The 10.00 V battery in the figure (Figure 1) is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A complex network Figure 2.00 2 10.00 V -ww+ a 1.00 25.00 V. 4.00 Ω b 10.00 Ω 1 of 1 3.00 Ω Part A Find the current through 3.00 resistor. Express your answer with the appropriate units. 13.00 = Submit Part B Value X Incorrect; Try Again; One attempt remaining Submit A Previous Answers Request Answer Find the current through 4.00 2 resistor. Express your answer with the appropriate units. 14.00 = Value Part C A 2 Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining ? ?

The 10.00 V battery in the figure (Figure 1) is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A complex network Figure 2.00 2 10.00 V -ww+ a 1.00 25.00 V. 4.00 Ω b 10.00 Ω 1 of 1 3.00 Ω Part A Find the current through 3.00 resistor. Express your answer with the appropriate units. 13.00 = Submit Part B Value X Incorrect; Try Again; One attempt remaining Submit A Previous Answers Request Answer Find the current through 4.00 2 resistor. Express your answer with the appropriate units. 14.00 = Value Part C A 2 Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining ? ?

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter18: Direct-Current Circuits

Section: Chapter Questions

Problem 18P: (a) Find the current in each resistor of Figure P18.18 by using the rules for resistors in series...

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In the circuit shown in (Figure 1) all meters are idealized and the batteries have no appreciable internal resistance. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A complex network. Figure 25.0+ V 100.0 Ω a 115.0 S V 36 V 75.0 A Ω b 1 of 1 impi pitivas par Part C With the switch closed, find the reading of the voltmeter. Express your answer in volts. V = Submit Part D I = ΑΣΦ With the switch closed, find the reading of the ammeter. Express your answer in amperes. V7 ΑΣΦ Submit Request Answer Request Answer ? V A
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The 10.00 V battery in the figure (Figure 1) is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A complex network. Figure 2.00 2 10.00 V, a ww+ 1.00 2 5.00 V 14.00 Ω 3.00 Ω www 10.00 Ω b 1 of 1 Find the current through 3.00 resistor. Express your answer with the appropriate units. 13.00 2 = Submit Part B 14.00 12 = Submit Part C Find the current through 4.00 2 resistor. Express your answer with the appropriate units. I10.00 = Submit Part D Value μA Request Answer Vab = · μA Value Request Answer Find the current through 10.00 resistor. Express your answer with the appropriate units. Value μÃ Request Answer Value Units μA C Units Units Find the potential difference Vab of point a relative to point b. Express your answer with the appropriate units. Units ? ? ? ?
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