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Wattless Current | What Is Wattless Current

Wattless Current | What Is Wattless Current :- The electric current that does not consume electrical power in a circuit is called wattless current or idle current.

We know that in an AC circuit, the average power is equal to the product of the RMS voltage (E_rms), the RMS current (I_rms), and the cosine of the phase angle (cos ϕ) between the voltage and the current; that is,

$\displaystyle {{P}_{avg.}}={{E}_{rms}}{{I}_{rms}}cos\phi$

Suppose that in an alternating current circuit, the rms voltage (E_rms) leads the rms current (I_rms) by a phase angle ϕ, as shown in the figure below :-

In the above figure, I_rms has been divided into two components :

I_rmscos ϕ which is in phase with E_rms and

I_rmssin ϕ which is perpendicular to E_rms

Now, since the angle between E_rms and I_rmscos ϕ is 0°, therefore, the average power consumed in one complete cycle due to I_rmscos ϕ is,

P_avg. = (E_rms) × (I_rmscos ϕ) × (cos 0°) = E_rmsI_rmscos ϕ

Similarly, since the angle between E_rms and I_rmssin ϕ is 90°, therefore the average power spent in one complete cycle due to I_rmssin ϕ is :

P_avg.‘ = (E_rms) × (I_rmssin ϕ) × (cos 90°) = 0

Therefore, the current component I_rmssin ϕ does not consume electric power in an AC circuit. For this reason, this component of the current is called idle current or wattless current.

Example 1.

(NCERT Example 7.7)

(a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.

(b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

Solution :

(a) We know that in an AC circuit the average power is given by P_avg. = E_rms I_rms cos ϕ. Now if we want to supply a fixed power (P_avg.) at a constant voltage (E_rms), then for a low power factor (cos ϕ) we must increase the current. However, this will cause greater power loss (I²R) during transmission.

(b) Suppose in an electric circuit the current I_rms lags behind the voltage E_rms by a phase angle ϕ; then the power factor of this circuit is,

$\displaystyle cos\phi =\frac{R}{Z}$

According to the above figure, the current can be divided into two components: I_rms cos ϕ (the power component, which is in phase with the voltage E_rms) and I_rms sin ϕ (the wattless component, which lags behind the voltage E_rms by 90°).

Now, if we want to increase the power factor, we must allow a current to flow in the opposite direction of the wattless component (I_rms sin ϕ). Since in a capacitor the current leads the voltage by 90°, the wattless component I_rms sin ϕ can be cancelled by using a capacitor of suitable value that provides the current I′. In this way, the power factor can be improved.

Example 2.

(NCERT Example 7.8)

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω , L = 25.48 mH, and C = 796 μF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.

Solution :

(a) Impedance of the circuit (Z)

$\displaystyle Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}=\sqrt{{{R}^{2}}+{{\left( 2\pi \nu L-\frac{1}{2\pi \nu C} \right)}^{2}}}$

$\displaystyle \Rightarrow Z=\sqrt{{{3}^{2}}+{{\left( 2\pi \times 50\times 25.48\times {{10}^{-3}}-\frac{1}{2\pi \times 50\times 796\times {{10}^{-6}}} \right)}^{2}}}$

$\displaystyle \Rightarrow Z=\sqrt{9+{{\left( 8-4 \right)}^{2}}}=5\Omega$

(b) Phase difference between voltage and current,

$\displaystyle \phi =ta{{n}^{-1}}\left( \frac{{{X}_{L}}-{{X}_{C}}}{R} \right)=ta{{n}^{-1}}\left( \frac{8-4}{3} \right)=ta{{n}^{-1}}\left( \frac{4}{3} \right)$

$\displaystyle \Rightarrow \phi ={{53}^{\circ }}$

That is, the current lags the voltage by a phase angle of 90°.

$\displaystyle P={{\left( {{I}_{rms}} \right)}^{2}}R={{\left( \frac{{{I}_{0}}}{\sqrt{2}} \right)}^{2}}R={{\left( \frac{{{E}_{0}}}{\sqrt{2}Z} \right)}^{2}}R$

$\displaystyle \Rightarrow P=\frac{283\times 283}{2\times 5\times 5}\times 3=4805.3Watt$

(d) Power factor

$\displaystyle cos\phi =cos53=0.6$

Example 3.

(NCERT Example 7.9)

Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs ? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.

Solution :

(a) Resonance frequency

$\displaystyle {{\nu }_{r}}=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\times 3.14\times \sqrt{25.48\times {{10}^{-3}}\times 796\times {{10}^{-6}}}}=\text{35}\text{.4}Hz$

(b) At resonance, impedance (Z) = resistance (R) so

Z = R = 3Ω

$\displaystyle {{I}_{rms}}=\frac{{{E}_{rms}}}{Z}=\frac{{{E}_{0}}}{\sqrt{2}Z}=\frac{{{E}_{0}}}{\sqrt{2}R}=\left( \frac{283}{\sqrt{2}} \right)\times \frac{1}{3}=66.7A$

Power dissipated at resonance,

$\displaystyle P={{\left( {{I}_{rms}} \right)}^{2}}R={{\left( 66.7 \right)}^{2}}\times 3=13.35kW$

Here we can see that, in this case at resonance, the power loss is greater than the power loss that occurred in Example 7.8.

Example 4.

(NCERT Example 7.10)

At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ?

Solution :

The metal detector at an airport operates on the principle of electromagnetic induction.

It works as follows:

(1).Generation of Electromagnetic Waves : The metal detector has a transmitter coil that produces a varying electromagnetic wave.

(2). Inducing Eddy Currents : If a person carrying a metallic object passes through the detector, the varying electromagnetic field generated by the transmitter coil induces eddy currents in the metallic object.

(3). Secondary Magnetic Field : The eddy currents generate their own secondary magnetic field, which affects the primary field produced by the transmitter coil.

(4). Detection and Alarm : A receiver coil in the metal detector senses these changes in the magnetic field. The altered signal is processed, and if metal is detected, the system triggers an alarm.

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