Circuits: Power, Voltage, Resistance

AI Thread Summary
The discussion focuses on calculating the resistance of a lamp operating at 12V and 24W, determining it to be 6 Ohms. To vary the voltage across the lamp from 0V to 12V, a potentiometer is suggested as the variable resistor. Participants discuss the configuration of the circuit, noting that the potentiometer creates a voltage divider with the lamp's resistance. The challenge lies in calculating the appropriate resistance value for the potentiometer to achieve the desired voltage across the lamp. The conversation emphasizes the need for a clear understanding of the circuit components and their interactions.
barnaby
Messages
17
Reaction score
0

Homework Statement



The circuit shown is used to produce a current-voltage graph for a 12V, 24W lamp:

http://img338.imageshack.us/img338/9139/circuitlu6.jpg
http://img338.imageshack.us/img338/9139/circuitlu6.jpg
a) Calculate the resistance of the lamp in normal operation.
b) Calculate the value for R which wold enable the voltage across the lamp to be varied between 0V and 12V.

Homework Equations



P = IV
V = IR

The Attempt at a Solution



a) P=IV
24 = 12I
I = 2A

V = IR
12 = 2R
R = 6 Ohms

b) Absolutely no idea how to go about doing this... it seems that the voltage across the resistor R ought to be 8V, but I don't know, since we haven't been told what the component with the arrow pointing into the 24 ohm resistor is... Thanks for your help.
 
Last edited by a moderator:
Physics news on Phys.org
barnaby said:

Homework Statement




b) Absolutely no idea how to go about doing this... it seems that the voltage across the resistor R ought to be 8V, but I don't know, since we haven't been told what the component with the arrow pointing into the 24 ohm resistor is...

This is a "potentiometer" A resistor with a sliding contact, forming a variable voltage divider.
you have to determine such that you get 12 V across the lamp if the contact is at the top and 0 V if the contact is at the bottom.
 
So if the contact is at the top, you get what is in effect two resistors in parallel - one at 24 ohms and the other at 6 ohms? I tried solving it like that... and got nowhere - calculated that the total resistance of the two strands of the circuit was 4.8 ohms... don't know if that's helpful at all, though.
 
Last edited:
barnaby said:
So if the contact is at the top, you get what is in effect two resistors in parallel - one at 24 ohms and the other at 6 ohms? I tried solving it like that... and got nowhere - calculated that the total resistance of the two strands of the circuit was 4.8 ohms... don't know if that's helpful at all, though.

That's certainly helpful. Now you have to find the R that makes the voltage divider formed by R and the 4.8 Ohm resistance produce 12 V
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top