There is an opportunity to write the KCL for the time before and the time after the switch is
closed.
Before the switch can is closed, the main power is shut on and we watch the capacitor fully
charge after some time.
The very instant the power (not the switch!) is shut on…The current is at a maximum, as if
the capacitor were simply a wire, (‘ the capacitor acts a s a short circuit…’)
𝑉𝑉
For an instant, 𝐼𝐼𝑡𝑡<0 = 𝐼𝐼0 = 𝑅𝑅𝑠𝑠 , if we use 1V and 1kΩ, this makes 1mA of current.
𝑠𝑠
But we also observe that this voltage begins to change right away!
Since the two components are in series we know that their currents are equal.
𝐼𝐼𝑅𝑅 = 𝐼𝐼𝐶𝐶 →
𝑉𝑉𝑅𝑅
𝑑𝑑𝑉𝑉𝐶𝐶
= 𝐶𝐶
𝑅𝑅𝑠𝑠
𝑑𝑑𝑑𝑑
Also, KVL would deduce that the Source voltage is the sum of the voltages across the loop.
(the left one in this case)
𝑉𝑉𝑠𝑠 = 𝑉𝑉𝑅𝑅𝑠𝑠 + 𝑉𝑉𝐶𝐶 → 𝑉𝑉𝑐𝑐 = 𝑉𝑉𝑠𝑠 − 𝑉𝑉𝑅𝑅𝑠𝑠 →
𝑑𝑑𝑉𝑉𝑅𝑅 𝑠𝑠 𝑉𝑉𝑅𝑅𝑠𝑠
+
=0
𝑑𝑑𝑑𝑑
𝑅𝑅𝑠𝑠 𝐶𝐶
𝑑𝑑𝑉𝑉𝑅𝑅𝑠𝑠
𝑑𝑑𝑉𝑉𝐶𝐶
= −
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Could we swap them around to solve for VC, yep, but this will do for now…
Assume a solution: 𝑉𝑉𝑅𝑅𝑠𝑠 = 𝑎𝑎𝑒𝑒 𝑏𝑏𝑏𝑏
𝑎𝑎𝑎𝑎𝑒𝑒 𝑏𝑏𝑏𝑏 +
Next,
𝑎𝑎𝑒𝑒 𝑏𝑏𝑏𝑏
1
= 0 → 𝑏𝑏 = −
𝑅𝑅𝑠𝑠 𝐶𝐶
𝑅𝑅𝑠𝑠 𝐶𝐶
−𝑡𝑡
−𝑎𝑎𝑒𝑒 𝑅𝑅𝑠𝑠 𝐶𝐶
𝑅𝑅𝑠𝑠 𝐶𝐶
−𝑡𝑡
𝑎𝑎𝑒𝑒 𝑅𝑅𝑠𝑠 𝐶𝐶
+ 𝑅𝑅 𝐶𝐶 = 0 → 𝑎𝑎 − 𝑎𝑎 = 0
𝑠𝑠
O_o
To solve for a, consider the conditions at t=0
𝑉𝑉
For an instant the current pushing through the components is 1ma or I0 = 𝑅𝑅𝑠𝑠 , in general.
−𝑡𝑡
𝑠𝑠
−𝑡𝑡
𝑑𝑑𝑉𝑉𝑅𝑅 𝑠𝑠 𝑉𝑉𝑅𝑅𝑠𝑠
𝑉𝑉𝑠𝑠
−𝑎𝑎𝑒𝑒 𝑅𝑅𝑠𝑠 𝐶𝐶 𝑎𝑎𝑒𝑒 𝑅𝑅𝑠𝑠𝐶𝐶
𝑉𝑉𝑠𝑠
+
=
→
+
=
→ 𝑎𝑎 − 𝑎𝑎 = 𝑉𝑉𝑠𝑠 ; 𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡 = 0
𝑑𝑑𝑑𝑑
𝑅𝑅𝑠𝑠 𝐶𝐶 𝑅𝑅𝑆𝑆 𝐶𝐶
𝑅𝑅𝑠𝑠 𝐶𝐶
𝑅𝑅𝑠𝑠 𝐶𝐶
𝑅𝑅𝑆𝑆 𝐶𝐶
We know, that the instant the power is shut ON, the voltage in the resistor is equal to Vs…
But the math says otherwise?
At this point we need to force the math to reveal what we know is true, so we will use truth
found within this equation to excavate what is happening for this circuit.
Maing a true statement is what we do now…
As long we don’t break the equality we will be fine..
𝑎𝑎 − 𝑎𝑎 = 𝑉𝑉𝑠𝑠 = 2𝑉𝑉𝑠𝑠 − 𝑉𝑉𝑠𝑠
1−1=
𝑉𝑉𝑠𝑠 2𝑉𝑉𝑠𝑠 − 𝑉𝑉𝑠𝑠
=
𝑎𝑎
𝑎𝑎
2𝑉𝑉𝑠𝑠
2𝑉𝑉𝑠𝑠
=
𝑎𝑎
𝑎𝑎
To which the only solution for ‘a’ is that 𝑎𝑎 = 𝑉𝑉𝑠𝑠
Finally, we have the voltage for the resistor
−
−𝑡𝑡
𝑉𝑉𝑅𝑅𝑆𝑆 = 𝑉𝑉𝑠𝑠 𝑒𝑒 𝑅𝑅𝑆𝑆 𝐶𝐶
To solve for the capacitor, plug this into the KVL.
−𝑡𝑡
𝑉𝑉𝑠𝑠 = 𝑉𝑉𝑅𝑅𝑠𝑠 + 𝑉𝑉𝐶𝐶 = 𝑉𝑉𝑠𝑠 𝑒𝑒 𝑅𝑅𝑆𝑆 𝐶𝐶 + 𝑉𝑉𝐶𝐶
−𝑡𝑡
𝑉𝑉𝐶𝐶 = 𝑉𝑉𝑆𝑆 (1 − 𝑒𝑒 𝑅𝑅𝑆𝑆 𝐶𝐶 )
As you can see, when we power
ON the circuit, the resistor voltage
drops as the capacitor voltage
rises. Next let’s see what happens
after we let the capacitor fully
charge and flip the switch.
When we flip the switch the right side of the circuit goes from one state to the next. As seen
above.
The capacitor has fully charged and will act as a ‘voltage-source’ until it depletes
completely.
We can model the current or the voltage, let’s model the current.
Since the components are in series, the current running through them is the same.
𝐶𝐶
𝐼𝐼𝐶𝐶 = 𝐼𝐼𝑅𝑅
𝑑𝑑𝑉𝑉𝐶𝐶 𝑉𝑉𝑅𝑅
=
𝑑𝑑𝑑𝑑
𝑅𝑅
Take note that at the instance the switch is flipped, we have 𝑉𝑉𝑠𝑠 potential for the circuit.
𝑉𝑉
𝑑𝑑𝑉𝑉
𝑉𝑉
Therefore, 𝐼𝐼0𝑅𝑅 = 𝑅𝑅𝑆𝑆 . Using KVL to replace 𝑉𝑉𝐶𝐶 . Leads to; −𝐶𝐶 𝑑𝑑𝑑𝑑𝑅𝑅 = 𝑅𝑅𝑅𝑅
𝐶𝐶
𝑑𝑑𝑉𝑉𝑅𝑅 𝑉𝑉𝑅𝑅
+
=0
𝑑𝑑𝑑𝑑
𝑅𝑅
𝑡𝑡
In similar fashion as before the solution for the resistor voltage is 𝑉𝑉𝑆𝑆 𝑒𝑒 −𝑅𝑅𝑅𝑅
Likewise, the voltage
for the capacitor is
𝑡𝑡
𝑉𝑉𝐶𝐶 = 𝑉𝑉𝑆𝑆 (1 − 𝑒𝑒 −𝑅𝑅𝑅𝑅 )
Modeling the switch
action requires the use
of piecewise functions.
Checkout the
DESMOS! And build the circuit in EveryCircuit.Com to see it in action.
https://www.desmos.com/calculator/bk6ogo3i6i