PHYS-102hw5soln

PHYS-102hw5soln - PHYS-102 Homework 5 Solutions Spring-08 _...

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PHYS-102 Homework 5 Solutions Spring -08 ____________________________________________________________ Chapter 21 Current and Direct Current Circuits 25, 29, 31 P21.25 (a) P = Δ V ( ) 2 R becomes 20.0 W = 11.6 V ( ) 2 R so R = 6.73 Ω . (b) Δ V = IR so 11.6 V = I 6.73 Ω ( ) and I = 1.72 A e = IR + Ir so 15.0 V = 11.6 V + 1.72 A ( ) r r = 1.97 Ω . FIG. P21.25
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P21.29 If we turn the given diagram on its side, we find that it is the same as figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first reduction is shown in (b). In addition, since the 10.0 Ω , 5.00 Ω , and 25.0 Ω resistors are then in parallel, we can solve for their equivalent resistance as: R eq = 1 1 10.0 Ω + 1 5.00 Ω + 1 25.0 Ω ( ) = 2.94 Ω . This is shown in figure (c), which in turn reduces to the circuit shown in figure (d). Next, we work backwards through the diagrams applying I = Δ V R and Δ V = IR alternately to every resistor, real and equivalent. The 12.94 Ω resistor is connected across 25.0 V, so the current through
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PHYS-102hw5soln - PHYS-102 Homework 5 Solutions Spring-08 _...

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