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AS PHYSICS!!!
- Thread starter libra94
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salam
This question doesn't seem to make much sense to me. Anyways, as far as i cn understand, part "A" shud be false. That's cos when the switch K is closed, the current in the lower circuit will by-pass R2 (since current always chooses the path of lowest resistance, so it will pass through the negligible resistance wire and not through R2). This implies that R2 won't have any effect on the circuit no matter what its resistance is, as the current is not passing through R2. If m not wrong, this also implies that change in resistance of R2 will have no change in balanced length "l."
Hop that helps
This question doesn't seem to make much sense to me. Anyways, as far as i cn understand, part "A" shud be false. That's cos when the switch K is closed, the current in the lower circuit will by-pass R2 (since current always chooses the path of lowest resistance, so it will pass through the negligible resistance wire and not through R2). This implies that R2 won't have any effect on the circuit no matter what its resistance is, as the current is not passing through R2. If m not wrong, this also implies that change in resistance of R2 will have no change in balanced length "l."
Hop that helps
- Messages
- 133
- Reaction score
- 47
- Points
- 38
o0Ops sorry @ i meant A shud b correct (NOT false)
- Messages
- 133
- Reaction score
- 47
- Points
- 38
For the last question (it was quite interesting btw...;p), the answer seems to be "C"
Working:
The keyword "ideal" in 'ideal voltmeter' implies that the voltmeter has infinite resistance i.e. it gives the p.d. b/w the two points it is connected to, without drawing any current 4m the circuit
Also as given in the question, the ideal voltmeter measures two different voltages across the battery with e.m.f. say E, for two different resistances connected to the battery which means that the battery must have some internal resistance say r, interfering the current passing through the battery.
Now, let's use the potential divider principle to set up 2 equations to find the value of r and E.
Using the 1st statement, [2/(2+r)]*E = 7 ..... [1]
Using the 2nd statement, [3/(3+r)]*E = 9 ..... [2]
V can solve [1] and [2] simultaneously to get E = 21V and r = 4 ohms
Finally when J1 and J2 are closed, the effective (or combined) resistance of the two resistors is 1/[1/2 + 1/3] = 1.2 ohms
We can again use the potential divider principle here to get:-
required voltage = [1.2/(1.2 + 4)]*21 = 63/13 ~ 4.85V
Hope that helps
Working:
The keyword "ideal" in 'ideal voltmeter' implies that the voltmeter has infinite resistance i.e. it gives the p.d. b/w the two points it is connected to, without drawing any current 4m the circuit
Also as given in the question, the ideal voltmeter measures two different voltages across the battery with e.m.f. say E, for two different resistances connected to the battery which means that the battery must have some internal resistance say r, interfering the current passing through the battery.
Now, let's use the potential divider principle to set up 2 equations to find the value of r and E.
Using the 1st statement, [2/(2+r)]*E = 7 ..... [1]
Using the 2nd statement, [3/(3+r)]*E = 9 ..... [2]
V can solve [1] and [2] simultaneously to get E = 21V and r = 4 ohms
Finally when J1 and J2 are closed, the effective (or combined) resistance of the two resistors is 1/[1/2 + 1/3] = 1.2 ohms
We can again use the potential divider principle here to get:-
required voltage = [1.2/(1.2 + 4)]*21 = 63/13 ~ 4.85V
Hope that helps
- Messages
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- Reaction score
- 47
- Points
- 38
Yeah sAm buddy i do remember it @ "the thing 2 notice was that 'fringe width which remains constant in an interfernce pattern such as this one'." Therefore, it's "A"