18_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

18_Ch 18 College Physics ProblemCH18 Direct-Current...

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Unformatted text preview: 116 CHAPTER 18 18.23 (a) We name the currents I1 , I 2 , and I 3 as shown. Applying Kirchhoff’s loop rule to loop abcfa , gives + ε 1 − ε 2 − R2 I 2 − R1 I1 = 0 or 3I 2 + 2I1 = 10.0 mA b I1 I2 e1 (1) R3 e2 70.0 V d I3 e3 80.0 V 60.0 V R2 Applying the loop rule to loop edcfe yields 3.00 kW 2.00 kW a + ε 3 − R3 I 3 − ε 2 − R2 I 2 = 0 or 3I 2 + 4 I 3 = 20.0 mA 4.00 kW c R1 f e (2) Finally, applying Kirchhoff’s junction rule at junction c gives I 2 = I1 + I 3 (3) Solving equations (1), (2), and (3) simultaneously yields I1 = 0.385 mA, I 2 = 3.08 mA, and I 3 = 2.69 mA (b) Start at point c and go to point f, recording changes in potential to obtain Vf − Vc = − ε 2 − R2 I 2 = −60.0 V − ( 3.00 × 10 3 Ω ) ( 3.08 × 10 −3 A ) = −69.2 V or ∆V 18.24 cf = 69.2 V and point c is at the higher potential 1.50 V (a) Applying Kirchhoff’s loop rule to the circuit gives + 3.00 V − ( 0.255 Ω + 0.153 Ω + R ) ( 0.600 A ) = 0 or R = 3.00 V − ( 0.255 Ω + 0.153 Ω ) = 4.59 Ω 0.600 A 0.255 W 1.50 V –+ 0.153 W R –+ I = 0.600 A ...
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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