# Get better at Circuit questions? (EMF, Internal resistance, potential dividers, etc)Watch

#1
It's all confusing and nothing makes sense.

Physics is the only subject that's currently pulling me down and I don't know how I can improve at all.

Some help would be nice.
0
1 year ago
#2
Just do a crap tonne of questions. The only way really.
0
1 year ago
#3
Emf = Vtpd + Ir( internal resistance)
or you can use it in the form
Emf= I(R + r)

Potential divider rule:

You find the Voltage through R2 (aka Vout1) by using Vout1= R2/R1+R2 x Vin(supply voltage)
Voltage through R1 can then be either found by same method Vout2= R1/R1+R2 x Vin(supply voltage) or by doing Vin - Vout1 (since we only had 2 resistors in this example)

It’s pretty straight forward, all you need is to understand the concepts, remember the formulae and then apply then practice a lot of questions. Good luck
0
#4
(Original post by shameful_burrito)
Emf = Vtpd + Ir( internal resistance)
or you can use it in the form
Emf= I(R + r)

Potential divider rule:

You find the Voltage through R2 (aka Vout1) by using Vout1= R2/R1+R2 x Vin(supply voltage)
Voltage through R1 can then be either found by same method Vout2= R1/R1+R2 x Vin(supply voltage) or by doing Vin - Vout1 (since we only had 2 resistors in this example)

It’s pretty straight forward, all you need is to understand the concepts, remember the formulae and then apply then practice a lot of questions. Good luck
What about questions with like internal resistance and etc.

Thanks for that btw.
0
1 year ago
#5
(Original post by RickHendricks)
What about questions with like internal resistance and etc.

Thanks for that btw.
Internal resistance r is usually used in the equation for lost volts found by V= Ir (current x internal resistance). They will probably either give you values for r and the current or a value for V(the lost volts) and the current and ask you to find the missing term in the equation. From there is just basic algebra and rearranging. Next step might be to find the Emf. Formula for that is Emf= Vtpd + Ir (lost volts)
0
#6
(Original post by shameful_burrito)
Internal resistance r is usually used in the equation for lost volts found by V= Ir (current x internal resistance). They will probably either give you values for r and the current or a value for V(the lost volts) and the current and ask you to find the missing term in the equation. From there is just basic algebra and rearranging. Next step might be to find the Emf. Formula for that is Emf= Vtpd + Ir (lost volts)
Could you explain the answer in a mark scheme for me?
0
1 year ago
#7
(Original post by RickHendricks)
Could you explain the answer in a mark scheme for me?
I can try, been a while since I did this topic back in high school. Post a picture
0
#8
(Original post by shameful_burrito)
I can try, been a while since I did this topic back in high school. Post a picture

Q:

The temperature of the thermistor is increased so that its resistance decreases. State and explain what happens to the pd across the 1200 Ω resistor.

A:
(voltage of supply constant) (circuit resistance decreases) (supply) current increases or potential divider argument (1)

hence pd across 540 Ω resistor increases (1)

hence pd across 1200 Ω decreases (1) or

resistance in parallel combination decreases (1)

pd across parallel resistors decreases (1)

pd across 1200 Ω decreases (1)

why does the current increase? I swear current always stays the same.
0
1 year ago
#9
(Original post by RickHendricks)

Q:

The temperature of the thermistor is increased so that its resistance decreases. State and explain what happens to the pd across the 1200 Ω resistor.

A:
(voltage of supply constant) (circuit resistance decreases) (supply) current increases or potential divider argument (1)

hence pd across 540 Ω resistor increases (1)

hence pd across 1200 Ω decreases (1) or

resistance in parallel combination decreases (1)

pd across parallel resistors decreases (1)

pd across 1200 Ω decreases (1)

why does the current increase? I swear current always stays the same.
If resistance decreases, current increases and the potential difference (the voltage) also decreases. If you look at ohm’s law V=IR and imagine R gets smaller then the value for V will also be smaller. And the value for I, the current, would get bigger if R decreases, as I=V/R
1
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