Retsek
Badges: 17
Rep:
?
#1
Report Thread starter 2 years ago
#1
Three identical cells, each of internal resistance R, are connected in series with an external resistor of resistance R. The current in the external resistor is I. If one of the cells is reversed I'm the circuit, what is the new current in the external resistor? (In terms of I and R)
_____________
Can someone please help me with this question? I have no idea how to deal with reversed cells, I know the total resistance is going to be 4R and the current is constant throughout the entire circuit, but that's as far as I've gotten.
0
reply
Eimmanuel
Badges: 13
Rep:
?
#2
Report 2 years ago
#2
(Original post by Retsek)
Three identical cells, each of internal resistance R, are connected in series with an external resistor of resistance R. The current in the external resistor is I. If one of the cells is reversed I'm the circuit, what is the new current in the external resistor? (In terms of I and R)
_____________
Can someone please help me with this question? I have no idea how to deal with reversed cells, I know the total resistance is going to be 4R and the current is constant throughout the entire circuit, but that's as far as I've gotten.
Name:  Cell_in_series.png
Views: 154
Size:  3.9 KB

The cells are in opposition, so the p.d. across XY is zero.
0
reply
chemlegend27
Badges: 0
Rep:
?
#3
Report 1 year ago
#3
Does anyone have the method to figure this out? I'm stuck on it as well
0
reply
i_do_maths
Badges: 5
Rep:
?
#4
Report 1 year ago
#4
sum of p.d. in a closed loop must equal to the sum of emf. in this case the sum of emf is 0 ∴there is no p.d. in the circuit. However, the internal resistance still adds up, because they're in series, ∴if there's another cell/battery in the circuit this must be considered.
1
reply
Eimmanuel
Badges: 13
Rep:
?
#5
Report 1 year ago
#5
(Original post by chemlegend27)
Does anyone have the method to figure this out? I'm stuck on it as well

When the cells are connected in series, the combined e.m.f across the battery is equal to the absolute value of the algebraic sum of the voltage change across each of the individual cells.

You can think of emfs of the cells as vector when they are connected in series and then sum the vector arrows to get the emf of the cells connected in series. For example,

Name:  Cell_in_series_01.JPG
Views: 186
Size:  49.5 KB

So the emf of the cells connected in series is 1.5 V. Note that we ignore the minus sign.
1
reply
Bullyhunter76
Badges: 5
Rep:
?
#6
Report 1 year ago
#6
When the cells are connected in series, the combined e.m.f across the battery is equal to the absolute value of the algebraic sum of the voltage change across each of the individual cells.

You can think of emfs of the cells as vector when they are connected in series and then sum the vector arrows to get the emf of the cells connected in series. For example,



So the emf of the cells connected in series is 1.5 V. Note that we ignore the minus sign.
1
reply
madhatter101
Badges: 3
Rep:
?
#7
Report 11 months ago
#7
If each cell has emf of 'v', initially the total current in circuit=total voltage/total resistance = 3v/4R.As people have explained before, reversing a cell will cause it and its neighbouring cell's voltage to effectively cancel out, but their resistances will remain affecting the circuit.So now the current = v/4RTherefore the current has divided by 3 as (3v/4R)/(v/4R)=1/3.Therefore the new current in terms of I(and R) is I/3.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Regarding Ofqual's most recent update, do you think you will be given a fair grade this summer?

Yes (344)
34.68%
No (648)
65.32%

Watched Threads

View All