-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Physics: Post your doubts here!
- Thread starter XPFMember
- Start date
- Messages
- 90
- Reaction score
- 34
- Points
- 28
Yes
But HOW ????
In the circuit shown, let the current be I1
I1=V/R=6/2R
Power dissipated in external resistor (P1)=I1*I1*resistance of external resisitor=(6/2R)*(6/2R)*R= 36R/(4R*R)=9/R
in the circuit with negligible internal resistance, let the current be I2
I2=V/R=6/R
so power dissipated in xternal circuit(P2)=I2*I2*Resistance of external resistor=(6/R)*(6/R)*R=36R/(R*R)=36/R
SO P2=4P1
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
HELP !! 
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
ans is D
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
but isnt it that anything earthed becomes neutral?
do u mean if the bottom plate was not earthed, it wud not get a +ve charge?
- Messages
- 924
- Reaction score
- 1,096
- Points
- 153
If the bottom plate was not earthed, there will be positive charge in upper part of this plate and negative charge in bottom half, because the elections are repelled and so move down. They are not able to escape anywhere. So top part would be positive and bottom part would be negative, within the bottom plate itself.
but isnt it that anything earthed becomes neutral?
do u mean if the bottom plate was not earthed, it wud not get a +ve charge?
By earthing it we allow electrons to move away so the plate is positively charged everywhere as a whole
- Messages
- 1,318
- Reaction score
- 1,374
- Points
- 173
Thankyou!
- Messages
- 1,318
- Reaction score
- 1,374
- Points
- 173
Thankyou! This is a really helpful blog!
- Messages
- 90
- Reaction score
- 34
- Points
- 28
can anyone answer plzzzz?????
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
is it B
- Messages
- 90
- Reaction score
- 34
- Points
- 28
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
basically here the voltmeter acts as a resistor. So consider it as a resistor
so when its resistance is 1K ohms , the voltage divides in 1:1 ratio
but when its resistance increases to 1000000 ohms, which is huge compared to 1000 ohms, therefore almost all the voltage goes to the voltmeter .. as in a series circuit current is constant and v is proportional to r
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
what does o.5mm/rev imply here? 
- Messages
- 8
- Reaction score
- 1
- Points
- 3
Since it is a parallel circuit, the pd across the (variable r. and a resistor in series with it) and the other single resistor(rightmost) is the same (i.e the same value as the emf of the battery). Resistance of the rightmost resistor is also unchanged. Hence V/R=I is unchanged.HELP !!
When the resistance of vr is increased, pd across it will increase and reading on the voltmeter will therefore decrease.
- Messages
- 1,318
- Reaction score
- 1,374
- Points
- 173
Help please!?
- Messages
- 1,318
- Reaction score
- 1,374
- Points
- 173
I thought X and Y is actually the plastic region??
as shown here:
But then the answer is D :/ ???
- Messages
- 1,229
- Reaction score
- 740
- Points
- 123
but doesn't I depend on the ratio of resistances? for example if the emf is 10
variable resistor & the resistor with voltmeter have a combined resistance of 50 while the resistor on the right 20
in that case I will be = 10/ 14= O.71
but if now the variable resistor's resistance has increased so that the combined resistance now is 70
now I= 10/ (70*20/20+70) ,I= 0.6 :/