EMF and internal resistance laboratory work

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isorounded
1. The problem statement
I did a laboratory work in which I had to find power and internal resistance. I used an ordinary battery so it's closed circuit. I did all tasks except writing a hypothesis and conclusion I can't really think of it. I don't know what to write since I just measure using voltmeter and ammeter and then calculate Electric power and internal resistance. Can someone suggest anything pls?

Homework Equations


Equation: Δr = r_ap ε_r, where Δr is the absolute error of the internal resistance, r_ap is the calculated internal resistance and ε_r is its relative error.

Handwritten formula for relative error of internal resistance: εr = (ΔE + ΔU)/(E_ap − U_ap) + ΔI/I_ap.
Equation: r_ap = (E_ap - U_ap) / I_ap, with subscript ap on each symbol.
Typeset error formulas: E = E_ap ± ΔE, ε_E = ...%; r = r_ap ± Δr, ε_r = ...%, with the blank percentage placeholders left unfilled.


E - [/B]emf, electromotive force r - internal resistance ap - calculated
U - voltage I - current

The Attempt at a Solution



Well from ammeter and voltmeter I got that Uap= 2V, Iap= 1,5 A and Eap= 4V. Then I calculated rap (rap=1,33 ohms). I found delta u(△u=+-0,25) , delta I (△I=+-0,03) and I know that electromotive force error is almost same as voltage error because power supply's resistance is usually low so R≥r and △E = △u = +-0.25 V. I found Er what led me to finding △r. △r = rap+Er = 1,33 * 0.27= 0.3591 ohms. Now I am bit confused we the results I got E = 4 +- 0.25 V and r = 1.33 +- 0,3591 ohms, please let me know if I did it wrong. Now the main thing I struggle with is writing hypothesis and conclusion. I was thinking to relate given formulas with ohm's law and formula but not sure if it's good. So can someone help me to write hypothesis and conclusion or atleast give me some ideas. Thank you.
 
Hi iso,

isorounded said:
I used an ordinary battery
isorounded said:
because power supply's resistance is usually low
So when did you use the one and when the other ?
I can follow your 1.33 Ohm calculation, but after that I am lost in your story. What do the ##\Delta## stand for ?
You write that you did all tasks (which is good), but I have hard time guessing what all these tasks were.
 
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First of all, thank you for a response. You probably would have guessed that english is not my native language so that might be a part of the reason why you don't understand me :D And that my laboratory work is described in my native language, it doesn't make it easier, but I will try to explain as much as I can. I have used one power supply which was battery. Here is a picture of circuit
Circuit schematic with a battery, voltmeter V connected across the battery, ammeter A in series, a rheostat with a sliding contact, and a switch; labelled 222 pav.


Now Δ stands for "error" if that makes sense.

Also I will try to translate laboratory tasks step by step and it's really simple so I hope you will understand atleast this one :D

Firsly what did I use for this work: battery, voltmeter, ammeter, toggle, connecting wires.

1. Measure electromotive force of power supply(a battery's in my case).

2. Mark ammeter's and voltmeter's results when a toggle is on. Calculate rap(there is a formula above). Calculate power supply's resistance and electromotive force absolute and relative error.

3. Write down r and electromotive force results:
Two printed formulas for absolute and relative error: E = E_ap ± ΔE, E_E = ...%; and r = r_ap ± Δr, E_r = ...%.


And I'm pretty sure that I did all calculations right but I just can't think of hyphotesis and conclusion.
 
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Also was thinking to relate with ohm's law because r=E-U/I is derived from E=U+Ir
 
OK, so under 1. you have measured a voltage of 4 V ?
Then, under 2, you press the switch and measure 2 V and 1.5 A ?

And the hypothesis is that the battery can be represented by a circuit consisting of an ideal voltage source in series with an internal resistance. And the calculated internal resistance that would then follow is (4 V - 2 V ) / 1.5 A = 1.33 ##\Omega## according to
Equation: r_ap = (ε_ap − U_ap) / I_ap, with subscript ap on each symbol.
.
So I don't understand the subscripts ap on the righthand side: I should think they stand for 'measured' (as under points 1 and 2), not for 'calculated' ; can you explain ?

Then:
I have the faint suspicion that ##\varepsilon_r## in
Equation for relative error: epsilon_r = (Delta E + Delta U)/(E_ap - U_ap) + Delta I / I_ap, with subscript ap on E and U and I.
stands for the relative error in the calculated internal resistance of 1.33 ##\Omega##. For your purposes that relative error (or inaccuracy, if you want) is the sum of the relative errors in the numerator and in the denominator. To estimate a value, you need the accuracy of the voltmeter and the accuracy of the current meter. Do you have those ? Are they provided on the instruments or given by teacher ?