physics ocr unit 2 help

Original post by dany_boy

When working out internal resistance, you increase the resistance of the variable resistor and this increases voltage as the share between it and the internal resistane is higher. Apparently "The current decreases becuase the resistance is higher", i really dont get this bit because i=v/r and since v is higher and r lower shouldnt I be the same.

Also coming to potential dividers, when you increase the resistance of a component why does its voltage increase, i understand the reason is because current is constant and therefore the coltage sher must change but then doent that contradict the previous concept of what an increase in resistance does.

Can someone help please because right now physics is one big mess when it comes to this.

Just imagine internal resistance is just another tiny resistor in series with the variable resistor, the only difference is that this resistor is 'inside' the battery.

So if you increased the resistance of the variable resistor, the total resistance of the circuit would increase. The emf of the cell is constant so according to I=V/R, the current will decrease.

About the potential divider. Voltage is 'energy dropped' across the resistance. So if you increased the resistance, then more 'energy' would be needed to push the current through. This mean voltage across this resistance has increased. The current still decreases as you would expect when increasing the resistance though so no contradiction here :smile:

(edited 8 years ago)

Reply 2

OP

Thanks a lot for your response, you cleared this up a lot more but i do have have querstion about what you said.

Original post by NDVA

So if you increased the resistance of the variable resistor, the total resistance of the circuit would increase. The emf of the cell is constant so according to I=V/R, the current will decrease.

About the potential divider. Voltage is 'energy dropped' across the resistance. So if you increased the resistance, then more 'energy' would be needed to push the current through. This mean voltage across this resistance has increased. The current still decreases as you would expect when increasing the resistance though so no contradiction here :smile:

Firstly, when using I=V/R, i agree with the fact that emf is constant but since your applying it to the variable resistor should you look at the voltage the variable resistor as.

Also, if i increased the resistance of a component of a potential divider, does that mean that the curent of the circuit as a whole would be lower.

once again thanks and apoligise for all the typing mistakes yesterday, i was pretty tired :smile:

Reply 3

Original post by dany_boy

Firstly, when using I=V/R, i agree with the fact that emf is constant but since your applying it to the variable resistor should you look at the voltage the variable resistor as.

We apply I=V/R to the whole circuit in this case, that is why the current will decrease. Have a look at this picture :smile:

Original post by dany_boy

Also, if i increased the resistance of a component of a potential divider, does that mean that the curent of the circuit as a whole would be lower.

Yes, the current would be lower because you increase the resistance of the whole circuit. However, this decrease is to a smaller extent than the increase in the resistance of component resistor (the current depends on both resistors), so the voltage drops across that resistor will increase.

(edited 8 years ago)

Reply 4

OP

Original post by NDVA

We apply I=V/R to the whole circuit in this case, that is why the current will decrease. Have a look at this picture :smile:

Thanks, again :smile:

, i understand the p.d stuff now but for the variable resistor stuff
i have a question.
As we change the variable resistor, we mesure the current and voltage aroundthe variable resistor. Like this right?

If this is the case dont we want to use V as the voltage of the resistor?, i dont get why we just know the V is the emf. Also if i did use V=IR on the variable resistorwhat would the value of current be for.

Reply 5

Original post by dany_boy

Thanks, again :smile:

, i understand the p.d stuff now but for the variable resistor stuff
i have a question.
As we change the variable resistor, we mesure the current and voltage aroundthe variable resistor. Like this right?

If this is the case dont we want to use V as the voltage of the resistor?, i dont get why we just know the V is the emf. Also if i did use V=IR on the variable resistorwhat would the value of current be for.

In this case, the ammeter measures the current in the whole circuit (which is also the current passing through the variable resistor, as two resistors R and r are in series). So

I=\frac{emf}{R_{total}}

with

R_{total}=r+R

is the total resistance of the circuit :smile:

Now we have the current

I

in the circuit. The voltage across the variable resistor will be

V=IR

and this is the value you would get on the voltmeter.

Reply 6

OP

Original post by NDVA

We apply I=V/R to the whole circuit in this case, that is why the current will decrease. Have a look at this picture :smile:

Yes, the current would be lower because you increase the resistance of the whole circuit. However, this decrease is to a smaller extent than the increase in the resistance of component resistor (the current depends on both resistors), so the voltage drops across that resistor will increase.

Thank YOUUUUUUU!!!! :smile:

So basically:
Resistance of the variable resistor increases its voltageshare between it and the internal resister increases. Just like in a normalpotential divider. Because the overall resistance is higher and the emf constant,the overall current (same throughout the variable resistor and battery ) willdecrease. When it comes to the potential divider, an increase inresistance means an increase in voltage just like before so the current decreases.But say if the voltage doubled, the current doesn’t decreases by double becausethe current is always determined by the total resistance which the othercomponent has a part of. :wink:

Reply 7

OP

Im right, right? :ahee:

Reply 8

Original post by dany_boy

Im right, right? :ahee:

Yes

Reply 9

OP

Original post by NDVA

Yes

Thanks a lot, owe you mate. If you ever need help with physics, which you probably wont, you can ask me. (Althought i wouldnt recommend it because it seems im evidently bad, at electricity atleast)

Reply 10

Original post by dany_boy

Thanks a lot, owe you mate. If you ever need help with physics, which you probably wont, you can ask me. (Althought i wouldnt recommend it because it seems im evidently bad, at electricity atleast)

No worries mate :biggrin:

I was really confused the first time I studied about this stuff as well, so I'm glad that I could help you :smile:

Good luck with your exams!

Reply 11

OP

Original post by NDVA

No worries mate :biggrin:

I was really confused the first time I studied about this stuff as well, so I'm glad that I could help you :smile:

Good luck with your exams!

Hi, its me again, im going to start doing past papers and i needed a word of advice and a little help. From now on when i look at a circuit, should i try to detemine the current of the circuit by looking at the net resistance and voltage.

I mean last time, with all that variable resisotor stuff, i kept on getting confused because i thought if i considered just that resistor i would be able to see the fact that current decrease which i couldnt. And from what you pretty much taught me, you cant really look at one component and find the resisance.

So i guess my question is when can i look at one component to find the current, and whats the exception?

Reply 12

Original post by dany_boy

Hi, its me again, im going to start doing past papers and i needed a word of advice and a little help. From now on when i look at a circuit, should i try to detemine the current of the circuit by looking at the net resistance and voltage.

I mean last time, with all that variable resisotor stuff, i kept on getting confused because i thought if i considered just that resistor i would be able to see the fact that current decrease which i couldnt. And from what you pretty much taught me, you cant really look at one component and find the resisance.

So i guess my question is when can i look at one component to find the current, and whats the exception?

I don't think there is a general way to solve problems involving circuits. However, my advice is that when we use

V=IR

, we should ask ourselves 'which V,I,R are we dealing with here?'. Also, remember that resistors in series would have the same current passing through, while resistors in parallel would have the same voltage drop across them.

Let's take this circuit as an example:

Say, we want to find the current in this circuit. We notice that current passing through

R_1

would be equal to total current passing through

R_2

and

R_3

, as

R_1

is in series with the

R_2R_3

system. Hence:

I_1 = I_{23} = I_{circuit} = \frac{emf}{R_{total}}

Now that we have

I_1

, we can calculate

V_1 = I_1 \times R_1

From this, we can calculate voltage across the

R_2R_3

system by using:

V_{23} =\ emf\ - V_1

as

R_1

and

R_{23}

are in series.

Now we notice

R_2

and

R_3

are parallel with each other. This means:

V_2 = V_3 = V_{23}

From this we can calculate the current in each branch:

I_2 = \frac{V_2}{R_2}\ ,\ I_3 = \frac{V_3}{R_3}

(It would be no surprise that

I_2 + I_3 = I_1 = I_{circuit}

as

R_1

and

R_2R_3

are in series)

:smile:

(edited 8 years ago)

Reply 13

OP