P28_031 - 31. (a) We first find the currents. Let i1 be...

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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 left-hand loop produces E1 − i3 R3 + i1 R1 = 0 and applied to the right-hand 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 ...
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