Rosa Electrie Cireuit Conceptual Evaluation uana answer shect, Answer questions

Question

Rosa Electrie Cireuit Conceptual Evaluation uana answer shect, Answer questions 1-16 by circling the letter corresponding to the correct choice. Also include written answers for Questions 2, 4, 5 and 6 in the boxes on the answer sheet. DIRECTIONS: Write your name, section and date on the top of the answer sheet. Answer all questions on.the g wires have no On this test. all batteries are ideal (they have n test de not change as the current the resistances of the bulbs on this test do not change as th 1. A balb and a battery are connected as shown below Which is true about the current at various points in this circuit? A. The current is largest at A. B. The current is largest at B. C. The current is largest at C. D. The current is largest at D. E. The current is the same everywhere. F. The current is the same between A and B and smaller than between C and D. G. The current is the same between A and B and larger than between C and D. H The current is the same everywhere except in the bulb. I. The current is the same everywhere except in the battery J. None of these is true. For Questions 2-4, a second identical bulb is added to the circuit in Question 1, as shown below Compare the current at A now to the current at A before with only one bulb. A. The current at A is now twice as large as before. B. The current at A is now larger than before but not twice as large. C. The current at A is the same as before D. The current at A is now half as large as before. E. The current at A is now smaller than before but not half as large. J. None of these is correct Briefly explain in the space below how you arrived at your answer to Question 2 Real Tiee Physics Electric Ceoui Conceptual Evaluation 01992-4 University of Oregon February 16, 1994

Explanation / Answer

1. this is series circuit.

hence current will be same everywhere.

(E)

2. now two resistance(bulb) are connected so total reistance of circuit will
decrease. hence current through battery will increase.

1/Req = 1/R + 1/R

Req = R/2

I = 2e/R that twice of erlier ( e/R)

Ans(A)

3.
current through battery will divide into two bulbs.
current throug bulb will be = I/2 = e/R

that is equal to initial value.

Ans(B)

4.
braches connected parallely have same PD across them.

hence PD across each bRANCh will be e.

Ans(C)

5. I = e/2R

ans(D)

6. V = IR = R(e/2R) = e/2

ans(D)

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