If the current for a closed switch is increasing exponentially (inversely) across an inductor, what is the voltage doing (decreasing the time constant multiple of an inverse exponential?)
e L(di/dt) write the current equation in exponential form, and take the derivative. +++++
Your question is a bit vague. If current is increasing exponentially then the relationship governing the current is not an inverse exponential. The voltage developed across the inductor is given by: v L (di/dt) So that if the current is exponentially increasing then so is the voltage.
If the inductor is a real inductor, with series resistance, not a superconducting or mathematically ideal, then a constant voltage connected across it initially has the current limited completely by the inductance. The formula for the relation between inductor current and voltage is that the rate of change of current is proportional to the voltage across the inductor. So, at the beginning, the current ramps linearly up from zero after voltage is connected across the inductor. But as soon as there is current, there is also voltage lost across the internal resistance in the wire that makes up the inductor. So as the current rises, there is a rising loss of voltage that remains across the inductance, since some of the total voltage is lost across the resistance of the inductor. This falling effective voltage across the inductance causes the rate of rise of the current to fall the current rises at a lower and lower rate.) Eventually, the current will have risen high enough that the resistance uses up all of the voltage with zero left to drive a current increase across the inductance and the current reaches an ultimate value of V/R amps. The series resistor inductor combination therefore has a current time constant under constant total voltage. When the voltage has been on for L/R seconds (with L being the inductance in henries and R, the resistance in ohms), the current will have risen to 62% of the ultimate (steady state current. In another L/R seconds, the current will reach 62% of what is left of the way to that ultimate current, etc. The link I included, below has the actual exponential formula for this rising current, that is difficult to write with text only. -- Regards, John Popelish
