Internal Resistance Help + Questions - Rep involved!

Me again

Alright, the topic of Internal Resistance, etc. wasn't explained that well at school and I'm teaching it to myself but I'm getting confused when we get to the questions. Could someone first of all, show me all the equations (AS bear in mind) I need to know and explain the basic concepts such as when each is used. Also lost volts and how to find the e.m.f and internal resistance and all that.

Also, could someone help me out with these questions. When you solve the question, could you explain which equation you use and how you worked it out:

1. A dry cell can deliver 0.15 A when connected to an 8 ohm resistor. When another 8 ohm resistor is connected in series with the first, the cell can only deliver 0.08 A to the combination.
(a) Calculate the e.m.f and internal resistance of the cell.
(b) The two resistors are re-connected in parallel with each other and then re-connected to the cell terminals. What is the current passing through each resistor now ?

My attempt:

Would you do V=IR = 0.08x16 = 1.28V and this would be your EMF? Right? :frown:

But what's confusing me is why wouldn't V=IR = 0.15x8 = 1.20V be right as well?!

2. To find the e.m.f. and internal resistance of an unmarked battery you are given two resistors each of value 5 ohm and an ammeter. When the resistors are connected in series the current drawn from the battery is 0.75A and when connected in parallel the current drawn from the battery is 2.0A. What is the e.m.f. and internal resistance of the battery ?

Attempt:

V=IR = 10x0.75 = 7.5V in series
V=IR = 1/5 + 1/5 = 2/5 = 5/2 = 2.5 ohms x 2.0 = 5V

But shouldn't they be the same :s:

3. Two cells, each of e.m.f. 1.5V and internal resistance 0.4 ohm are connected (i) in series and then (ii) in parallel.
(a) What is the e.m.f and internal resistance of the battery formed in each of these ways ?
(b) What current would be driven by each of these batteries through an external resistance of 1.6 ohm ?

Please help :s:

Reply 1

teachercol

Hmm -

All sources of emf can be treated as an emf in series with a resistance usually lablled lower case r.

Emf = I x total resistance in circuit

Lost volts correspond to the energy converted to heat by the source resistance inside the cell.

Lost volts = I x r

Terminal PD = emf - lost volts = E - i r = energy per coulomb available in rest if circuit
= I x total external resistance

Reply 2

teachercol

1. Emf = Ix total resistance in circuit

E = 0.15 ( r + 8 )
E = 0.08 ( r +16 )

Should be able to solve these

Reply 3

teachercol

2. Same basic technique

Set up two equations which correspond to

E = I ( total external resisatnce + r)

Solve for E and r.

Reply 4

teachercol

3. Cells in series : add emf, add r ; E = 3.0v r = 0.8ohm
Cells in parallel ; same emf, combine r in parallel : E = 1.5v r = 0.2 ohm.

Reply 5

Cataclysm

teachercol

1. Emf = Ix total resistance in circuit

E = 0.15 ( r + 8 )
E = 0.08 ( r +16 )

Should be able to solve these

That's the bit that's a bit confusing because would you use:

0.15 X 8 = 1.20 Volts

or

0.08 x 16 = 1.28 Volts.

:s:

Reply 6

phys981

Neither. In doing either of those you're ignoring r which is crucial.

They're simultaneous equations.

So

E = 0.15 ( r + 8 ) and
E = 0.08 ( r +16 )

which means that

0.15 ( r + 8 ) = 0.08 ( r +16 ) becasue E = E

Multiply out and solve for r then put that back in and solve for E.