# Issac Physics: Current between ResistorsWatch

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Thread starter 3 weeks ago
#1
Hi guys,

I was doing this question and I'm really frustrated since I have no idea how to approach it, I managed to form the equations in the hints, but I'm still stuck on how to use them to find the answer.

Image: https://imgur.com/I2GIiqi

Link to the question: https://isaacphysics.org/questions/c...f-aed4a9188241

Help along with an explanation is much appreciated!
0
3 weeks ago
#2
You should be able to get the total current by treating it as the circuit in hint 3 first

The ideal ammeter acts like a conductor so the potential on the left hand side of the 2 right hand resistors is the equal and the potential on the right hand side of those resistors is equal... the resistors have equal value and the current in them both has got to add up to the total current you found before... so you can find the current in each of those resistors.

using a similar attack on the left resistors we can calculate the current in each of those resistors - they have the same potential difference across them but in this case the resistances are different so more current will flow through one than the other.
0
3 weeks ago
#3
(Original post by Navboi)
Hi guys,

I was doing this question and I'm really frustrated since I have no idea how to approach it, I managed to form the equations in the hints, but I'm still stuck on how to use them to find the answer.

Image: https://imgur.com/I2GIiqi

Link to the question: https://isaacphysics.org/questions/c...f-aed4a9188241

Help along with an explanation is much appreciated!

It may be better that you show your equations and how you set up the equations.
0
Thread starter 3 weeks ago
#4
(Original post by Joinedup)
You should be able to get the total current by treating it as the circuit in hint 3 first

The ideal ammeter acts like a conductor so the potential on the left hand side of the 2 right hand resistors is the equal and the potential on the right hand side of those resistors is equal... the resistors have equal value and the current in them both has got to add up to the total current you found before... so you can find the current in each of those resistors.

using a similar attack on the left resistors we can calculate the current in each of those resistors - they have the same potential difference across them but in this case the resistances are different so more current will flow through one than the other.
Thanks for the reply man,

I initially got the current in the two right-hand resistors to be equal like you said because my teacher explained that since their the same resistance they must be drawing the same current. But I don't understand this, or how because the "ammeter acts like a conductor" makes the current flowing in them equal. From my understanding of Kirchoff's First Law, shouldn't the current from the 2R resistor be the same as the current in the resistor under it to ensure that the same amount of current is going through the two right resistors? I'm assuming the vertical wire messes things up, but I don't understand how they both can have equal currents with one resistor being having double the resistance of the other (from the left side)
0
3 weeks ago
#5
(Original post by Navboi)
Thanks for the reply man,

I initially got the current in the two right-hand resistors to be equal like you said because my teacher explained that since their the same resistance they must be drawing the same current. But I don't understand this, or how because the "ammeter acts like a conductor" makes the current flowing in them equal. From my understanding of Kirchoff's First Law, shouldn't the current from the 2R resistor be the same as the current in the resistor under it to ensure that the same amount of current is going through the two right resistors? I'm assuming the vertical wire messes things up, but I don't understand how they both can have equal currents with one resistor being having double the resistance of the other (from the left side)
KCL says that the current entering the T junctions either side of the ammeter must be equal to the current leaving... but they're T junctions so there could be 2 different currents in the 2 left resistors with some current crossing vertically through the ammeter to make the currents in the right two resistors equal...

The ammeter acting like a conductor has a couple of useful consequences
current can flow vertically through it
the potential at the mid points between the top 2 resistors is the same as the potential at the mid point between the bottom 2 resistors.
so the potential difference across the top left resistor is equal to the PD across the bottom left resistor - and what can we say about the current through a 2R resistor compared to a 1R resistor if they both have the same potential difference across them?
0
Thread starter 3 weeks ago
#6
(Original post by Joinedup)
KCL says that the current entering the T junctions either side of the ammeter must be equal to the current leaving... but they're T junctions so there could be 2 different currents in the 2 left resistors with some current crossing vertically through the ammeter to make the currents in the right two resistors equal...

The ammeter acting like a conductor has a couple of useful consequences
current can flow vertically through it
the potential at the mid points between the top 2 resistors is the same as the potential at the mid point between the bottom 2 resistors.
so the potential difference across the top left resistor is equal to the PD across the bottom left resistor - and what can we say about the current through a 2R resistor compared to a 1R resistor if they both have the same potential difference across them?
Alright, I think I get it now. So since the resistance in 2R is double of R, half the current should pass through it to keep the p.d the same right?
based on that idea ive drawn this crude diagram:

https://imgur.com/t71uluZ

am I on the right track?
0
3 weeks ago
#7
(Original post by Navboi)
Alright, I think I get it now. So since the resistance in 2R is double of R, half the current should pass through it to keep the p.d the same right?
based on that idea ive drawn this crude diagram:

https://imgur.com/t71uluZ

am I on the right track?
1
Thread starter 3 weeks ago
#8
(Original post by Joinedup)
Nah I just didn't work out the resistances yet. But now I plugged in the values and I got it correct!
Thank you so much for the assistance and for being patient with me 0
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