Alternating Current Test - 38

Answer : A

$$ switch\quad in\quad position\quad 2:\\ { I }_{ max }=\dfrac { 150 }{ \left| L\omega -\dfrac { 1 }{ C\omega  }  \right|  } =3\quad \Rightarrow \left| L\omega -\dfrac { 1 }{ C\omega  }  \right| =50\\ open\quad switch\quad condition\quad :\\ \hat { Z } =R+j(L\omega -\dfrac { 1 }{ C\omega  } )=Z\angle -60\quad \Rightarrow \dfrac { { R }^{ 2 } }{ { R }^{ 2 }+{ \left| L\omega -\dfrac { 1 }{ C\omega  }  \right|  }^{ 2 } } ={ cos }^{ 2 }(-60)\quad \Rightarrow \dfrac { { R }^{ 2 } }{ { R }^{ 2 }+{ 50 }^{ 2 } } =\left( \dfrac { 1 }{ 2 }  \right) ^{ 2 }\Rightarrow R=\dfrac { 50 }{ \sqrt { 3 }  } A\\  $$ 

In ideal choke coil all voltage drop occur at  inductor. Resistance is neglected in case of ideal choke coil. so there is only inductance.
Let inductance of choke coil $$= L$$
given value of AC voltage $$(V) = 125\ volt$$,  frequency $$(f) = 50\ Hz$$
Current in inductor  $$=\dfrac { V }{ \omega L }$$
$${\omega L} = $$Resistance due to inductor,
$$\omega $$ = $$2\pi f$$
Current in inductor  $$=\dfrac { V }{ \omega L } = \dfrac { V }{ 2\pi fL } = \dfrac { 125 }{ 2\pi \times 50\times L } $$
Value of current is given in question$$ =10\ A$$.
$$\dfrac { 125 }{ 2\pi \times 50\times L } = 10\Longrightarrow \quad L= \dfrac { 125 }{ 1000\pi  } \ H$$

Let  the value of resistance is $$R$$
So in pure resistance case
Current  $$=\dfrac { V }{ R }$$ , current given $$= 12.5\ A$$
$$\dfrac { 125 }{ R } = 12.5 \Longrightarrow \ R =10\ \Omega $$

Now this $$R\ \&\ L$$ is connected in series to another AC source of $$V = 100\sqrt { 2 }$$ volt,
 frequency $$(f) = 40\ Hz$$
so impedance of combination $$(Z)  = \sqrt { { R }^{ 2 }+{ \omega L }^{ 2 } } = \sqrt { { R }^{ 2 }+{ 2\pi fL }^{ 2 } }\ \Omega $$
$$Z=\sqrt { { 10 }^{ 2 }+\left( { 2\pi \times 40\times \dfrac { 125 }{ 1000\pi  }  } \right) ^{ 2 } }$$ , putting value of $$R,L,$$ and new $$f$$.
$$Z=10\sqrt { 2 } \ \Omega $$
Current in new combination $$= \dfrac { V }{ Z }= \dfrac { 100\sqrt { 2 }  }{ 10\sqrt { 2 }  } = 10\ A$$
So, A is correct Answer.

The current in inductor and capacitor is always at an phase difference of $$180^o$$
for $$V=V_osin\omega t$$
Capacitor, $$i=i_osin(\omega t  -  \dfrac{\pi}{2})$$
Inductor, $$i=i_osin(\omega t  +  \dfrac{\pi}{2})$$
So, the current from both branches will be $$0.9 + (-0.4) = 0.5A$$

$$ \omega =2 \pi f =100 \pi $$ ;  $$ f= 50Hz$$

Let the same reading be $$x V $$,
then let, $${ Z }_{ total }=R+j({ X }_{ L }-{ X }_{ C })=Z\angle \theta \\ I=\dfrac { { V }_{ S } }{ Z } \angle -\theta \\ therefore,\quad { V }_{ R }=\left| \dfrac { { V }_{ S } }{ Z } R\angle -\theta  \right| \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ L }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ L }\angle 90-\theta  \right| \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ C }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ C }\angle -90-\theta  \right| \\ given,\quad { V }_{ R }={ V }_{ L }={ V }_{ V }=x\\ \quad \Rightarrow \quad R={ X }_{ L }={ X }_{ C }\\ \Rightarrow { Z }_{ total }=R=Z\angle 0=Z\quad \quad \\ \Rightarrow \quad \quad { V }_{ R }=\left| \dfrac { { V }_{ S } }{ Z } R\angle -\theta  \right| ={ V }_{ S }\\ \quad \quad \quad \quad { V }_{ L }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ L }\angle 90-\theta  \right| =\left| j{ V }_{ S } \right| ={ V }_{ S }\\ \quad \quad \quad \quad { V }_{ C }=\left| \dfrac { { V }_{ S } }{ Z } { X }_{ C }\angle -90-\theta  \right|=\left| -j{ V }_{ S } \right| ={ V }_{ S }\\ \Rightarrow x={ V }_{ S }=100V $$

The impedance inductor and capacitor are $$2\pi fL\quad \& \quad 1/2\pi fC$$ respectively, since both of them have the quantity 'f' in their expression the change in frequency will effect their impedance and hence the current in the circuit, so the circuit only with Resistance can be independent of frequency in terms of current.

$$\omega =\dfrac { 1 }{ \sqrt { LC }  } $$
For, $$\omega = constant$$ $$\rightarrow \sqrt { LC } = constant$$
Therefore, $$C$$ is inversely proportional to $$L$$
So, if $$C$$ is made $$4C$$ then $$L$$ should be reduced to $$L/4$$ to keep $$\omega$$ constant.

Using Ohm's law

We know that in an L-C-R circuit impedance is given by,
$$Z=\sqrt { { R }^{ 2 }+{ ({ X }_{ L }-{ X }_{ C }) }^{ 2 } } $$
When $${ X }_{ L }=\omega L$$ is less than $${ X }_{ C }=1/\omega C$$
With increase in $$\omega$$ impedance decrease and become minimum when $${ X }_{ L }={ X }_{ C }$$
After that $${ X }_{ L }=\omega L$$ will be greater than $${ X }_{ C }=1/\omega C$$

$$L=10\ \text{mH}={10}^{-2}\ \text{H}$$
$$f=1\ \text{MHz}={10}^{6}\ \text{Hz}$$
$$f=\cfrac {1}{2\pi \sqrt {LC}} $$
$$f^2=\cfrac {1}{ 4 \pi ^2LC } $$
$$\Rightarrow C=\dfrac { 1 }{ 4\pi ^2 f ^2 L } =\dfrac { 1 }{ 4\times

10\times 10^{12}\times 10^{-2}} =\dfrac {10^{- 11}}{4} =2.5\ \text{pF}$$