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# Inductance Circuit, Electrostatics, Physics Assignment Help

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Electrostatics - Inductance Circuit, Electrostatics, Physics

**Physics Assignment Help >> Electrostatics >> Inductance Circuit**

Let a source of alternating be connected to a circuit containing a pure inductance L only suppose the alternating supplied is represented by

**E = E**_{0} sin wt

If si/st is the rebate of change of current through L at any instant then induced in the inductor at the same instant **is - L s**_{i}/s_{t}. the negative sign indicates that induced opposes the change of current.

Usually inductors have some resistance in their windings. By we shall assume that resistance of this inductor is negligible. The circuit is therefore a purely inductive circuit. To maintain the flow of current in this circuit applied voltage the flow of current in this circuit applied voltage must be equal and opposite to the induced voltage.

E = - (- L dI/dt) = E_{0} sin wt

**Or dI = E**_{0} / L sin wt dt

Integrating both sides we get

**I = E**_{0} / L ∫sin w t dt

**I = E**_{0}/L (- cos wt/w) + constant

The integration constant has the dimensions of current, and is time-independent. As of the source oscillates symmetrically about zero, the current it sustains must also oscillate symmetrically about zero. Therefore no component of current which is time indecent exists. Therefore integration constant is zero.

**∴ form I = (- E**_{0}/wL) cos wt

I = (- E_{0}/wL) [sin (π/2)] – wt

I= E_{0} / wL sin [wt – (π/2)

The current will be maximum** I = Iw**. When **sin (wt = π/2) = maximum = 1**.

From above eqn, **I0 = (E**_{0} /wL) × 1

Putting in (eq.), we get

I = I_{0} sin [wt – (π/2)]

This is the form of alternating current developed.

Comparing with we find that in an circuit containing L only alternating current I lags behind the alternating voltage E by a phase Only of 90° by one fourth of a period.

Conversely voltage across L leads the current by a (c) phase angle of 90° this shown in

Figure (b) represents the vector diagram or the phasor diagram of circuit containing L only. The vector representing **E**_{0} makes an angle** (wt)** with **OX**, As current lags behind the by **90°** therefore, phasor representing Io is turned clock wise through **90°** from the direction of **E**_{0} the projections of these phasor on** YOY** give instantaneous values E and I as shown then **E**_{0} and Io rotate with frequency w curves in (c) are generated.

Comparing eqn. with Ohm’s law equation viz. **current = voltage / resistance** we find that **(wL)** represents the effective resistance offered by inductance L. this is called inductive reactance and is denoted by **X**,.

Thus ,** XL = wL = 2π vL**, where **v **is the frequency of supply.

The inductive reactance limits the current in a purely inductive circuit in the same way as resistance limits the current in a purely resistance circuit. Clearly the inductive reactance **(XL)** is directly proportional to the inductance** (L)** and to the frequency **(v) **of the current.

In dc circuit **v = 0 ∴ XL = 0**

A pure inductance offers zero resistance to dc. It means a pure inductor cannot reduce dc. The units of inductive reactance

**XL = w L » 1 / sec (Henry) = 1 / sec volt amp/ sec = ohm.**

Hence XL is measured in ohm just as resistance is measured in ohm. The dimensions of inductive reactance are the same as those of resistance.

It showily be clearly understood, that in a circuit containing, R opposition arises on account of obstruction to the passage of electrons through are resistor. In an inductive circuit it is the self induced that opposes the growth of current.

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**Popular Tags:** Inductance Circuit Assignment Help, Inductance Circuit Homework Help, Inductance Circuit Tutors, Inductance Circuit Solutions, Inductance Circuit Tutors, Electrostatics Help, Physics Tutors, Inductance Circuit Questions Answers

**Physics Assignment Help >> Electrostatics >> Inductance Circuit**

**E = E**

_{0}sin wtIf si/st is the rebate of change of current through L at any instant then induced in the inductor at the same instant

**is - L s**

_{i}/s_{t}. the negative sign indicates that induced opposes the change of current.

Usually inductors have some resistance in their windings. By we shall assume that resistance of this inductor is negligible. The circuit is therefore a purely inductive circuit. To maintain the flow of current in this circuit applied voltage the flow of current in this circuit applied voltage must be equal and opposite to the induced voltage.

E = - (- L dI/dt) = E

E = - (- L dI/dt) = E

_{0}sin wt**Or dI = E**

_{0}/ L sin wt dtIntegrating both sides we get

**I = E**

_{0}/ L ∫sin w t dt**I = E**

_{0}/L (- cos wt/w) + constantThe integration constant has the dimensions of current, and is time-independent. As of the source oscillates symmetrically about zero, the current it sustains must also oscillate symmetrically about zero. Therefore no component of current which is time indecent exists. Therefore integration constant is zero.

**∴ form I = (- E**

I = (- E

I= E

_{0}/wL) cos wtI = (- E

_{0}/wL) [sin (π/2)] – wtI= E

_{0}/ wL sin [wt – (π/2)The current will be maximum

**I = Iw**. When

**sin (wt = π/2) = maximum = 1**.

From above eqn,

**I0 = (E**

_{0}/wL) × 1Putting in (eq.), we get

I = I

I = I

_{0}sin [wt – (π/2)]This is the form of alternating current developed.

Comparing with we find that in an circuit containing L only alternating current I lags behind the alternating voltage E by a phase Only of 90° by one fourth of a period.

Conversely voltage across L leads the current by a (c) phase angle of 90° this shown in

Figure (b) represents the vector diagram or the phasor diagram of circuit containing L only. The vector representing

**E**

_{0}makes an angle

**(wt)**with

**OX**, As current lags behind the by

**90°**therefore, phasor representing Io is turned clock wise through

**90°**from the direction of

**E**

_{0}the projections of these phasor on

**YOY**give instantaneous values E and I as shown then

**E**

_{0}and Io rotate with frequency w curves in (c) are generated.

Comparing eqn. with Ohm’s law equation viz.

**current = voltage / resistance**we find that

**(wL)**represents the effective resistance offered by inductance L. this is called inductive reactance and is denoted by

**X**,.

Thus ,

**XL = wL = 2π vL**, where

**v**is the frequency of supply.

The inductive reactance limits the current in a purely inductive circuit in the same way as resistance limits the current in a purely resistance circuit. Clearly the inductive reactance

**(XL)**is directly proportional to the inductance

**(L)**and to the frequency

**(v)**of the current.

In dc circuit

**v = 0 ∴ XL = 0**

A pure inductance offers zero resistance to dc. It means a pure inductor cannot reduce dc. The units of inductive reactance

**XL = w L » 1 / sec (Henry) = 1 / sec volt amp/ sec = ohm.**

Hence XL is measured in ohm just as resistance is measured in ohm. The dimensions of inductive reactance are the same as those of resistance.

It showily be clearly understood, that in a circuit containing, R opposition arises on account of obstruction to the passage of electrons through are resistor. In an inductive circuit it is the self induced that opposes the growth of current.

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**Inductance Circuit Assignment Help, Inductance Circuit Homework Help, Inductance Circuit Tutors, Inductance Circuit Solutions, Inductance Circuit Tutors, Electrostatics Help, Physics Tutors, Inductance Circuit Questions Answers**

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