A S R

+
Vs - L

𝑑𝑖
𝑉𝑠 = 𝑖𝑅 + 𝐿
𝑑𝑡
𝑑𝑖
0 = 𝑖𝑅 + 𝐿
𝑑𝑡
=− i

Root S=−

𝑖 (𝑡) = 𝐴𝑒

𝑖 (𝑡) = 𝐴𝑒
𝑉𝑠
𝐼𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑎𝑐𝑡𝑠 𝑎𝑠 𝑠ℎ𝑜𝑟𝑡 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑎𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑖 (𝑡) =
𝑅
𝑖 (𝑡) = 𝑖 (𝑡) + 𝑖 (𝑡)
𝑉𝑠
𝑖 (𝑡) = + 𝐴𝑒
𝑅
Calculating Value of constant “A” by using initial condition at t=0 sec
×
𝑖(0) = + 𝐴𝑒 =0 𝐴=−

𝑖(𝑡) = 1−𝑒 1

Voltage across Inductor 𝑉

𝑑𝑖
𝑉 =𝐿
𝑑𝑡
 𝑉 = 𝑉𝑠𝑒

Initial current at t=0 is 10A then

𝑉𝑠 −𝑅×0
𝑖(0) = + 𝐴𝑒 𝐿 = 10
𝑅
𝑉𝑠
𝑖(0) = + 𝐴 = 10
𝑅

𝑉𝑠
𝐴 = 10 −
𝑅
𝑉𝑠
𝑖(𝑡) = 1−𝑒 + 10𝑒
𝑅

Voltage across Inductor 𝑉

𝑑𝑖
𝑉 =𝐿
𝑑𝑡
10𝐿
𝑉 = 𝑉𝑒 + 𝑒
𝑅

In input is Sinusoidal wave input applied voltage Vrms = Vs then

𝑉𝑠
𝑖(𝑡) = < −𝜃
𝑅 + (𝜔𝐿)

√2 × 𝑉𝑠𝑥𝑠𝑖𝑛(𝜔𝑡 − 𝜃)
=
𝑅 + (𝜔𝐿)

√2 × 𝑉𝑠𝑥𝑠𝑖𝑛(𝜔𝑡 − 𝜃) 𝑅
− 𝑡
+ 𝐴𝑒 𝐿 = 0
𝑅 + (𝜔𝐿)

√2 × 𝑉𝑠𝑥𝑠𝑖𝑛(−𝜃)
− =𝐴
𝑅 + (𝜔𝐿)

√2 × 𝑉𝑠𝑥𝑠𝑖𝑛(𝜔𝑡 − 𝜃) √2 × 𝑉𝑠𝑥𝑠𝑖𝑛(−𝜃)
𝑖 (𝑡) = − 𝑒
𝑅 + (𝜔𝐿) 𝑅 + (𝜔𝐿)
 A S R

B
+
Vs L
-

𝑑𝑖
0 = 𝑖𝑅 + 𝐿
𝑑𝑡

=− i

S=−

𝑖(𝑡) = 𝐴𝑒
𝑖(𝑡) =0

𝑖(𝑡) = 𝐴𝑒
𝑉𝑠 ×
𝑖(0) = = 𝐴𝑒
𝑅
𝑉𝑠
𝑖(0) = =𝐴
𝑅
𝑉𝑠
𝑖(𝑡) = 𝑒
𝑅
𝑑𝑖
𝑉 =𝐿
𝑑𝑡

𝑉 = −𝑉𝑠𝑒

RC Circuits
 A S R

+ B
Vs - C

Switch A is closed
𝑉𝑠 = 𝑖𝑅 + ∫ ∝ 𝑖𝑑𝑡 always add initial value ∫ ∝ 𝑖𝑑𝑡 of voltage in integrator

1 1
𝑉𝑠 = 𝑖𝑅 + 𝑖𝑑𝑡 + 𝑖𝑑𝑡
𝐶 ∝ 𝐶

𝑉𝑐(𝑡 = 0) = ∫ ∝ 𝑖𝑑𝑡 = 0V

1
𝑉𝑠 = 𝑖𝑅 + 𝑖𝑑𝑡
𝐶
By taking derivative at both sides to make it differential Equation
𝑑𝑖
0 = 𝑖/𝐶 + 𝑅
𝑑𝑡
=− i

Root S =−

𝑖(𝑡) = 𝐴𝑒
Initial current at t=0 is 𝑉𝑠/𝑅 then

𝑖(𝑡) (𝑡 = 0) = 𝐴𝑒 = 𝑉𝑠/𝑅
𝑖(𝑡) =0

𝑖(𝑡) = 𝑒

Voltage across Inductor 𝑉 (𝑡)

 𝑉𝑐(𝑡) = ∫ 𝑖𝑑𝑡 = Vs(1-𝑒 )

A S R

Vs=0 B C

Switch is connected with point B

1
0 = 𝑖𝑅 + 𝑖𝑑𝑡
𝐶 ∝

1 1
0 = 𝑖𝑅 + 𝑖𝑑𝑡 + 𝑖𝑑𝑡
𝐶 ∝ 𝐶

𝑉𝑐(𝑡 = 0) = ∫ ∝
𝑖𝑑𝑡 = Vs

0 = 𝑖𝑅 + 𝑉𝑠 + ∫ 𝑖𝑑𝑡

By taking derivative at both sides to make it differential Equation

𝑑𝑖
0 = 𝑖/𝐶 + 𝑅
𝑑𝑡

=− i

S=−

𝑖(𝑡) = 𝐴𝑒
𝑖(𝑡) =0

Initial current at t=0 is −𝑉𝑠/𝑅 then

𝑖(𝑡) (𝑡 = 0) = 𝐴𝑒 = −𝑉𝑠/𝑅

𝑉𝑠
𝑖(𝑡) = − 𝑒
𝑅
 1 0 1 𝑡
𝑉𝑐 (𝑡) = 𝑖𝑑𝑡 + 𝑖𝑑𝑡
𝐶 −∝ 𝐶 0
1 𝑡
𝑉𝑐 = 𝑉𝑠 + ∫0 𝑖𝑑𝑡 = -Vs(1-𝑒 )+Vs
𝐶

= 𝑉𝑠 𝑒

A S L

+ B
Vs - C

𝑖
0 = 𝐿𝑑 /𝑑𝑡 +
𝐶
𝑆 = −1/𝐿𝐶
𝑆1,2 = ± 𝑗/𝐿𝐶

𝑖 (𝑡) = 𝐴𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑠𝑖𝑛𝜔𝑡

I(0) = 𝐴𝑐𝑜𝑠𝜔0 + 𝐵𝑠𝑖𝑛𝜔0 = 0
A=0

𝑖 (𝑡) = 𝐵𝑠𝑖𝑛𝜔𝑡
𝑑 𝑑 (𝐵𝑠𝑖𝑛𝜔𝑡 )
𝑖 (𝑡) = = 𝜔𝐵𝑐𝑜𝑠𝜔𝑡
𝑑𝑡 𝑑𝑡
𝑉𝑠
𝜔𝐵𝑐𝑜𝑠𝜔0 =
𝐿
𝑉𝑠
𝐵=
𝐿𝜔
 𝑉𝑠
𝑖 (𝑡) = 𝑠𝑖𝑛𝜔𝑡
𝐿𝜔
1 𝑉𝑠
𝑉𝑐 (𝑡) = 𝑠𝑖𝑛𝜔𝑡 + 𝑉𝑐 (𝑡 = 0)
𝐶 𝐿𝜔
1 𝑉𝑠
𝑉𝑐 (𝑡) = 𝑠𝑖𝑛𝜔𝑡 + 𝑉𝑐 (𝑡 = 0)
𝐶 𝐿𝜔
𝑉𝑠 1
𝑉𝑐 (𝑡) = |−𝑐𝑜𝑠𝜔𝑡| 0 𝑡𝑜 𝑡
𝐿𝜔 𝜔𝐶

Current passing through inductor

Vs ,L
1
𝑖 (𝑡) = 𝑉𝑠 𝑑𝑡 + 𝐼(𝑡 = 0)
𝐿
𝑉𝑠
𝑖 (𝑡) = 𝑡
𝐿
Inductor is short circuited

𝑖 (𝑡) = ∫ 0 𝑑𝑡 + 𝐼(𝑡 = 0)=10A

𝑉 (𝑡) = 𝐿 =𝐿 = -α

Equations for defining RC and L

V = IR, V𝜶𝑰
Vs, C
Ic = cdv/dt, Ic 𝜶 dv/dt
 Vs, L
VL = L di/dt, VL 𝜶 di/dt