Step 1: Understanding the Square Wave Input

The input voltage  is a square wave defined as:

This pattern repeats every T seconds.

Step 2: Charging Phase (  )

At t=0, the source switches to +1V, and the capacitor begins charging from its initial voltage  (where  ).

The charging equation for the capacitor voltage is:

Here:

•  = +1V (target voltage).
•   (starting voltage).

Substituting these values:

At , the capacitor voltage reaches +V (due to steady-state symmetry):

Step 3: Solving for V

Rearranging the equation:

For high RC(  ) , the exponential term can be approximated using the Taylor series:

Substituting this approximation:

Simplifying:

Neglecting the small term  :

The exact solution is:

For , so .

Step 4: Discharging Phase

At , the source switches to , and the capacitor begins discharging toward  from .

The discharging equation is:

At , the capacitor voltage reaches , completing the cycle.

Final Expression for Capacitor Voltage

For  :

For  :

Conclusion

The capacitor voltage  oscillates between  and , where:

This is because the capacitor cannot fully charge/discharge due to the high RC time constant. The waveform resembles a clipped exponential curve, oscillating below the input square wave amplitude.

For t<0 sec,

Under steady-state,

For ,

= 25V

Under steady state (when capacitor is fully charged and current becomes zero),

Let  time takes to charge the capacitor to 50 V.

Current source → OC

Since the switch was closed for a long time before t = 0 circuit is in steady state at

∴ L is short circuited & C is open circuited

At t = 0+ S is opened & inductor is replaced by a current source & capacitor is replaced by a voltage source.

By KCL, current in capacitor branch = 4A

By KVL, 6R1 + VL = 4R2 + 12

VL = 4 × 2 – 6 × 2 + 12 = 8 volt

In steady state capacitor acts as an open circuit for DC supply.

The voltage across 6 Ω is

At t = 0-

For t > 0

At t = ∞

For 1st half cycle, V0 = 5 – VC

For 2nd half cycle, V0 = -Vc

Vp-p = (5 – Vc min) – (-Vc max)

6.2 = 5 + Vc max – Vc min

Vc max – Vc min = 1.2 ………..(α)

For first half cycle i.e.

………(i)

………(ii)

 ……….(iii)

From (ii),

Using equation (iii),

As

And

If initial charge polarities on the capacitor is opposite to the supply voltage then only the capacitor voltage crosses the zero line.

Vc(t) ⇒  Final value + (Initial value - Final value)

Now,

% change in voltage

at                 

For maximum transient,  

Initial capacitor voltage,

Due to discharge capacitor

0.1msec

0.46msec

Voltage across capacitor =

Inductor current

The locus of this equation is a circle centered at (5,0)

The circuit enters steady state at .

So inductor is shorted

After opening the switch, Req across inductor

For source free RL circuit

The given circuit can be redrawn as.

The network of resistors is a balanced bridge, so the network is reduced to

Current in RC circuit will be =

Power supplied by source =

When \(S_1\) is open and \(S_2\) is closed, the circuit enters steady state so Inductor will be shorted. The system is shown below,

Current through inductor

When switch is closed at then inductor

current at that instant will not change i.e.

It will be 1.5 A; it can be replaced by a current source at that instant.

Let V→ voltage at node A = voltage across inductor

KCL at A,

The current in an sourced RL circuit is given by,

Switch across current source is opened at t = 0

Since, Voltage across capacitor cannot change instantaneously

For t > 0

 =  

So, curve is a straight line passing through origin as shown.

When the switch is connected to position ‘1’ for a long time, the circuit enters steady state,

Now when the switch is moved to position ‘2’ the steady state voltage is,

If we reverse the polarity of capacitor voltage, we get

Impedance of capacitor =  

Impedance of inductor = s

Series equivalent =  

This equivalent is connected in parallel with capacitor. Assuming equivalent of parallel combination is Z1

Driving Point Impedance,  

For , the switch is shorted

Since capacitor behaves as an open circuit

for steady state

Since voltage across a capacitor doesn’t change suddenly

For

For t < 0, the circuit configuration is shown below,

Since voltage across capacitor doesn’t change immediately

At , the circuit configuration is shown below,

Change stored in  before t=0

Since, there is no resistor in the circuit, the charge shall remain conserved.

So, charge stored after t=0 in the combination of capacitor

Since voltage across both capacitor will be identical as both are in parallel

From equation (1) & (2)

For finding time constant, current source is opened and finding equivalent resistance & equivalent capacitance. The process of simplification is shown below,

Time constant

At , \(SW_1\) is OPEN and \(SW_2\) is CLOSED

Applying KCL at

Since no switching action takes place, so the circuit is in steady state.

In steady state capacitor acts as open circuit while inductor acts as short circuit.

So redrawing the circuit:

On applying KCL at node A,

Applying KVL in 2nd loop

Current in

Current in voltage source=4A

Voltage across capacitor will discharge through inductor up to voltage across capacitor becomes zero. In this period electrostatic energy stored in capacitor will transfer to electromagnetic energy in inductor. After that the inductor starts charging the capacitor and magnetic energy of inductor is converted to electrostatic energy again.

Voltage across capacitor=V(s)=Voltage across inductor

Let

Taking inverse Laplace

At steady state capacitor acts as open circuited

At steady state circuit is shown in the figure,

From potential divider rule

Before t=0, switch is opened and circuits at steady state

So inductor will be short circuited

Since current through inductor doesn’t change instantly

For t<0, the switch is connected in Position 1 as shown below,

Since the circuit is at steady state the inductor behaves

as short circuit

Current at t<0,

Current doesn’t change abruptly in inductor

At  , the inductor behaves as constant current source.

Voltage across coil at =Voltage across  

For t > 0, the circuit in Laplace Domain can be represented as shown below,

Inductor Voltage Source,

Initial stored energy in inductor

If 95% energy is dissipated in Inductor only 5% energy remains

Remaining energy in inductor

Before t=0

Since voltage across capacitor doesn’t change abruptly

After t=0

At steady state,

Capacitor behaves as open circuit

At t=1sec

[Peak]

Before t=0

Circuit was not energized

So  and

Since current through inductor and voltage across capacitor doesn’t change abruptly

 and

i.e., Inductor will behave as open circuit and capacitor will behaves as short circuit

At

Current from source at ,

Voltage across inductor at

Since voltage pulse is applied to . So capacitor will change up to . So capacitor voltage will be at peak and then the capacitor will discharge upto

Before applying voltage pulse,

Equivalent Resistance across Capacitor Terminals,

At steady state capacitor acts as open circuit, so voltage can be determined by Potential Divider.

Capacitor voltage as a function of time is,

;  

This equation has been written assuming a constant 11V source at the terminals but this equation is only valid upto 10μs till the source is connected and after that discharging RC circuit will be used.        

For peak voltage

At steady state inductors behave as short circuit while capacitors will behave as open circuit

Circuit diagram at steady state is shown below,

Voltage across  resistor

By potential divider rule,

When the switch is closed for a long time the entire current flows through the short circuit and hence initial inductor current,

When the switch is opened current gets divided into

resistance and inductance,

         ---------(1)

Since, both R and L are connected in parallel,

From (1)

Solving the differential as a linear DE, we get

The steady state current in Inductor is,

Hence, energy stored after a long time or in Steady State,

Note: This circuit can also be simplified by using source transformation and converting the current source to a voltage source and then standard equation of charging RL circuit can be applied.

The circuit before closing the switch and under steady state is shown below,

Before t = 0,

Current in an Inductor is continuous,

At steady state inductor will be short circuited.        

The current as a function of ‘t’ is given by,

Hence current is exponential = option (A) & (D) are wrong.

Since at  current through resistor inductor combination is zero.

So voltage across resistor at = 0 x R

Since current at steady state is and is non – zero.

So the energy stored at steady state in non – zero.

Hence option (c) is not correct.

Current cannot change instantaneously in an Inductor.

For t > 0

Since,

c = 0

The rectangular pulse would be applied to series RC circuit as shown below,

During the interval 0 < t < T

Supply Voltage v(t) = V

The equation of capacitor voltage can be determined on the basis of standard charging RC circuit.

At t = T,

For the interval t > T, v(t) = 0

So, the circuit behaves as discharging RC Circuit.

Thus, the voltage exponentially decays after t=T and hence maximum voltage occurs at t = T.

The given circuit configuration can be represented as,

Current through voltage source, l =

Various waveforms for the above circuit are shown below,

The network given is shown below,

Initially, capacitor is charged to 12V

Since, there is no active device to supply energy in the circuit.

For equivalent resistance across capacitor, the capacitor terminals are opened.

Since

i(t) = constant DC current = step current

where u(t) is unit step function.

Voltage across ideal inductor =

 = Ideal Impulse

The given system is shown below,

L = 2H                        

(a) at t<0:- Since switch is connected to 12V for long time. Hence, the circuits enters steady state. So ‘L’ behaves as short circuit while capacitor behaves as open circuit.

This condition is shown below,

(b) For t>0, the switch is moved to other position

Using KVL

Putting

(c) At steady state

Inductors becomes short circuited and capacitor becomes open circuited.

Converting the circuit into laplace domain        

Using KVL,

By Partial Fraction,

Solving for A, B & C, we get   A = 1.25, B = -2.5, C = 2.5

Taking Inverse Laplace Transform

at t = 4sec,

At t < 0 : switch is open so the capacitor is uncharged.

Since, voltage across capacitor cannot change instantaneously.

Just after t = 0, current in capacitor

Applying KVL at

Putting the value of  &

Time constant of RC Network is =

Here, equivalent resistance is Thevenin resistance

across capacitor terminals,        

To find , deactivate voltage source,

Time Constant,

Assuming time constants of the two coils are,

Coil 1,

Coil 2,

When both coils are connected in series

Time constant =

Given,

=average of time constants of individual coils

The circuit in Laplace Domain is shown below,

Steady state value,

 is voltage across capacitor

Steady state value

As polarity is not mentioned in question, only magnitude will be the answer.