Let  be the Load pf angle.

Applying nodal analysis to super node

∴ v(t) = 12.5 sin (1000t)

At t = 5 msec v(t) = 12.5 sin 5 = -11.986

Before t = t1 Kcs open & is closed & circuit reaches steady state

The current of 5V dc source keeps circulating in a loop & does not affect capacitor.

at t = t1,

after t = t1 k closed so ac current source is shorted &  is open so a series RC circuit with 5V dc source is formed.

Initial capacitor voltage

Steady state capacitor voltage = 5V

Second term represents transient

For transient-free response

When a coil a capacitor are connected in series then overall Q-factor is

⇒ 

⇒ Xc = 10 Ω

= 20 μF

For maximum power transfer

I = 1 ∠ 10° + 1 ∠ 70°

I = 1.732 ∠ 40°

The reading of ammeter is 1.732 A.

For finding power, constant current to sin

In this problem  leads both  & hence current will flow from  towards

Hence,

Reactance of capacitor

Parallel combination of R & C

Total impedance

Since current & voltage have same phase. Imaginary part of impedance should be zero

Hence,

For calculation of power, same frequency terms must be used in both voltage & current.

Assume equivalent impedance = Z

Total instantaneous power

The given system is shown below with two parallel LC branches marked as A and B.

In branch A:

Resulting parallel combination

In branch B:

Resultant parallel combination

Hence, both branches A and B will act as Open Circuit.

Now, equivalent circuit will be:

Peak value of current through

For admittance to have only conductance, imaginary part of admittance should be zero.

For resonance at all frequency is variable, so,

Note: For the circuit to be independent of frequency the time constant of both branches should be equal.

Let,

is the voltage across resistance of coil

is the voltage across inductance of coil

Since,

Option A:

Given,

                       …………….(1)

Applying KVL in loop,

        ………….…(2)

Putting the values of in equation (2)

Putting this value in equation (1)

This means voltage across inductance = 0

This is not possible and hence option is incorrect.

Option B,

Not possible so option B is also wrong.

Option C:

Putting these values in eq (2)

Voltage across resistance can’t be negative so option C is also wrong.

Option D:

Putting in eqn (2)

This is only possible when coil is ideal but in question coil is not ideal, so this is also wrong.

Note: There is something wrong in the question and marks to all were given in this question by the GATE committee.

Current have several frequency components but voltage has only one frequency component. Average power will be due to only same frequency components.

W

Equivalent Impedance of Parallel RC circuit shown below is,

When  (inductive)

When  (capacitive)

At , impedance goes to infinity as the net admittance goes to zero. Hence, the plot of impedance versus frequency is shown below,

This problem can be solved by superposition principle.

Solving with DC source alone by short circuiting the AC source:

In steady state, inductor is short – circuited,

Solving with AC source alone by short circuiting the DC Source:

=1,

v(t) = 12 sin t

By superposition theorem,

RMS value of i(t)

The currents in various branches of the circuit is shown,

By KVL,

Assuming the supply voltage is represented by V (peak)

Fourier series of square wave:

 ms

 rad/sec

So, we are interested in fundamental.

Since , circuit is in resonance. Hence, fundamental current can be obtained as,

Using super position theorem,

First, open circuit current source

By Potential Division,

 =   =

Therefore,  ;  

Now, short circuit voltage source

By Current Division,

Hence, =1.98 , = 78.69°

Thus, the capacitor voltage is given by,  

Power loss in 1Ω =  

Real Power is only consumed in resistances. Hence,

Power loss in R = 1 kW – 4W = 996W

Assuming no current goes into voltmeter i.e. ideal voltmeter,  

R = 9.96Ω

,

Given,  and  

Since both the current & voltage are in phase, it is resonance condition. Under resonance condition,

Net Impedance, Z = R

R == 50 Ω

Resonance frequency, =

   -----------(1)        

At = 1krad/s, current is leading by an angle of 31°    

So, net impedance is capacitive.

= 0.6

= 50 × 0.6 = 30

From (1)

Since,

C = 25 μF

From (1), L = 10mH

The configuration of the given system is shown below,

Since, the current is coming out of the positive terminal of load the load is said to be delivering power.

Complex power delivered by the load is =

Both P & Q are negative means load absorbs active power as well as reactive power from the source.

For  disconnect the load as shown below,

For Coil 1,

Similarly for Coil 2,          

When connected in series,

Equivalent inductance =

Equivalent resistance =

quality factor =

This implies that the bridge circuit is in balanced condition.        

So points A & C will be at same potential.

Due to this 1Ω resistor and j1Ω are connected in parallel.

So applying current division rule at node B.

The current source can be represented in phasor form as,

Here, we are considering maximum value as the

magnitude of phasor.

Here, current is leading the voltage as the impedance has negative imaginary part which means that the impedance is capacitive.

Average Power,

Finding thevenin’s equivalent circuit of circuit  - B.

Open circuiting the terminals of circuit B.

Short the voltage source to determine the Thevenin Resistance.

Representing the circuit B by its Thevenin Equivalent.

                        ----------(1)

Power transferred from A to B

For P to be maximum

Since  is the voltage across a resistor so, it will be in same phase as current but voltage across RL load will lead the voltage by power factor angle.

The circuit and phasor diagram are shown below,

By Phasor Sum,

Given,

Power consumption in load impedance is equal to

The circuit with each element represented as Impedance as,

1F capacitor

1H inductor

Inductor and Capacitor cancel the impedance of each other. So,

the upper branch act as Short Circuit.

RMS value of current

Taking V as reference phase

Since, current lags the voltage by an angle  radians

Note: The magnitude of phasor is always rms value of the quantity.

It is given that, ,  

The current in RL branch lags the applied voltage by 45°

Power dissipated in R resistance

P=1W

Since         

Let us applying a voltage source at input side as shown below,

Applying KVL in input loop

Input impedance

Negative imaginary part of impedance shows capacitive reactance

Since impedance has non-zero real part

So it has resistance

The impedance is having negative imaginary part. So it has overall reactance as capacitive. It may be purely capacitive or may have both capacitor and inductor having value of capacitive reactance greater than inductive reactance.

But for minimal realization of impedance a resistor and a capacitor will realize the Thevenin impedance.

Representing each element in terms of its impedance, the circuit can be redrawn as shown below,

Impedance of parallel branch,

For resonant frequency imaginary of impedance should zero

Admittance of the circuit

At

Real part of admittance

Real part will always be positive

At

 [Maximum value]

Imaginary part of admittance

At

For

For

So the locus of admittance looks like as shown above.

Reactance of

Applying KVL

When         

When         

Hence, the locus looks like as shown above.

For Open the XY terminals as shown,

Open circuit the current source

Various resistances and impedances have been marked in the circuit shown below,

At resonance,         

I is in phase with

At resonance

 is also in phase with  and less than

Hence the phasor diagram is,

RMS of 3 volts=3

RMS of

RMS of

Let inductive reactance at 100Hz is

Since voltage drop across R & L is equal at 100Hz

Magnitude

RMS voltage across

Now at 50Hz

Inductive reactance=        

Current at 50Hz

RMS voltage drop across R at 50Hz=

Now from equation (4) & (2)

For parallel resonance to be happen, imaginary part of admittance of parallel network should be zero.

Since inductor is already exists. So, Z should be capacitive

Admittance of parallel network, Y=

=

For imaginary part of admittance of zero

Load impedance

Where  &

Balanced  voltage supply

Current in the balanced  load,

Similarly,

Instantaneous power in  load ‘p’

Constant

Total admittance of parallel combination

Current in resistor        

Current in capacitor

Using KCL at node ‘0’

Voltage across inductor=

For thevenin impedance short circuiting the voltage source

Given,

By using KVL in circuit

In a series RLC circuit, the inductor and capacitor voltage are out

of phase.

Hence,

So the circuit is at resonance

Hence

Quality factor of circuit

Since

Admittance of capacitor                

Admittance of combination

Total admittance of parallel combination

For unity power factor at source end

Imaginary part of admittance should be zero.

Resonance frequency

Hence

For transient free response

Power factor angle

This is based on the concept that if we switch ON at the time corresponding to lagging angle between voltage and current then current starts from 0 and it follows steady state current so there is no transient.

If we switch ON when the steady state current is non-zero then transients are induced.

To determine the Thevenin Voltage, the load terminals must be opened as shown below,

Applying KVL in loop

To determine Thevenin Impedance, both voltage sources must be shorted as shown below,

Thevenin equivalent circuit:

= 0A

Current in series RLC circuit is,

Under Resonance Condition,

         (in phase with voltage source)

Voltage across capacitor

if ; then = so  can be greater than voltage source and  is  phase behind the source voltage.

Real Power will only be absorbed by Load and not by Capacitor.

Assuming currents and voltages on both sides the system is shown below

Since,         ---------(1)

                ---------(2)

Since output side is short circuited, so

                        ---------(3)

From equation 1 & 3

When

Resonance Frequency

Quality factor

The given system is shown below,

Referring load impedance

‘’ to primary side

Equivalent circuit becomes as shown below,

Impedance seen by source,

To determine Thevenin equivalent across the 10Ω resistor, the resistor must be open circuited as shown below,

By voltage divider rule

Zth across  resistor, the voltage source is short circuited,

So, Thevenin equivalent circuit is shown below,

Since,

but voltage across capacitor are opposite in phase.

So, net effect of capacitor and inductor cancels out. Circuit is under resonance.

The arrangement in the question is shown below,

Admittance         

When R = 0

When R = 0.01

= 50 S

When         

So the locus of Y looks like as shown below,

It is a semi-circle of diameter j100 and center at zero.

Given voltage source,

Supply Current,

Complex power

Current

Since ‘X’ is fixed and ‘R’ is variable

When R = 0;

When R = X;

When ;

So the locus of I will ok like as shown below

So the locus of I is a semicircle with diameter

Note: In such type of questions, the process followed is similar to drawing Polar Plot in Control Systems.

Let initially resistance, capacitance & inductance is R, C & L respectively and resonant frequency is .

Now, new resistance, capacitance & inductance

R’ = 2R

C’ = 2C

L’ = 2L

New resonant frequency

Hence,

So, resonance frequency is halved.

Quality Factor,

Voltage across inductor at resonance= 20 x 10= 200Volt

Power consumed by boiler =

where V = voltage across boiler =

Here,

Note that 230V is rms value of the supply voltage

R = resistance of Boiler circuit coil

Where

Hence frequency of P is 100Hz

Note: The instantaneous power in a single phase AC Circuit oscillates at twice the supply frequency.

Since Q- factor of series RL circuit =

When frequency is doubted,

Hence, Q is doubled.

Active power,

Current in a RL Series Circuit is given by,

Since,         ,

Since coil is having high quality factor. So,         

So P decreased by 4 times.

Since resonant frequency,

Quality factor of series RLC circuit =

The given network is,

Applying KCL at node ‘’

Current through