Physics: Post your doubts here!

[Ahmed Aqdam]

But the fact that the overall resistance in the circuit increases, doesn't that change the current in the circuit?

The overall current in the circuit does decrease, but the effect is only on the portion with the variable resistor. Since it is a parallel circuit, the current in the two parts of the circuit will not necessarily be same.

 

[qwertypoiu]

But the fact that the overall resistance in the circuit increases, doesn't that change the current in the circuit?

The current next to the battery is the sum of the currents in the two branches. THIS will decrease. Because the branch with the variable resistor has less current.
This is explained by the fact that overall resistance increases.

 

[fantastic girl]

Phoenix Blood

 

[Phoenix Blood]

Phoenix Blood

why did u bring me here ?

 

[fantastic girl]

why did u bring me here ?

I thought u had to check xP
I got an alert :3)

 

[Phoenix Blood]

I thought u had to check xP
I got an alert :3)

Hahah u got an alert that i hve to check?

 

[fantastic girl]

Hahah u got an alert that i hve to check?

nooooo
thot u'll find it important :3)

 

[Phoenix Blood]

nooooo
thot u'll find it important :3)

aww allie
thanks anyway for tagging :3)

 

[fantastic girl]

aww allie
thanks anyway for tagging :3)

ur welcome

 

[princess Anu]

The current next to the battery is the sum of the currents in the two branches. THIS will decrease. Because the branch with the variable resistor has less current.
This is explained by the fact that overall resistance increases.

The overall current in the circuit does decrease, but the effect is only on the portion with the variable resistor. Since it is a parallel circuit, the current in the two parts of the circuit will not necessarily be same.

is it correct to say that the ammeter reading stays unchanged because increase in effective resis of the circuit will cause I in the circuit to decrease
plus due to an increase in the resis of variable resistor, there will now be a greater share of I to the resistor in series with Ammeter. Hence Ammeter reading will neither decrease nor increase & stays unchanged

 

[ritx]

It'll be a such a favour if someone explained these screenshotted MCQ's for me, they are just last year's, the freshest. Thanks tons!

 

[jesse d]

9702 s08 p2 question 6, i dnt understand how they got the answer for the total power. I have been on it, sumone should please help me

 

[Ahmed Aqdam]

is it correct to say that the ammeter reading stays unchanged because increase in effective resis of the circuit will cause I in the circuit to decrease
plus due to an increase in the resis of variable resistor, there will now be a greater share of I to the resistor in series with Ammeter. Hence Ammeter reading will neither decrease nor increase & stays unchanged

Yes and the decrease in share of I to the section with variable resistor is the same as the decrease in overall I. So there is no effect on the ammeter reading.

 

[jesse d]

help me with my question peeeaaaseee!!!!!

 

[69ermalik007]

Need help with this question....

 

[nehaoscar]

Why A? Wouldn't it be D?

after collision, mass = 2m
so 0.5 x 2m x v^2 = mv^2

and momentum
= 2mv^2?

But then why is it A?

 

[Ahmed Aqdam]

View attachment 54321
Why A? Wouldn't it be D?

after collision, mass = 2m
so 0.5 x 2m x v^2 = mv^2

and momentum
= 2mv^2?

But then why is it A?

Velocity does not remain same. Momentum does.
Hence mass will be 2m so v will be v/2. So KE is 1/4 mv^2.

 

[nehaoscar]

P = I^2 R

As W is removed , Total R increases so I would increase
So then wouldn't the power across Y and Z also increase?

But it's C... Why would the power across Y and Z decrease?

 

[nehaoscar]

Has not everyone done paper 2 and practicals yet? :/

 

[Ahmed Aqdam]

Has not everyone done paper 2 and practicals yet? :/

Practical 34 is on 04 June...