हिंदी

A Voltage V = V0 Sin ωt is Applied to a Series LCR Circuit. Derive the Expression for the Average Power Dissipated Over a Cycle. - Physics

Advertisements
Advertisements

प्रश्न

A voltage V = V0 sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition (i) no power is dissipated even though the current flows through the circuit, (ii) maximum power is dissipated in the circuit?

Advertisements

उत्तर

Voltage V=V0sinωt is applied to an series LCR circuit.

Current is  π2" role="presentation" style="position: relative;">π2

`I_0=V_0/Z`

`phi=tan^(-1)((X_C-X_L)/R)`

Instantaneous power supplied by the source is

P=VI

=(V0sinωt)×(I0sin(ωt+ϕ)

`=(V_0I_0)/2[cosphi-cos(2omegat+phi)]`

The average power over a cycle is average of the two terms on the R.H.S of the above equation. The second term is time dependent; so, its average is zero.

 `P=(V_0I_0)/2cosphi`

`=(V_0I_0)/(sqrt2sqr2)cosphi`

 =VIcosϕ

P=I2Zcosϕ

cosϕ  is called the power factor.

Case I

For pure inductive circuit or pure capacitive circuit, the phase difference between current and voltage is `pi/2`

`:.phi=pi/2,cosphi=0`

Therefore, no power is dissipated. This current is sometimes referred to as wattless current.

Case II

For power dissipated at resonance in an LCR circuit,

`X_C-X_L=0, phi=0`

∴ cos ϕ = 1

So, maximum power is dissipated.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2013-2014 (March) All India Set 2

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

In a series LCR circuit connected to an a.c. source of voltage v = vmsinωt, use phasor diagram to derive an expression for the current in the circuit. Hence, obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum


Why does current in a steady state not flow in a capacitor connected across a battery? However momentary current does flow during charging or discharging of the capacitor. Explain. 


(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase : current or voltage ?

(ii) Without making any other change, find the value of the additional capacitor C1, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.


A coil of resistance 40 Ω is connected across a 4.0 V battery. 0.10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil.


An LR circuit with emf ε is connected at t = 0. (a) Find the charge Q which flows through the battery during 0 to t. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time t. (e) Verify that the results in the three parts above are consistent with energy conservation.


Answer the following question.
Draw the diagram of a device that is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.  


In series LCR circuit, the phase angle between supply voltage and current is ______.


Define Impedance.


A series LCR circuit containing a resistance of 120 Ω has angular resonance frequency 4 × 105 rad s-1. At resonance the voltage across resistance and inductance are 60 V and 40 V respectively. At what frequency the current in the circuit lags the voltage by 45°. Give answer in ______ × 105 rad s-1.


Select the most appropriate option with regard to resonance in a series LCR circuit.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×