Recent content by Adriano25

Then it would just be the difference between QO and QP, so the voltage between P and O would be 0.02 V?

Ups.. -Rx*Ii - Ri * Ii + E = 0 Ri = 0.04 Ω

But we don't know Ri. I was thinking that since Rx and Rv are in parallel, they share the same voltage, so V1 = 1.46. Then, using Kirchhoff's voltage law: -V1 * Ii - Ri * Ii + E = 0 so, Ri = 4.7 Ω ?

The potential difference between P and O would be V = Ii * Ri

I'm sorry, I don't follow you. Could you please reexplain it? Thank you.

From method 1, would that mean that I can find an equivalent resistance from Rv and Rx using the parallel circuit formula? Also, from method 2, I could find an equivalent resistance from Ri and Rx using the series circuit formula? I'm still not sure how that would help me in solving the problem.

Homework Statement [/B] Calculate the internal resistance of V & A based on methods A & B below: Method A Voltmeter = 1.46 V Ammeter = 0.24 A E (emf of battery) = 1.48 V Method B Voltmeter = 1.48 V Ammeter = 0.24 A E (emf of battery) = 1.48 VHomework Equations Ohm's Law: V = I*R The...

Thank you. It's very clear now. I have two more questions if you would be kind to answer them: 1) If we take the left side path from c to a, why do we follow the current path like if we were starting from a to c? Looking at the diagram, if we start form c to a, I would be putting negative...

Homework Statement I = 0.47 A ε1 = 16.0 V ε2 = 8.0 V R1 = 5.0 Ω R2 = 9.0 Ω r1 = 1.6 Ω r2 = 1.4 Ω a) Find the terminal voltage Vab of the 16.0 V Battery b) Find the potential difference Vac of point a with respect to point c Homework Equations Vab = Va - Vb = ε1 - Ir1 Vac = Va - Vc = IR1 + ε2...

Great. Thank you. It's probably a typo then on the answer sheet. Again, thanks and thank you all the people on this forum for helping me get through my physics course.

Homework Statement In the drawing below, a rod of length L and mass M is pivoted a distance L/4 from one end. The pivot attaches the rod to a smooth horizontal table, allowing the rod to rotate frictionlessly in a horizontal plane (so that gravity does not affect the motion). The end furthest...

Great explanation. Thanks.

Oh I get it now. Thanks!

Thanks for your replies. Yes, I'm assuming it's rolling without slipping. I get that the spool is rotated clockwise, but when we compute the torque for tension, why is it making it rotate counterclockwise? Is it because of the way that a spool works?

Right. I'm just a little confused in why the torque from the tension force is opposite to the rotation of the spool.