Maximum Power Transfer Theorem

A choke coil is an inductance coil of very small resistance used for controlling current in an a.c. circuit. If a resistance is used to control current, there is wastage of power due to Joule heating effect in the resistance. On the other hand there is no dissipation of power when a current flows through a pure inductor.
 

Construction

It consists of a large number of turns of insulated copper wire wound over a soft iron core. A laminated core is used to minimise eddy current loss as shown in figure.

Working

The inductive reactance offered by the coil is given by

X_L = ωL

In the case of an ideal inductor the current lags behind the emf

by a phase angle π/2 .

Thus,the average power consumed by the choke coil over a complete cycle is

P_av = E_rms I_rms cos π/2 = 0

However in practice, a choke coil of inductance L possesses a small resistance r. Hence it may be treated as a series combination of an inductor and small resistance r. In that case the average power consumed by the choke coil over a complete cycle is

P_av = E_rmsI_rmscos φ

P_av = E_rmsI_rms [r/(√r²+ω²L²)]        …... (1)

Here [r/(√r²+ω²L²)]  is the power factor. From equation (1), the value of average power dissipated works out to be much smaller than the power loss I²R in a resistance R.

Chokes used in low frequency a.c. circuit have an iron core so that the inductance may be high. These chokes are known as audio – frequency (A.F) chokes. For radio frequencies, air chokes are used since a low inductance is sufficient. These are called Radio Frequency (R. F) or High Frequency (H.F) chokes and are used in wireless receiver circuits (Fig. 4.24a and Fig. 4.24b).

Choke coils can be commonly seen in fluorescent tubes which work on alternating currents.

Problem (JEE Advanced)

A resistance is connected to a capacitor in AC and the phase difference is π/4 between current and voltage. When the same resistance is connected to an inductor, phase difference becomes tan^-1(2). Power factor of the circuit when both capacitor and inductor are connected to the resistance will be,

(a) 1

(b) 1/√2

(c) 1/√3

(d) ½

Solution:

We know that, tan π/4 = X_L/R

So, X_L = R

tan^-1 (2) = X_C/R

Or, X_C/R = 2

So, X_C = 2R

Now, in L-C-R circuit

Z = √[R²+(X_C – X_L)²] =√2R

Thus, power factor = cos θ = R/Z = 1/√2

Therefore, from the above observation we conclude that, option (b) is correct.

Problem:

A copper wire having a cross-sectional area of 0.5 mm² and length of 0.1 m is initially at 25^oC and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire,

(a) find the time in which the wire starts melting. [The change of resistance of the wire with temperature may be neglected.]

(b) What will this time be, if the length of the wire is doubled?

[Density of Cu = 9 × 10³ kg m^–3, specific heat of Cu = 9 × 10^–2 cal kg^–1 oC^–1, M.P. (Cu) = 1075 ^oC and specific resistance =

1.6 × 10^–8 Wm].

Solution:

(a) Mass of Cu = Volume × density = 0.5 × 10^–6 × 0.1 × 9 ×

10³ = 45 × 10^–5 kg.

Rise in temperature = q = 1075 – 25 = 1050 ^oC. Specific

heat = 9 × 10^–2 kg^–1 oC × 4.2 J

Þ l²Rt = mSq

Þ t = mSq/l².R

But R = pL / A = 1.6 × 10^–8 × 0.1 / 0.5 × 10^–8 = 0.558 s

(b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, the wire will start melting in the same time.

Question 1