# Voltage across inductance

## Main Question or Discussion Point

ok, lets say we have a battery connected to a resistor and an inductance in series

-|battery|+______ 00000(inductor)______resistance(R)

if the rate of change of current is decreasing , then the induced EMF in the inductor would be towards the right? because they would want to strengthened and thus oppose the change.

so since the induced emf is towards the right, why is the potential at the right higher? shouldn't the left be higher? so that the induced emf is from higher(left) to lower(right) ?

thanks

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I not sure I understand the description you give. But I think you can answer your own question this way:

At t = 0-, the voltage across the inductor and the resistor is zero....right?? when a switch is closed and the battery connected, at t = infinity what is sign the voltage across the resistor....plus to battery plus right??....

so now you can tell how the voltage must have been changing across the inductor....increasing on the right.....in other words, part of the voltage drop during the period between zero and infinity is across the inductor and part across the resistor...totalling the battery voltage...so voltage on the inductor is plus on the right.

At t = 0-, the voltage across the inductor and the resistor is zero....right?? when a switch is closed and the battery connected, at t = infinity what is sign the voltage across the resistor....plus to battery plus right??....
er, i am not sure what you mean by resistor plus to battery plus?

i think it should be -ve end of battery connected to the right side of the resistor. the left of the resistor is connected to the inductor

so basically, i rephrase my question,
-if current is decreasing,
-then an induced emf will be set up in the inductor to oppose the change in the drop of current
-so, it would act in a direction towards the right to strengthen the decreasing current
-with higher potential at left side of inductor and lower potential at right side of inductor?
-so shouldn't the left side of the inductor be at a higher potential? and right side lower, so that the EMF will be pointing towards the right?

i quote the following abstract from masteringphysics.com

"In sum: when an inductor is in a circuit and the current is changing, the changing magnetic field in the inductor produces an electric field. This field opposes the change in current, but at the same time deposits charge, producing yet another electric field. The net effect of these electric fields is that the current changes, but not abruptly. The "direction of the EMF" refers to the direction of the first, induced, electric field."

ok , so lets say current is increasing, di/dt >0 ,

-battery+ _____ (left)inductor(right)

ok, so since current is towards right, -->, and increasing, an induced emf will be set up on the inductor to oppose this change, by the magnetic field due to the changing current, so it will be directed to the left <--- , so shouldn't the (right) side of the inductor be at a higer potential compared to the (left) side of the inductor? since the direction of EMF refers to the direction of the first , induced, electric field?