Simple DC circuits with R, L & C components | Electricity & Magnetism - Physics PDF Download

Circuit Analysis by Classical Method

Whenever a circuit is switched from one condition to another, either by a change in the applied source or a change in the circuit elements, there is a transition period during which the branch currents and element voltages change from their former values to new ones. This period is called the transient. After the transient is passed, the circuit is said to be in the steady state. Now, the linear differential equation that describes the circuit will have two parts to its solution. The complementary function corresponds to the transient state and the particular solution corresponds to the steady state. The v-i relation for an inductor or capacitor is a differential equation. A Circuit containing an inductor L or a capacitor C and resistor R will have current and voltage variables given by differential equations of the same form. It is a linear first order differential equation with constant coefficients when values of R , L and C are constant.  L and C are storage element. Circuits have two storage elements like one L and one C are referred to as second order circuits.  

The circuit changes are assumed to occur at time t = 0 and represented by a switch.

t = 0⁻ ; the instant prior to t = 0 and 

t = 0⁺ ; the instant immediately after switching

Switching on or off an element or source in a circuit at t = 0 will not disturb the storage element so that i_L(0^-) = i_L(0⁺) and  v_c(0^-) = v_c(0⁺)

Circuit element: Resistance

Resistance R = V/I

Power absorbed by the resistor P=I²R = V²/R

and energy lost in the resistance in form of heat is

Circuit element: Inductance

Voltage across inductance
In a pure inductive circuit with applied voltage v,

Power absorbed by inductor  

Circuit element: Capacitance 

Differential Equations 

Type I-(First order Homogeneous Differential Equation)

  where P is any constant

 take k ' =ln k ⇒ ln y(t) = −Pt + ln k

⇒ y(t) = ke^−Pt  where k is a constant

Type II-(First order Non Homogeneous Differential Equation)

  where P is a constant and Q may be a function of independent variable 

t or a constant.

If Q is constant, then
 

Initial Conditions in Circuits

Number of initial conditions required is equal to the order of the differential equation for an unique solution. 

Transient Response of Series R-L Circuit having DC Excitation

If switch S is closed at t = 0 . Then the current through the circuit is i (t ) . Applying KVL around the loop, we will get

Now L behaves as open circuit (O.C.) at switching,

Thus

New Transient Condition

As circuit reaches at steady state (at t = ∞ ), suddenly switch is open and new condition     at t = 0 is shown.

Now

at t = 0⁺ , the inductor keep the steady state

Thus

Corresponding voltages across the resistor and inductor are

Transient Response of Series R-C Circuit having DC Excitation

If switch S is closed at t = 0 . Then the current through the circuit is i (t ) . Applying KVL around the loop, we will get  

Initially capacitor was uncharged (0⁻) = 0

Differentiating above equation, we will get

At

New Transient Condition

As circuit reaches at steady state (at t = ∞ ), suddenly switch is open and new condition     t = 0 is shown.  

Now

Transient Response of Series RLC Circuit having DC Excitation

Consider series RLC circuit VS = 2V , L =2H , R = 6 Ω and C = 0.25F . At t = 0 switch S is closed then in the transient state find current i(t ).

 Applying KVL,

⇒ p² + 3p + 2 = 0 ⇒ p₁ = −1, p₂ = − 2

⇒ i(t)  = k₁e^−t + k₂e⁻² ………………… (3) 

At t = 0 , i (0⁺) =0 (inductor is open circuited)  
⇒ k₁ + k₂ = 0

From (1)


Thus
k₁ = 1, k₂ = -1
⇒ i(t)= e^−t − e^−2t
and From (2)

Transient Response of Parallel RLC Circuit having DC Excitation

Consider Parallel RLC circuit I₀ = 2A, L = 1/16H, R = 1/16 Ω and C = 4F . At t = 0 switch  is closed then in the transient state find voltage v(t ).

 Applying KCL,