Find the current through 3.00 N resistor. Express your answer with the appropriate units. HẢ ? I300 1 = Value Units Submit Request Answer Part B Find the current through 4.00 2 resistor. Express your answer with the appropriate units. ? I400 2 = Value Units Submit Request Answer Part C Find the current through 10.00 2 resistor. Express your answer with the appropriate units. I10.00 n = Value Units Submit Request Answer Part D Find the potential difference Vas of point a relative to point b. Express your answer with the appropriate units.

Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter7: Parallel Circuits
Section: Chapter Questions
Problem 3PP: Using the rules for parallel circuits and Ohmslaw, solve for the missing values....
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Question
Find the current through 3.00 2 resistor.
Express your answer with the appropriate units.
?
I3.00 1 =
Value
Units
Submit
Request Answer
Part B
Find the current through 4.00 n resistor.
Express your answer with the appropriate units.
?
I400 2 =
Value
Units
Submit
Request Answer
Part C
Find the current through 10.00N resistor.
Express your answer with the appropriate units.
?
I10.00 2 =
Value
Units
Submit
Request Answer
Part D
Find the potential difference Vab of point a relative to point b.
Express your answer with the appropriate units.
?
Vab =
Value
Units
Transcribed Image Text:Find the current through 3.00 2 resistor. Express your answer with the appropriate units. ? I3.00 1 = Value Units Submit Request Answer Part B Find the current through 4.00 n resistor. Express your answer with the appropriate units. ? I400 2 = Value Units Submit Request Answer Part C Find the current through 10.00N resistor. Express your answer with the appropriate units. ? I10.00 2 = Value Units Submit Request Answer Part D Find the potential difference Vab of point a relative to point b. Express your answer with the appropriate units. ? Vab = Value Units
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
< 1 of 1
2.00 Ω 10.00 V
a 3.00 N
1.00 N 5.00 V,
4.00 N
ww ww
10.00 2
Transcribed Image Text: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 < 1 of 1 2.00 Ω 10.00 V a 3.00 N 1.00 N 5.00 V, 4.00 N ww ww 10.00 2
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