Formulas and Rules
Capacitor
Capacitors in Series vs Parallel
Series
−1
Equivalent
Parallel
𝐶𝑒𝑞 = 𝐶1 + 𝐶2 + ⋯
Current
1
1
+ +⋯)
𝐶1 𝐶2
(𝑡)
𝑖𝑒𝑞
= 𝑖1 (𝑡) = 𝑖2 (𝑡)
𝑖𝑒𝑞 (𝑡) = 𝑖1 (𝑡) + 𝑖2 (𝑡)
Voltage
𝑉 = 𝑉1 + 𝑉2
𝑉 = 𝑉1 = 𝑉2
𝐶𝑒𝑞 = (
Current of Capacitor 𝒊(𝒕)
𝑑𝑣
𝑖𝑐 (𝑡) = 𝐶 ( 𝑑𝑡 )
Case 1: Solve for i(t)
Given: Diagram of 𝑖(𝑡) source, cap and resis. in parallel
(Series is just all the currents of the elements equal to each other)
1) Convert to base units. Label nodes, GND Caps and Resis.
2) Use KCL (same rules).
Resis. still uses Ohms the only difference is voltage is now an equation
wrt 𝑡.
3) Simplify
Capacity Voltage-Current Relationship
Capacitor Voltage Formula
𝑡
𝑣𝐶 (𝑡) = 𝐶 ∫𝑎 𝑖(𝜏)𝑑𝜏 − 𝑣(0) ∗ 𝐵𝑎𝑠𝑒 𝑢𝑛𝑖𝑡 ∗ Piecewise
1
Let 𝒂 = defined lower bound of time interval
𝒕 = undefined upper bound, 𝒗(𝟎) = voltage from previous interval.
Case 1: 𝑖𝑠 − 𝑡 graph (Piecewise)
1) Convert to base units.
**See DGD 8
𝑖(𝑡1 ) , 𝑎 < 𝑡 < 𝑏
𝑖𝑐 (𝑡) = { 𝑖(𝑡2 ) , 𝑏 < 𝑡 < 𝑐
…
2) Look at graph and list:
‼ Don' t forget base unit ‼
1
3) Set aside 𝐶 ∗𝑏𝑎𝑠𝑒 𝑢𝑛𝑖𝑡∗ × ∗ 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑓𝑜𝑟 𝑏𝑎𝑠𝑒 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑖(𝑡) ∗
(optional)
4) Start with first interval (given). Sub all known into general form and solve eq. should have variable 𝒕.
[2]
Look list and state which interval you are cal.
{ 𝑖(𝑡) =∗ 𝐵𝑎𝑠𝑒 𝑢𝑛𝑖𝑡 ∗ ___ , 𝑎 ∗ 𝐵𝑎𝑠𝑒 𝑢𝑛𝑖𝑡 ∗< 𝜏 < 𝑏 ∗ 𝐵𝑎𝑠𝑒 𝑢𝑛𝑖𝑡 ∗ }.
5) Sub in upper bound value 𝒃 ∗ 𝐵𝑎𝑠𝑒 𝑢𝑛𝑖𝑡 ∗ of that interval as the 𝑡 in [2] to
get "𝑣(0)" for next interval.
6) Repeat steps 2 and 3 until all intervals are complete.
Current
Current Divider Rule
𝑅
2
𝑖1 = 𝑖𝑡𝑜𝑡𝑎𝑙 (𝑅 +𝑅
)
1
2
Parallel resistors.
Ratio must be < 1 or else the R values must be reanalyzed.
Current wrt time graph (Capacitor)
**See Capacity Voltage-Current Relationship
Elements
Two elements are in:
Series: if shared only ONE common node between them. (No other current paths
(junctions) leading from that node to other parts of the circuit st same current must
flow through both components.
Parallel: if share two or more common nodes (one at each end) st they have exact
same voltage across both components.
Frequency (w/ L,C,R circuits – only AC circuits)
!! SET CALCULATOR TO DEGREES !!
**See DGD 8 and 9
**CASIO fx-991EX:
*Menu* > *2* “Complex”
Calculator can handle mixed forms as one equation but:
Syntax:
𝑖 comes after value ; × between brackets;
best to do it in chunks.
Phasor:
When writing, put rectangular form before phasor
where possible.
If neg. on X in gen. form. of phasor,
make pos. by adding 180˚ to angle.
or
covert reg. form as positive by the flipping all signs;
or
if a factor then put the neg. on the reg. form like normal alg.
If forms are different:
Must keep forms separate when solving and inputting.
Polar form rules are different then rect. Form.
Can’t mix and match place holders without thinking.
If different forms but same denom. then can split sum of num.
Can: (Both Forms) Collect, change sides and factor out.
ONLY Rectangular form: Follows same rules as ℝ algebra BUT it’s
actually ℂ st cannot + or – phasors.
Note: Phasor and Polar forms contain complex numbers ( j2 = −1).
𝑎 of both can be scaled w/ the base. Ie: mAmp  𝑎 × 10−3 gives Amp.
Ohm’s (freq domain)
𝑽(𝝎) = 𝑍 𝑰(𝝎)
Kirchhoff Same but in phasor form.
Impedance 𝛀 (time ⇔ frequency domain – only AC circuits)
Hint: Sometimes helpful to use a placeholder until the final equation has
been found.
Resistor 𝑣(𝑡) = 𝑅𝑖(𝑡) → 𝑉(𝜔) = 𝑅 𝑰(𝝎):
Inductor
𝑑𝑖
𝑍𝑟 = 𝑅
𝑣(𝑡) = 𝐿 𝑑𝑡 → 𝑉(𝜔) = 𝑗𝜔𝐿 𝑰(𝝎):
𝑍𝐿 = 𝑗𝜔𝐿
Capacitor 𝑖(𝑡) = 𝐶 𝑑𝑡 → 𝐼(𝜔) = 𝑗𝜔𝐿 𝑽(𝝎):
𝑍𝐶 = 𝑗𝜔𝐶
𝑑𝑣
1
Series: 𝑍𝑒𝑞 = 𝑍1 + 𝑍2 + ⋯
1
1
1
2
Parallel: 𝑍𝑒𝑞 = 𝑍 + 𝑍 + ⋯
Voltage Divider (freq domain)
𝑍
2
𝑉2 = 𝑉𝑖𝑛 (𝑍 +𝑍
)
1
2
**Also see “Voltage Divider” (general)
Only if two impedances are in series.
Current Divider
𝑍
2
𝐼1 = 𝐼𝑖𝑛 (𝑍 +𝑍
)
1
2
Steady-State: Sinusoidal ⇔ Phasors (General formula for conversion)
𝑥(𝑡) = 𝑋 cos(𝜔𝑡 + 𝜙) → 𝑋(𝜔) = 𝑋∠𝜙𝑋
Only when all sources are in cos w/ same 𝜔 (angular freq.)
𝑋𝑠𝑖𝑛(𝜔𝑡 + 𝜙) = 𝑋𝑐𝑜𝑠(𝜔𝑡 + 𝜙 − 90°) → 𝑋(𝜔) = 𝑋∠(𝜙 − 90°)
Steady-State: Current Source: Sinusoidal ⇔ Phasors
𝑖(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜙𝑖 ) → 𝐼(𝜔) = 𝐴∠𝜙𝑖
(conversion)
Steady-State: Voltage Source: Sinusoidal ⇔ Phasors
𝑣(𝑡) = 𝐵 cos(𝜔𝑡 + 𝜙𝑉 ) → 𝑉(𝜔) = 𝐵∠𝜙𝑉
(conversion)
Phasors ⟹ Polar ⟹ Euler/Trig ⟹ Rectangular
(conversion to add Phasors)
𝑋 = |𝑋| ∠𝜃 ⟹ 𝑋 = |𝑋|(𝑐𝑜𝑠𝜃 + 𝑗𝑠𝑖𝑛𝜃) ⟹ 𝑋 = 𝑎 + 𝑗𝑏
Where 𝑎 = |𝑋|(𝑐𝑜𝑠𝜃)
;
𝑗𝑏 = 𝑗|𝑋|(𝑠𝑖𝑛𝜃)
Casio: Converting a phasor fxn
Enter X value in phasor form > *Shift* **ENG* to get "∠" > add value for phi >
*Equal*
Casio: To add Rectangular
Group ℝ > + > *ENG* to get 𝑖 > Group VALUES of ℂ
Rectangular ⟹ Polar (conversion)
𝑋 = 𝑎 + 𝑗𝑏 ⟹ 𝑋 = |𝑋| ∠𝜙
𝑏
Using: |𝑋| = √𝑎2 + 𝑏 2
,
𝜙 = tan−1 (𝑎) 𝑎 = 𝑥 − 𝑎𝑥𝑖𝑠 , 𝑏 = 𝑦 − 𝑎𝑥𝑖𝑠. Angle
must match quad.
Casio: Rect form to polar
Input rect. Form like normal (no brackets needed) > *OPTN* > *down arrow* > *1*
for 𝑟∠𝜃 > *=*