Battery terminal voltage?
With a fresh 9V battery, a voltage of 9.2V is measured without load (idle). When drawing 100mA, the voltage at the terminals is only 8.6V.
What was the internal resistance of the battery? I was able to do that; it was 6 ohms.
What terminal voltage do you expect when loaded with an external load resistance of 18 ohms? I would add the resistances: 6 + 18 = 24 ohms, but how do I proceed? I hope someone can help me. Thanks in advance.
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In the first load, there must also be a load resistance so that the 100 mA is set. The internal resistance drops 0.6 V, so Ri=6 Ohm, right. The load resistance RL, on the other hand, was RL=8.6V/0.1A=86 Ohm.
This resistance is now replaced with an 18 ohm resistance. You get a total of 24 ohms. The current of the circuit is then: I=9.2V/24Ohm=383 mA
The terminal voltage is then:
U_KL = 18 ohms * 0.383A = 6.894V
What do you mean… this resistance you're replacing with an 18 ohm resistance?
The load resistance before, ie in the first task part, was 86 ohms. Because it was only possible that 100 mA could flow. If there was no load resistance, the internal resistance of the source would be the only resistance in the circle and a much larger current would flow. The total voltage of 9.2 V would only drop at the internal resistance.
The new load resistance in the second task is, on the other hand, 24 ohms.
Correction: The new load resistance is 18 ohms. The sum of the circuit resistance is 24 ohms, as described in the main response.
0.6V ± 4 = 0.15V
That would be wrong, according to Löser, 6.9 V would get out. Why do you share through 4
What is the computational path for this?