Please help with the explanation as well. I thought it was option B that it would read negative then return to 0, but I got it wrong. Consider two copper coils. One is connected to a battery and the other is connected to a galvanometer (as shownbelow).Part A(C) ((!!!))batterySubmitIf the switch is closed, what will happen to the reading on the galvanometer?Current (A)The galvanometer will read a postive current.The galvanometer will read a negative current for a brief moment, then return to 0.The galvanometer will read a positive current for a brief moment, then return to 0.The galvanometer will read a negative currentPrevious Answers Request Answergalvanometer* Incorrect; Try Again; One attempt remaining

Let's assume the current is flowing initially from the positive terminal of the battery to the…

Consider two copper coils. One is connected to a battery and the other is connected to a galvanometer (as shown below). Part A (XXX)) ((!!!!!)) battery Current (A) If the switch is closed, what will happen to the reading on the galvanometer? Submit Previous Answers Request Answer galvanometer The galvanometer will read a postive current. The galvanometer will read a negative current for a brief moment, then return to 0. The galvanometer will read a positive current for a brief moment, then return to 0. The galvanometer will read a negative current X Incorrect; Try Again; One attempt remaining

Consider two copper coils. One is connected to a battery and the other is connected to a galvanometer (as shown below). Part A (XXX)) ((!!!!!)) battery Current (A) If the switch is closed, what will happen to the reading on the galvanometer? Submit Previous Answers Request Answer galvanometer The galvanometer will read a postive current. The galvanometer will read a negative current for a brief moment, then return to 0. The galvanometer will read a positive current for a brief moment, then return to 0. The galvanometer will read a negative current X Incorrect; Try Again; One attempt remaining

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter24: Magnetic Fields

Section: Chapter Questions

Problem 75A

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TEST ACTIVITY #5: DIRECT CURRENT CIRCUITS Name: Score: Section: Date: PROBLEM 2: Three resistors of 25 N, 30 N and an unknown resistance R are connected in parallel and the combination is connected to a 16 V source. If the total current delivered by the source is 1.5 A, | determine the following: A. The value of R. B. The current in each resistor. C. Which resistor dissipates the highest power? Show your solution. Based on your answer, what conclusion can be made? SOLUTION:
t Jumper Cables and Ohm's Law 3 of 20 > Review | Constants Part B Learning Goal: To learn to apply the concept of current density and microscopic Ohm's law. What electric field E' would have been required to create current of 30.0 A in an aluminum wire of the same diameter? Express your answer in V/m. Use three significant figures. A "gauge 8" jumper cable has a diameter d of 0,326 cm. The cable carries a current I of 30.0 A. The electric field E in the cable is 0.062 V/m • View Available Hint(s) Below is a list of some common metals with their resistivities. ? Resistivity p (N. m) Material E' = V/m copper 1.72 x 10-8 Submit silver 1.47 x 10-8 aluminum 2.75 x 10 8 Part C Complete previous part(s) gold 2.44 x 10-8 Part D Complete previous part(s) steel 20 x 10-8 Next > Provide Feedback
(1) show the figure, given, formula and substitution. thank you! promise i'll rate.
B, C, and D pls
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In the circuit shown in the figure (Figure 1), the current in the 20.0-V battery is 5.00 A in the direction shown and the voltage across the 8.00-SN resistor is 18.0 V, with the lower end of the resistor at higher potential. Part D Find the direction of the current I through the 200.0-V battery. The current through the 200.0 V battery is in the direction from the - to the + terminal. The current through the 200.0 V battery is in the direction from the + to the - terminal. Request Answer Submit Figure < 1 of 1 Part E Find the resistance R. 20.0 N R 30.0 N VO AZ R ? 18.0 N 8.00 N R= 200.0 V- - 20.0 V Submit Request Answer 5.00 A
solve these qsn a,b,c and also d. D) How much charge accumulated in t=1.8 sec??