Complex voltage and current
calculations
There are circumstances when you may need to
analyze a DC reactive circuit when the starting values of
voltage and current are not respective of a fully
"discharged" state. In other words, the capacitor might
start at a partiallycharged condition instead of starting
at zero volts, and an inductor might start with some amount
of current already through it, instead of zero as we have
been assuming so far. Take this circuit as an example,
starting with the switch open and finishing with the switch
in the closed position:
Since this is an inductive circuit, we'll
start our analysis by determining the start and end values
for current. This step is vitally important when
analyzing inductive circuits, as the starting and ending
voltage can only be known after the current has been
determined! With the switch open (starting condition), there
is a total (series) resistance of 3 Ω, which limits the
final current in the circuit to 5 amps:
So, before the switch is even closed, we
have a current through the inductor of 5 amps, rather than
starting from 0 amps as in the previous inductor example.
With the switch closed (the final condition), the 1 Ω
resistor is shorted across (bypassed), which changes the
circuit's total resistance to 2 Ω. With the switch closed,
the final value for current through the inductor would then
be:
So, the inductor in this circuit has a
starting current of 5 amps and an ending current of 7.5
amps. Since the "timing" will take place during the time
that the switch is closed and R_{2} is shorted past,
we need to calculate our time constant from L_{1}
and R_{1}: 1 Henry divided by 2 Ω, or τ = 1/2
second. With these values, we can calculate what will happen
to the current over time. The voltage across the inductor
will be calculated by multiplying the current by 2 (to
arrive at the voltage across the 2 Ω resistor), then
subtracting that from 15 volts to see what's left. If you
realize that the voltage across the inductor starts at 5
volts (when the switch is first closed) and decays to 0
volts over time, you can also use these figures for
starting/ending values in the general formula and derive the
same results:

 Time  Battery  Inductor 
Current 
(seconds)  voltage  voltage 


 0 
15 V  5 V 
5 A 

 0.1  15 V
 4.094 V  5.453 A 

 0.25  15 V 
3.033 V  5.984 A 

 0.5  15 V
 1.839 V  6.580 A 

 1 
15 V  0.677 V  7.162 A 

 2 
15 V  0.092 V  7.454 A 

 3 
15 V  0.012 V  7.494 A 

