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chp18solutions

chp18solutions - Direct-Current Circuits Problem Solutions...

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Chapter 18 Direct-Current Circuits Problem Solutions 18.1 From ( ) V I R r Δ = + , the internal resistance is 9.00 V 72.0 4.92 0.117 A V r R I Δ = - = - Ω = Ω 18.3 For the bulb in use as intended, ( ) ( ) 2 2 120 V 192 75.0 W bulb V R Δ = = = Ω Now, presuming the bulb resistance is unchanged, the current in the circuit shown is 120 V 0.620 A 0.800 192 0.800 eq V I R Δ = = = Ω + Ω + Ω and the actual power dissipated in the bulb is ( ) ( ) 2 2 0.620 A 192 73.8 W bulb I R = = Ω = 99
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100 CHAPTER 18 18.5 (a) The equivalent resistance of the two parallel resistors is 1 1 1 4.12 7.00 10.0 p R - = + = Ω Ω Ω Thus, ( ) 4 9 4.00 4.12 9.00 17.1 ab p R R R R = + + = + + Ω = Ω (b) ( ) 34.0 V 1.99 A 17.1 ab ab ab V I R Δ = = = Ω , so 4 9 1.99 A I I = = Also, ( ) ( ) ( ) 1.99 A 4.12 8.18 V ab p p V I R Δ = = Ω = Then, ( ) 7 7 8.18 V 1.17 A 7.00 p V I R Δ = = = Ω and ( ) 10 10 8.18 V 0.818 A 10.0 p V I R Δ = = = Ω 18.7 If a potential difference is applied between points a and b , the vertical resistor with a free end is not part of any closed current path. Hence, it has no effect on the circuit and can be ignored. The remaining four resistors between a and b reduce to a single equivalent resistor, 2.5 eq R R = , as shown below: Δ Ω
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