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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