0
$\begingroup$

Say we had a circuit with a cell (with internal resistance)and an external resistor of 10 ohms as shown:

enter image description here

The 10 ohm resistor then increases in temperature, leading to an increase in resistance in this resistor. My thinking was that since the 10 ohm resistor now has a value higher than 10 ohms, more voltage drops across this resistor. Since your emf is constant, less voltage must drop across the internal resistor. Less voltage drop means less resistance. Where is the flaw in my logic?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Welcome to SE.Physics! Since you're asking about why your calculation for voltage changes, how are you calculating it? What information/measurements do you have, and what equations are you using? $\endgroup$ – Nat May 18 at 6:28
  • $\begingroup$ Ignore the small resister initially. In this case, the voltage drop across the $10\ \Omega$ resistor is the same as the $V$ supplied by the battery. Since $V=IR$, this implies $I=V/R$ where $R=10\ \Omega$ and $V$ is constant. If $R$ increases because of higher temperature, then $I$ decreases since $V$ is constant. Adding the small resistor in series with $R$ increases the total resistance which decrease the current further. $\endgroup$ – Cinaed Simson May 18 at 7:21
0
$\begingroup$

Your logic is not flawed ( assuming the internal resistance is constant). However, if the internal resistance is much smaller than the load resistance, the increase in voltage across the load resistance may be difficult to measure. This is because a small change in the load resistance will have little effect on the cell’s terminal voltage. In other words the battery will approach being a constant (ideal) voltage source (terminal voltage = emf).

Hope this helps.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.