I am trying to graph the equation VbackEMF = L di/dt
When SW1 is off and has been off since negative infinity, I realize that there is no current in the system.
Now I imagine SW1 being pressed. The switch can probably make a good connection in 0.1 seconds, and current in the circuit without the inductor is 1A.
So here is what I am stuck on. Back EMF = 1 Volt at the first instant.
Well, how does the current ever start moving through the loop if the first instant doesn't allow any current through?
simulate this circuit – Schematic created using CircuitLab
Best Answer
When the series current is zero, the voltage across L1 must equal V1 (by KVL and Ohm's Law).
But, for an (ideal) inductor, we have:
\$v_L = L \dfrac{di_L}{dt}\$
Thus, by KVL and the definition of an ideal inductor, at the moment SW1 closes, the time rate of change of current is:
\$\dfrac{di_L}{dt} = \dfrac{1V}{1H} = 1 \dfrac{A}{sec} \$
So, the crucial insight here is this: there is no current at the moment the switch closes but, at that very moment, the current begins to change.
By solving the differential equation that describes the circuit. By KVL and Ohm's Law, we have:
\$v_L = L \dfrac{di_L}{dt} = v_1 - i_L R \rightarrow \dfrac{di_L}{dt} + \dfrac{R}{L}i_L = v_1\$
This is an easy 1st order ordinary differential equation for the series current \$i_L\$.
The solution, for zero initial condition, is:
\$i_L(t) = \dfrac{v_1}{R}(1 - e^{-\frac{t}{\tau}}) \$
Where
\$\tau = \dfrac{L}{R} \$
When t is "small enough", i.e., right after the switch closes, we have:
\$i_L(t) \approx \dfrac{v_1}{L}t \$
So, in the early moments, the resistance has negligible effect and the current is approximately a ramp. The current begins to significantly deviate from a ramp only after the current becomes large enough such that the voltage drop across the resistor is significant compared to the voltage source.