9780199339136, Chapter 1, Problem 16P

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Chapter Problem 16P Step step solution Step Refer Figure the text book Conside Figure +5 V Path-1 Path-2 Figure Step (a) Apply Kirchhoff's voltage law through path 10-7,(12k) (1) (2k) Anonymous shouldnt this be path 2 Anonymous Yes Step Apply voltage law through path Substitute 10-7,(12k) (2k) 10-7,(12k) (2k) 7,(80k)=60 60 =0.75 mA Therefore the value the current through the resistor 0.75 mA Anonymous shouldnt this be path Step Recall equation (1) 10-7,(12k) (2k) Substitute 0.75 mA for 10-(0.75 m)(12k) (2k) 10-9 (2k) (2k) the value of current through the resistor R, The current through the resistor Substitute for and for =1.25 mA Therefore the value current through the resistor 1.25 mA The voltage at the common Substitute 1.25 mA for and for =2.5 V Therefore node voltage 2.5 V Step (b) Apply current law common node V V-5 V-10 5k 10k 10 10k 5k 10k 10+10 10k 10k 10k 10k =2.5 V the value the node voltage 2.5 V Step The current through the resistor 5-V Substitute 5-2.5 5k 2.5 5k Therefore the value the current through the resistor Step 10 The urrent through the resistor 10-V 10k Substitute for 10-2.5 10k 7.5 10k =0.75 mA Therefore the value the current through the resistor 0.75 mA Step 10 The urrent through the resistor Substitute V 2.5 =1.25 mA the current through the resistor R. Step 10 Therefore Node equation the simplest since the unknown quantities required only single equation compared loop equation which requires two equations to solve the unknown prefer Node Equation method Anonymous paths were mixed up but good solution