Ask your own question, for FREE!
Physics 18 Online
OpenStudy (anonymous):

@.Sam.

OpenStudy (anonymous):

I have a couple of questions about batteries. I guess the V written next to the battery is actually the potential difference across the battery. This potential difference is allowing for the flow of electrons through a circuit. Let's say we have a capacitor, light bulb and battery in a circuit. The potential of the battery doesn't change whether we have the switch open or closed correct?

OpenStudy (anonymous):

Ohhhh.....would the internal resistance of the battery cause a change in the battery's V?

OpenStudy (anonymous):

Why does a battery only have internal resistance after the switch is closed and current is flowing? or does a battery have internal resistance at all times?

OpenStudy (anonymous):

I understand that the internal resistance of a battery increases over time, as the battery ages and eventually the battery becomes useless. But is that internal resistance of the battery obvious even before the switch is closed?

OpenStudy (anonymous):

by obvious I mean, if we were to measure V using some instrument like a DMM

sam (.sam.):

Any batteries will have internal resistance, they used the materials which are not zero resistivity, so that real batteries are not just pure voltage sources, remember that \[V=E-Ir\] where E is the e.m.f of the battery, and r is the resistance

OpenStudy (anonymous):

Is that what is meant by "the voltage drop across resistors is equal to the batteries output when the switch is open" So V=IR when the switch is open or does \(V=E-Ir\) hold true at all times?

OpenStudy (anonymous):

I guess \(V=E-Ir\) true at all times. But what is meant by this statement though? "the voltage drop across resistors is equal to the batteries output when the switch is open"?

sam (.sam.):

When in a closed circuit, then there's voltage drop when current passes though resistors, so that can't be equal to the emf of battery because there's increase in current, ohm's law, with that said, an open circuit with no current flow the voltage drop across resistors have to equal to the emf of battery

OpenStudy (anonymous):

why is there voltage drop across resistors when there is no flow of electrons (when the circuit is open)? Let's say the resistor was a battery.

OpenStudy (anonymous):

gosh I'm sorry about my silly questions, but I'm trying to learn this material :S

sam (.sam.):

That's why I said that thee's no voltage drop when the circuit is open, so the voltage in resistors is equal to e.m.f. of battery

OpenStudy (anonymous):

oh ok, makes sense.

OpenStudy (anonymous):

So, as the capacitor is being charged when the circuit is closed, the Potential across the plates increases until it reaches a maximum.

OpenStudy (anonymous):

(the plates of the capacitor)

sam (.sam.):

yes

OpenStudy (anonymous):

Yes! Thank you Sam! That really helped.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!