Question

The equation for the voltage drop accross an inductor is L(dI/dt) and accross a resistance is R*I. We know that a positive voltage drop accros a resistance causes a current. But according to the eq. an increase in current will cause a positive voltage drop across the inductor, how come this doesn't increase the current?

Answer 1

Actually the current does increase. If you solve the differential equation for an inductor so that you can see how current changes with time. you have \[I(t)=I(0)\left( 1- e ^{-\frac{ t }{\tau}}\right) \] where \[\tau = R/L\] where R is the resistance in the circuit since all circuits have some resistance. So you see the current starts off zero because the inductor initially opposes the change in current a self induced voltage across the inductor. This is gradually overcome and the current builds exponentially until it reaches a value governed by Ohms law for the voltage applied and the resistance. Tau (called the time constant) is the time it takes the current to reach about 63% of its max.

Answer 2

Well, so how does it opose the current at the begging, if it generates a positive voltage which will cause an electric field in the same direction the battery was pushing it?

Answer 3

It doesn't. The induced voltage opposes the applied voltage that's why the current starts off low. If you apply 10 volts to a coil with 1 henry inductance and 1 ohm resistance and you measure of voltage across the coil. when you close the switch at that instant the voltage across the coil is zero why not 10 V? Because the induced voltage is opposite the applied voltage. After 1 second (tau=L/R= 1sec.) the voltage across the coil is now ~ 6.3V, after 2 sec. it ~8.6V. after 3 sec V~9.5V. and eventually it will be 10 V as you would expect for a 1 ohm resistor.

Answer 4

Thank you, that was really helpful! And would it make a difference if the resistor was separate from the coil (and the coild had 0 resistance)? I suppose not, so does that mean that the voltage between the coil and resistor in series be 0 at the begging?

Answer 5

It would not make a difference if the resistor is apart from the coil. At the beginning the current is zero , so for a separate resistor the voltage across is is zero to start.