A resistor is connected to a battery of certain potential difference, it has now $V$ volt potential difference across its ends, after some time a resistor is connected in parallel with it. How does the potential difference across 1st resistor would change after the addition of 2nd resistor in that way? My physics text book says that potential difference across first would decrease,but according to me it should remains same since they are in parallel. can anyone correct me of i am wrong?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ You are correct but in practice a real battery has internal resistance from all the metal paths or conductors inside the battery. $\endgroup$– PhysicsDaveDec 10, 2018 at 14:42
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
For an ideal battery, for which the voltage is independent of the load current, you are correct: the battery and the two resistors are all in parallel and must have the same voltage drop.
For a real battery, that's not the case. A common model for the behavior of a battery whose output changes under load is an ideal voltage source in series with a small "internal resistance." As the current drawn from such a battery increases, the voltage drop across the internal resistance increases as well, so the voltage drop across the external terminals decreases. In this model, a "dying" or "drained" battery is one whose internal resistance has become large.
$\begingroup$
$\endgroup$
Unless this question is targeted at the decreasing voltage of a battery while it is discharging ("after some time") or the fact that the voltage of the battery depends on the load, the voltage across the resistors should be constant.
The other two aspects I mentioned would lead to a decrease of voltages, but all three (resistor 1, resistor 2, battery) voltages will be the same.
