This discussion is now closed.

Check out other Related discussions

Internal resistance calculations

18

1) A cell has an internal resistance r, of 0.8 ohms and is connected to a circuit of a 10 ohm resistor, R. If a current of 0.14 A flows, what is the emf of the cell?

2) When a current of 2.0A flows the pd across a battery's terminals falls from 12.0V to 9.0V. What is the internal resistance of the battery?

3) In a circuit there are;
a 9 volt battery of 9.0V and internal resistance 1.3 ohms.
a 6 ohm and 21 ohm resistor which are connected in parallel with the battery.
If the current supplied by the battery is 1.50 A and that through the 6 ohm resistor is 1.17A, what is the current passing through the 21 ohm resistor?

Reply 1

Christophicus

OP

18

Any physicists in the land of TSR?

Reply 2

BCHL85

12

Widowmaker

1) A cell has an internal resistance r, of 0.8 ohms and is connected to a circuit of a 10 ohm resistor, R. If a current of 0.14 A flows, what is the emf of the cell?

2) When a current of 2.0A flows the pd across a battery's terminals falls from 12.0V to 9.0V. What is the internal resistance of the battery?

3) In a circuit there are;
a 9 volt battery of 9.0V and internal resistance 1.3 ohms.
a 6 ohm and 21 ohm resistor which are connected in parallel with the battery.
If the current supplied by the battery is 1.50 A and that through the 6 ohm resistor is 1.17A, what is the current passing through the 21 ohm resistor?

1. R + r = 0.8 + 10 = 10.8 ohms
-> V = I(R+r) = 0.14*10.8 = ... V

2.Ir = dV
2*r = 12 - 9
-> r = 3/2 = 1.5 ohms

3. Can't get your question

Reply 3

john !!

14

Widowmaker

1) A cell has an internal resistance r, of 0.8 ohms and is connected to a circuit of a 10 ohm resistor, R. If a current of 0.14 A flows, what is the emf of the cell?

2) When a current of 2.0A flows the pd across a battery's terminals falls from 12.0V to 9.0V. What is the internal resistance of the battery?

3) In a circuit there are;
a 9 volt battery of 9.0V and internal resistance 1.3 ohms.
a 6 ohm and 21 ohm resistor which are connected in parallel with the battery.
If the current supplied by the battery is 1.50 A and that through the 6 ohm resistor is 1.17A, what is the current passing through the 21 ohm resistor?

1. see it as two resistors in series, add them to get a total reistance of 10.8 Ohm. V = IR = 10.8*0.14 V I think!! I suck at internal resistance.. missed a few lessons.

2. 12-9 = 3 lost volts. R = V/I = 3/2 Ohm. agian not sure

3. do it later got ddinner

Reply 4

Christophicus

OP

18

BCHL85

1. R + r = 0.8 + 10 = 10.8 ohms
-> V = I(R+r) = 0.14*10.8 = ... V

2.Ir = dV
2*r = 12 - 9
-> r = 3/2 = 1.5 ohms

3. Can't get your question

Exactly what I got. For question 3, I think it is a really badly worded question. My rubbish physics teacher typed it up (he got a 3rd at Bradford, judge for yourself).
I was thinking of kirchhoff's first law.
Sum of currents flowing into a point = sum of currents flowing out of a point
I might be wrong.

Reply 5

BCHL85

12

Do you have the diagram of it?

Reply 6

Christophicus

OP

18

BCHL85

Do you have the diagram of it?

No diagram I'm afraid.

Reply 7

BCHL85

12

So is it 6ohms is in series with 21ohms? or they are in parallel?

Reply 8

Christophicus

OP

18

Widowmaker

1) A cell has an internal resistance r, of 0.8 ohms and is connected to a circuit of a 10 ohm resistor, R. If a current of 0.14 A flows, what is the emf of the cell?

2) When a current of 2.0A flows the pd across a battery's terminals falls from 12.0V to 9.0V. What is the internal resistance of the battery?

3) In a circuit there are;
a 9 volt battery of 9.0V and internal resistance 1.3 ohms.
a 6 ohm and 21 ohm resistor which are connected in parallel with the battery.If the current supplied by the battery is 1.50 A and that through the 6 ohm resistor is 1.17A, what is the current passing through the 21 ohm resistor?

parallel.

Reply 9

BCHL85

12

Widowmaker

parallel.

If it's like that, I think it doesn't matter how the battery is, you just need to know the voltage across 6ohm = voltage across 21 ohm cuz they are in parallel.
So I(21) = 6/21.I(6) = 6*1.17/21 A

Reply 10

Christophicus

OP

18

BCHL85

If it's like that, I think it doesn't matter how the battery is, you just need to know the voltage across 6ohm = voltage across 21 ohm cuz they are in parallel.
So I(21) = 6/21.I(6) = 6*1.17/21 A

Your calculation isn't very clear. Can you go through it?

Reply 11

BCHL85

12

Ah, I did check ..
Here is what its mean
V = 9V, r = 1.3ohms, I = 1.5V
R1 = 6ohms, R2 = 21ohms
-> V(R1) = V - I.r = 9 - 1.5*1.3
-> I1 = 1.175A
So V(R2) = 9 - 1.5*1.3
-> I2 = 0.33..
which is similar to what I did before

Reply 12

BCHL85

12

Widowmaker

Your calculation isn't very clear. Can you go through it?

I just used V(6ohms) = V(21ohms) because they are in parallel
So I(6).6 = I(21).21

Internal resistance calculations

Related discussions

Articles for you

How The Student Room is moderated