AC Circuits: Series LR Circuits

Impedance Z = sqrt(R^2 + (2pif*L)^2). R = 10, L = 2, f = 60. XL = 2 * 3.14 * 60 * 2 = 753.6. Z = sqrt(100 + 567913) approx 754. Current I = V/Z = 120 / 754 approx 0.16 A.

Given R = 100. When C is removed (LR circuit), tan(60) = XL/R => XL = 100 * sqrt(3) = 173.2. When L is removed (RC circuit), tan(60) = XC/R => XC = 100 * sqrt(3) = 173.2. Since XL = XC, the circuit is at resonance when all elements are connected. At resonance, Z = R = 100. Current I = V/R = 200 / 100 = 2 A.

Given, $L = 0.04 , H, R= 12\Omega$
$V= 220$ volt and $f= 50 Hz$
The value of current
$I=\dfrac {V}{Z}$
or or $ I=\dfrac {V}{\sqrt{R^2+(\omega L)^2}}$
or $I=\dfrac {V}{\sqrt{R^2+(2\pi fL)^2}}$
or $I=\dfrac {220}{\sqrt{144+(2\pi50\times 0.04)^2}}$
$\Rightarrow I= 12.7, A$

For DC, R = V/I = 100/1 = 100 ohms. For AC, Z = V/I = 100/0.5 = 200 ohms. Since Z^2 = R^2 + (wL)^2, then 200^2 = 100^2 + (2pi50L)^2. Solving for L gives sqrt(30000)/(100pi) = sqrt(3)/pi H. The unit Hz in the option is a typo for H.

Voltage across capacitor V_c = I_rms * X_c = I_rms * (1/(w*C)). Given I_rms = 1.5mA, w = 300, C = 0.5uF, V_c = 0.0015 / (300 * 0.5 * 10^-6) = 0.0015 / 0.00015 = 10V. Impedance Z = sqrt(R^2 + X_c^2). With R = 10k ohms and X_c = 1/(300 * 0.5 * 10^-6) = 6666 ohms, Z is approx 12k ohms. Option B is the closest match.

Impedance of the circuit $\sqrt { { R }^{ 2 }+{ \omega  }^{ 2 }{ L }^{ 2 } } $
Amplitude of voltage$={V} _{0}$
$\therefore$ Amplitude of current $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }-{ \omega  }^{ 2 }{ L }^{ 2 } }  } $

${ X } _{ L }=\cfrac { 100 }{ 8 } ;R=\cfrac { 100 }{ 10 } =10\Omega $
$L\times 100\pi =\cfrac { 100 }{ 8 } $
$L=\cfrac { 1 }{ 8\pi  } H$
$Z=\sqrt { { \left( \cfrac { 1 }{ 8\pi  } \times 2\pi \times 40 \right)  }^{ 2 }+{ 10 }^{ 2 } } =10\sqrt { 2 } $
$I=\cfrac { E }{ Z } =\cfrac { 100 }{ 10\sqrt { 2 }  } =\cfrac { 10 }{ \sqrt { 2 }  } =5\sqrt { 2 } A$

$V=\sqrt{V _R^2+V _L^2}\ \therefore V _L=\sqrt{V^2-V _R^2}\=114.5V\V _R=IR\ \Rightarrow I+\cfrac{36}{90}=0.4A\ \therefore V _L=IX _L\=I\omega L\ \therefore L=\cfrac{V _L}{I\omega}=\cfrac{114.5}{0.4\times2\pi\times60}\=0.76H$

Z = sqrt(R^2 + (wL)^2). w = 2pi50 = 314 rad/s. wL = 314 * 0.19 approx 60 ohms. Z = sqrt(80^2 + 60^2) = 100 ohms. I = V/Z = 220/100 = 2.2A. Phase angle tan(phi) = wL/R = 60/80 = 3/4. Thus phi = tan^-1(3/4).

Power P = I^2 * R. 1000 = 100 * R, so R = 10 ohms. Z = V/I = 200/10 = 20 ohms. Z^2 = R^2 + (wL)^2, so 400 = 100 + (2pi50L)^2, 300 = (100piL)^2, L = sqrt(3)/(10pi). For phase angle pi/4, tan(pi/4) = 1 = w'L/R, so w' = R/L = 10 / (sqrt(3)/(10pi)) = 100pi/sqrt(3). f' = w'/(2*pi) = 50/sqrt(3).

Given that,

e. m. f = $18\ V$

${{E} _{rms}}=36V$

Internal resistance $r=9\Omega $

Current ${{I} _{rms}}=4\,A$

Now, the external resistance is

We know that,

  $ R=\dfrac{E _{rms}}{I _{rms}} $

 $ R=\dfrac{36}{4} $

 $ R=9\,\Omega  $

When the inductor coil is connected to a 18 V battery with 9 $\Omega$ internal resistance: 

Now, net resistance is

  $ {{R} _{net}}=R+r $

 $ {{R} _{net}}=9+9 $

 $ {{R} _{net}}=18\,\Omega  $

Now, the current is

  $ I=\dfrac{e.m.f}{{{R} _{net}}} $

 $ I=\dfrac{18}{18} $

 $ I=1\,A $

 Hence, the current is $1\ A$ in the circuit

A choke coil is designed to limit current in AC circuits with minimal power dissipation. Since power loss P = I^2 * R, we need R to be as small as possible compared to the inductive reactance X_L.

The magnetic flux through an inductor cannot change instantaneously. Phi = L * I. If L changes to L/n, the current I must change to I' such that L * I = (L/n) * I'. Therefore, I' = n * I. Since the initial current was I = E/R, the new current is nE/R.

$ X _L = L 2 \pi f = 0.21 \times 314 \Omega $
$ R = 12 \Omega $
$ I = \dfrac{ 220}{ \sqrt{ X _L^2 + R^2 } } = 3.28 = \dfrac{22}{3\sqrt{5}} A  $

$V=5\cos { \left( 1000t \right) V } $