physics 2 chapter 24-27 test 11. In the diagram, the current in the 3- resistor

Question

physics 2 chapter 24-27 test 11. In the diagram, the current in the 3- resistor is 2.4 A. The voltage points 1 and 2 is: Answer 0.75 V B) 0 D) 12 V E) 20 V Answer 12. Consider the three identical light bulbs shown in the circuit. Then a) All three bulbs are equally bright. b) Bulbs B and C are equally bright, but less bright than bulb A. c) Bulbs B and C are equally bright, but bulb A is less bright. d) Bulbs A and B are equally bright, but bulb C is less bright. e) Only bulb A is illuminated 13. A tota!resistance of3.0 is to be produced by combining an unknown resistor R wi resistor. What is the valueofB and how is it to be connected to the 12 resistor? Answ A) 2.4 , parallel B) 2.4 , series C) 4.0 , parallel D) 4.0 , series E) 9.0 , series 14. A particle with charge q is moving with speed v through a constant magnetic fiel the magnetic field will do on the charged particle after time t is: ) qvBt, )qv-Bt, where f-qvb, and d-vt 0, because the magnetic field is a vector and work is a scalar quantity 0, because the magnetic field is conservative 0, because the magnetic force is a velocity-dependent force

Explanation / Answer

11) Total resistance = R = 3 + 2 = 5 ohm

As current remains same in series resistances, so I = 2.4 A

Then voltage across points 1 and 2 is = V = I*R = 2.4*5 = 12 V

12) Option b) is correct.

As current in Bulb B and C will be half than the current through bulb A.

13) Option C) is correct.

Here as resultant resistance is lower than 12 ohm, so the unknown resistance will be connected in parallel.

So, 1/3 = 1/R + 1/12   ====> R = 4.0 ohm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.