
The self-inductance of a choke coil is $10{\text{ }}mH.$When it is connected with a $10{\text{ }}V$DC source, then the loss of power is $20 W$. When it is connected with $10 V$ AC source loss of power is $10 W$. The frequency of AC source will be-
(A) $80Hz$
(B) $100Hz$
(C) $120Hz$
(D) $220Hz$
Answer
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Hint
Self-inductance is defined as the property of an electric conductor to oppose the change in the electric current flowing in the conductor. It is usually represented by L. Whenever the current in the coil changes the E.M.F is induced in the circuit. Its S.I unit is usually taken as Henry. Power is defined as the rate of doing work.
Complete step by step solution
In DC power loss is only due to resistance we can write
Power loss in DC = $\dfrac{V^2}{R}$
Where $V$ is voltage and $R$ is resistance,
Putting the value-
$R = \dfrac{{\mathop V\nolimits^2 }}{P}$
$R = \dfrac{{\mathop {10}\nolimits^2 }}{{20}} = \dfrac{{100}}{{20}} = 5\Omega $
Now power loss in AC = $\dfrac{{\mathop V\nolimits_{rms}^2 \times R}}{{\mathop Z\nolimits^2 }}$
Where $Z$ is impedance.
Putting the values -
$\mathop Z\nolimits^2 = \dfrac{{\mathop V\nolimits_{rms}^2 \times R}}{P}$
$ = \dfrac{{10 \times 10 \times 5}}{{10}} = 50$
Now, $\mathop Z\nolimits^2 = \mathop {50}\nolimits_{} $
$Z = \sqrt {50} $
$\mathop {Z = R + X}\nolimits_L $ (where XL is reactance)
Squaring both sides
$\mathop {\mathop {50 = 25 + X}\nolimits_L }\nolimits^2 $
$\mathop X\nolimits_L^2 = \mathop {25}\nolimits_{}^{} $
$\mathop X\nolimits_L = \mathop 5\nolimits^{} $
$\mathop X\nolimits_L = \mathop \omega \nolimits_L = \mathop {2\pi f \times L}\nolimits_{} $ ( where f is frequency )
$\mathop {5 = 2\pi f \times 10 \times 10}\nolimits^{ - 3} f$
$ = {\text{ }}80Hz$.
Option (A) is the correct option.
Note
Impedance is usually defined as the opposition to the current in the electric circuit. It is equal to the sum of resistance and reactance. Reactance is the non – resistive component of impedance. Impedance is represented by $\mathop {Z.Z = R + X}\nolimits_L $ ( where $\mathop X\nolimits_L $is reactance and $R$ is resistance).
Self-inductance is defined as the property of an electric conductor to oppose the change in the electric current flowing in the conductor. It is usually represented by L. Whenever the current in the coil changes the E.M.F is induced in the circuit. Its S.I unit is usually taken as Henry. Power is defined as the rate of doing work.
Complete step by step solution
In DC power loss is only due to resistance we can write
Power loss in DC = $\dfrac{V^2}{R}$
Where $V$ is voltage and $R$ is resistance,
Putting the value-
$R = \dfrac{{\mathop V\nolimits^2 }}{P}$
$R = \dfrac{{\mathop {10}\nolimits^2 }}{{20}} = \dfrac{{100}}{{20}} = 5\Omega $
Now power loss in AC = $\dfrac{{\mathop V\nolimits_{rms}^2 \times R}}{{\mathop Z\nolimits^2 }}$
Where $Z$ is impedance.
Putting the values -
$\mathop Z\nolimits^2 = \dfrac{{\mathop V\nolimits_{rms}^2 \times R}}{P}$
$ = \dfrac{{10 \times 10 \times 5}}{{10}} = 50$
Now, $\mathop Z\nolimits^2 = \mathop {50}\nolimits_{} $
$Z = \sqrt {50} $
$\mathop {Z = R + X}\nolimits_L $ (where XL is reactance)
Squaring both sides
$\mathop {\mathop {50 = 25 + X}\nolimits_L }\nolimits^2 $
$\mathop X\nolimits_L^2 = \mathop {25}\nolimits_{}^{} $
$\mathop X\nolimits_L = \mathop 5\nolimits^{} $
$\mathop X\nolimits_L = \mathop \omega \nolimits_L = \mathop {2\pi f \times L}\nolimits_{} $ ( where f is frequency )
$\mathop {5 = 2\pi f \times 10 \times 10}\nolimits^{ - 3} f$
$ = {\text{ }}80Hz$.
Option (A) is the correct option.
Note
Impedance is usually defined as the opposition to the current in the electric circuit. It is equal to the sum of resistance and reactance. Reactance is the non – resistive component of impedance. Impedance is represented by $\mathop {Z.Z = R + X}\nolimits_L $ ( where $\mathop X\nolimits_L $is reactance and $R$ is resistance).
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