R1 + R3 ) − E1 R1 R1 R2 + R1 R3 + R2 R3 (1.00 V)(5.00 Ω + 4.00 Ω) − (3.00 V)(5.00 Ω) = −0.158 A . (5.00 Ω)(2.00 Ω) + (5.00 Ω)(4.00 Ω) + (2.00 Ω)(4.00 Ω) We substitute into the expression for i3 to obtain i3 = Finally, i1 = −i2 − i3 = −(−0.158 A) − (0.421 A) = −0.263 A . Note that the current in R1 is actually downward and the current in R2 is to the right. The current in R3 is also to the right. The power dissipated in R1 is P1 = i2 R1 = (−0.263 A)2 (5.00 Ω) = 0.346 W. 1 (b) The power dissipated in R2 is P2 = i2 R2 = (−0.158 A)2 (2.00 Ω) = 0.0499 W. 2 (c) The power dissipated in R3 is P3 = i2 R3 = (0.421 A)2 (4.00 Ω) = 0.709 W. 3 (d) The power supplied by E1 is i3 E1 = (0.421 A)(3.00 V) = 1.26 W. (e) The power “supplied” by E2 is i2 E2 = (−0.158 A)(1.00 V) = −0.158 W. The negative sign indicates that E2 is actually absorbing energy from the circuit. E1 − i2 R1 3.00 V − (−0.158 A)(5.00 Ω) = = 0.421 A . R1 + R3 5.00 Ω + 4.00 Ω ...
View
Full
Document
 Fall '08
 SPRUNGER
 Physics, Current

Click to edit the document details