current electricity question ....the most u wud probably need is ohm's law n internal resistance of batteries

Question

a high resistance voltameter is connected to a high resistance battery. it reads 9.0 V. a 24 Ohm resistor is connected across the terminals of the battery and the voltameter readin becomes 7.2V
calculate the internal resistance of the battery
the ANSWER is: 6 Ohms...
but i am not getting how we got it.....so please EXPLAIN the answer ,THE IMPORATANT THING IS TO EXPLAIN THE ANSWER N NOT DO IT...(of course i wudnt mind if u did it--it wud be a greater help)

2)factore affecting the electromotive force of a ell:
material of electrodes,electrolyte used ....
i dont understand why the distance between the two electrodes does not matter....it shud matter,,,it does change the resistance after all

plz answer even if u have got 1 answer...any help wud be appreciated

no,this is not a 12th standard practical question but a 10th standard,theory question......i have this question already solved in my book---its just that i dont understand how they got it

why did we do 7.2/24 n not 7.2/(24+internal resistance)

sorry,that wasnt high resistance battery...it wud be 'dry cell battery'

thanks,andy but i was asking for a qualitative explanation ,not this formula one...this one is very direct....i wanted the big one...thanks anyways...good luck!

Answer 1

The 9 volt across the battery has reduced to 7.2 V

The difference of (9-7.2) = 1.8 V is the potential difference across the internal resistance r.

Current in the circuit (or through the internal resistance and external resistance) will remain the same.

Current = Potential difference/ resistance.

Current = 7.2 / 24 = 1.8/ r

r = 6 ohm.

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OR

The current in the resistance 24 ohm is 7.2 / 24 = 0.3 ohm.

E = i (R+r)

9 = 0.3 (24 +r)

(24 +r) = 30

r = 6 ohm.
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The internal resistance of a cell or battery does not depend upon the DISTANCE between the electrodes of the cell.

When external resistance is connected to the battery the internal resistance varies according to the resistance used externally.

In short, the internal resistance depends upon the external resistance.

By drawing a graph between voltage and current for various external resistances, we can use it to determine the internal resistance of a cell or battery.

Vs: is the potential difference across the battery when no resistance is connected.
V: is the potential difference across the resistance R connected externally.
r: is the internal resistance
I is the current in the circuit.

Then
Vs = I (R +r)

Therefore

r = (Vs / I) - R

or

r = (Vs - V) / I




http://www.ectechnic.co.uk/exp7.html
http://www.personal.psu.edu/bqw/physics_151/lab/lab151_5.html

Answer 2

The way you all came up with 6 ohms is:
First measuring voltage at battery terminals without a load connected:
Then you all connected the load which is 24 ohms. And measured the voltage arcoss the battery terminal with the 24 ohm resistor connected to the battery and your voltage was 7.2 Volts.
Using the 7.2 volts you measured with the load connected and the 24 ohm resistor rating you all found the current total current.

7.2 volts/ 24 ohms = 0.3 or 300milli amps.
Now after getting your current in order to find out the resistance of the battery you have to know what the voltage drop across the battrery is. To find that you subtract the 9.volts you measured with no load on the battery from the 7.2 volts you measured with the 24 ohm resistor in the circuit.

9.0 volts - 7.2 volts = 1.8 volts. This is your voltage drop across the battery from the internal resistance of the battery.
Again using ohms law you and 1.8 volts which is the calculated voltage drop across the battey. you can find the internal resistance of the battery.

1.8 volts (voltage drop across the battery)
0.3 amps or 300milli amps. current flowing in circuit.
applying ohm's law.

1.8volts/0.3 amps = 6 ohms which is the internal resistance of your battery.

Distance between the two plates does not matter because the electrolyte is between the two plates filling in the void creating a path for the ions to flow. (internally inside a battery it is ion flow instead of electron or hole flow) It is the chemical reaction between the plates and the electrolytes that determines the voltage of the cell with no load on it not the distance between the two plates.
You can varify about it being the chemical reaction which determines the voltage of a cell at home if you want to.
Get a piece of stiff copper wire, a standard nail and a grapefruit or organge or lemon.

Stick the nail in one side of the grapefruit and as far apart as you can from the nail stick the copper wire in the grapefruit. Then measure the voltage with your Fluke or dmm. It has to be a high resistance meter so it doesn't load your make shift battery down!
Now take the nail out and stick it back in closer to the copper wire and measure your voltage again. You will see that the voltage ramains the same even though the nail and piece of copper are closer together.

Hope that helps you some !!!
I think what you was missing was that when you read the voltage with the resistor in the circuit the 7.2 volts was the voltage drop across the resistor so by ohms law you have to use 7.2 volts instead of 9,0 volts to calculate your total current for the circuit.

Answer 3

look any voltmeter measures the voltage across its terminals .if it is a ideal voltmeter then it will not drown any current .and the voltage ac cross its terminals is the emf of the cell.(there is a difference between the emf and the voltage of the cell , the voltage is the emf minus the multiplication of the current through the battery and the internal resistance of the battery.
the resistance of the ideal battery is zero.
for non ideal cases replace the battery with the an ideal battery with a ideal battery and the internal resistance in series .
the non ideal voltmeter could be replaced with the voltmeter resistance in parallel with an ideal voltmeter. so the ideal voltmeter will measure the voltage across the internal resistance of the voltmeter.and it is the reading of the voltmeter .
the emf of the cell is a standard one ,it's value depends on the concentration of the electrolyte ,nature of the electrolyte and the distance between the electrodes and the nature of the electodes. when cell gets used the concentration of the electrolytes decreases, and and voltage decreases but the emf remains the same because it is a standard one.the distance between the electrodes doest get altered in the usage of the cell.

Answer 4

Answer 1)

Physics approximates a value w.r.t. other.
Example 1000006 meters is approximated to 1000000 meters and it is fairly accurate with an error of 0.0006 %.
What ever may be internal resistance of a battery that is of the order of few ohms and resistance of voltmeter is of the order of mega ohm.
That is when you measure voltage directly on battery terminal using voltmeter it actually measures emf of the battery with negligible error.
But when you connect a 24 ohm resistance to the battery that may be comparable and can not be neglected. But again you measure the voltage across the 24 ohm resistance with the high resistance voltmeter the case is same as above.

Neglecting small error you can take current x 24 = 7.2.
Or current in 24 ohm resistance is 7.2/24 = 0.3 amp.

Open circuit voltage is 9 volt (as I explained above ).
Drop across battery = 0.3 r = 9-7.2 = 1.8.
r = 1.8/0.3 = 6 ohms
Approximation is necessary in real world calculation.
But take care what u r approximating w.r.t. what.
I think u got it.

Answer 2)

When we measure emf with a high resistance voltmeter virtually it measures the open circuit voltage less than a micro ampere current drawn from battery. So, what ever may be internal resistance of battery(or separation of plates of battery ) it will not affect the emf measurement but certainly it will afect in real work when we draw amperes of current.

Answer 5

Hi there, I'm in the Xth too n have done this ques. recently.

For the 1st ques., remember 1 formula:
Int. res. = (Ext. res. x Drop in potential) / p.d. across Ext. res.

=> r = [R x (E-V)] / V

where:
r --> Int. res.
R --> Ext. res.
E --> e.m.f. of cell
V --> p.d. at its ends when current is flowing.

Derivation:

Applying Ohm's law to ext. circuit only:
I = V / R
Applying Ohm's law to int. circuit only:
I = E / (R+r)

Comparing above both :
V / R = E / (R+r)
=> VR + Vr = ER
=> Vr = ER - VR
=> r = [R x (E - V)] / V

So, by ur readings:
r = [24 x (9 - 7.2)] / 7.2 = (24 x 1.8) / 7.2 = 43.2 / 7.2 = 6 ohms.

For 2nd ques., YES, dis. between electrodes does matter. The e.m.f. of a cell :
1. Is inversely proportional to surface area of electrodes.
2. Is directly proportional to dis. between electrodes.
3. Depends on (i) nature (ii) concentration (iii) temp. of electrolyte n electrodes.

Have a nice day...

Answer 6

There is a resistance in real battery called internal resistance.
When we connect the Volta meter to a battery in open state (means no other resistance is connected to it, which is equivalent to infinite resistance ) so then the whole potential difference is across the two terminals of vol meter as in Ur prob it is 9.0V. when we connect a external resistance there is the voltage drop across the ext resistance decreases which we record in voltmeter as in Ur prob 7.2 . in the prev case the current flowing is approx zero therefore there is no pot. diff. across int. res.In second case some current flows there fore there is some pot across int res. and hence the pot diff across ext res. is less

Answer 7

Regarding the internal resistance of a battery, john s's answer is correct.

Answer 8

current squared x internal resistance = joules lost in heat per second.

Answer 9

i hope this is a 12th std practical ques, i suggest 2 go n check out pradeeps 3rd chap about resistence