Some physics qn
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Another qn. This one shld be big enuf. took me 15 mins to ensure the correct size.
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Originally posted by davidche:
Want to ask. If the rheostat resistance is increased, more current flows through the bulb thus would the brightness of bulb increase?
Note that the voltage across the bulb does not change in this case.
I = V/R does not change as well.Hence, no matter how the rheostat resistance changes, the brightness of the bulb should remain the same.
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The current doesnt determine the brightness? If you cahnge the R, the current gets changed too.
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eagle is right. As he always is.
Brightness is determined by power. In a parallel circuit, voltage is ALWAYS constant no matter what (basic concept).
Hence, to determine Power, we will use P=V^2/R. If you want to use P=VI or P=I^2 x R, by all means, but you must be given specified values to allow you to determine current. Or you can sub in some logical values.
Everything will be the same in the end.
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The second question, if galvanometer has no reading, I think it means that the current flowing in the wire is equal to the R and 6 ohm resistors. Not sure, let me think ah.
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Hi,
With regards to your second question, the galvanometer registering a zero reading actually means that there is no potential difference between the two ends of the galvanometer, thus no current flow.
To solve the question, you need to understand that the potential difference across AB is the same for R, and BC for the 6ohm resistor. Concept tested here is that pd varies with length (of the resistance wire).
Haha. That's about all I remember. Can't seem to recall the formula. =X
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Originally posted by davidche:
Another qn. This one shld be big enuf. took me 15 mins to ensure the correct size.
Not sure if you have learned the wheat stone bridge, but something similar...
No current means that the potential difference between the two points is zero, i.e. the potential is the same.
Because potential drop across AC must be the same as potential drop across R to have the same potential at those points. Which means...
Resistance of AC / CB = R / 6
or
1.6/2 (by length) = R/6
R = 4.8 ohms -
Originally posted by eagle:
Not sure if you have learned the wheat stone bridge, but something similar...
No current means that the potential difference between the two points is zero, i.e. the potential is the same.
Because potential drop across AC must be the same as potential drop across R to have the same potential at those points. Which means...
Resistance of AC / CB = R / 6
or
1.6/2 (by length) = R/6
R = 4.8 ohmsEither you type wrong, see wrong or calculate wrong. Care to clarify?
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yup I was wrong :D
1.6 / 0.4 = R / 6
R = 24 ohms -
Omg i have tons of physics qn after doing a cedar girl paper. Thanks of the help.
Which position of a conducting retangular coil would result in most induced current. WHen the coil is moving through the field.
When the coil is halfway inside the magnetic field? or when the coil is totally inside the magnetic field(very big field). Note taht the motion is moving thro the field.
2) How does the ratings affect the real voltage.
A specific example: if the lamp in a parallel circuit has ratings (24v 80w) while the battery is 12v. Whats the result of the power and stuffs of the lamp. Whats the whole relationship.
3) What is the work done in moving a box up a ramp. The height of the ramp is 1.5m and the horizontal distance is 2m. The box is 500N weight.
4)Why cant work done be used to calculate net force?
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Which position of a conducting retangular coil would result in most induced current. WHen the coil is moving through the field.
When the coil is halfway inside the magnetic field? or when the coil is totally inside the magnetic field(very big field). Note taht the motion is moving thro the field.
Quite obvious. More magnetic flux linkage cut = to more induced current in the external circuit.
2) How does the ratings affect the real voltage.
A specific example: if the lamp in a parallel circuit has ratings (24v 80w) while the battery is 12v. Whats the result of the power and stuffs of the lamp. Whats the whole relationship.
I think Power is the more important factor here when deciding how to connect. Current is the other important factor when selecting a fuse. Voltage is just a in between link-connector.
Btw the above circuit doesn't work because
Emf of source = total p.d. of all components of the circuit
What is the work done in moving a box up a ramp. The height of the ramp is 1.5m and the horizontal distance is 2m. The box is 500N weight.
Length of ramp = Sq root 2^2 + 1.5^2 = 2.5
W=Fxd = 500x2.5 = 1250 J
4)Why cant work done be used to calculate net force?
Direction of force may not be the same as the direction of motion, a.k.a. the force may not produce a motion in the same direction as the force.
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thanks garrick
i trust your explanations... no need to look thru... many things to do at the moment :(
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Quite obvious. More magnetic flux linkage cut = to more induced current in the external circuit.
Ans does not imply so. Thats why i asked.
Length of ramp = Sq root 2^2 + 1.5^2 = 2.5
W=Fxd = 500x2.5 = 1250 J
No need to consider mgh?
Direction of force may not be the same as the direction of motion, a.k.a. the force may not produce a motion in the same direction as the force.
The force used in WD calculation is the force used to move the thing in the direction. Your definition seems to be wrong btw.
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Originally posted by eagle:
thanks garrick
i trust your explanations... no need to look thru... many things to do at the moment :(
The qn, esp the work done is rather complicating. I think i need your look-thru. XD
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Originally posted by davidche:
The qn, esp the work done is rather complicating. I think i need your look-thru. XD
Ok, I just looked throughYou only have to consider mgh. 500N * 1.5m = 750 J
Horizontal distance can be ignored.You only consider the length of the ramp if it has friction, which means work done against friction would have to be included.
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Ans does not imply so. Thats why i asked.
Doesn't a stronger magnetic field = more magnetic lines of force = more magnetic flux cutting?
No need to consider mgh?
Work done and G.P.E. are two different things.
Work is done when a force produces motion.
G.P.E. is the energy a body possesses due to its position relative to the ground.
I can confuse you even more, but never mind.
The force used in WD calculation is the force used to move the thing in the direction. Your definition seems to be wrong btw.
???
Work done = Force x Distance moved in the direction of force.
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Originally posted by eagle:
Ok, I just looked throughYou only have to consider mgh. 500N * 1.5m = 750 J
Horizontal distance can be ignored.You only consider the length of the ramp if it has friction, which means work done against friction would have to be included.
Ok fully understood as usual XD. Thanks.