Current Electricity and Circuits

${ 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 = 2*pi*50 = 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 + (2*pi*50*L)^2, 300 = (100*pi*L)^2, L = sqrt(3)/(10*pi). For phase angle pi/4, tan(pi/4) = 1 = w'L/R, so w' = R/L = 10 / (sqrt(3)/(10*pi)) = 100*pi/sqrt(3). f' = w'/(2*pi) = 50/sqrt(3).

Given that,

L = 10 H

R = 2 ohm

V = 10 volt

Now, the maximum current is

  $ {{i} _{0}}=\dfrac{V}{R} $

 $ {{i} _{0}}=\dfrac{10}{2} $

 $ {{i} _{0}}=5A $

The maximum energy is

  $ {{E} _{0}}=\dfrac{1}{2}Li _{0}^{2} $

 $ {{E} _{0}}=\dfrac{1}{2}\times 10\times 5\times 5 $

 $ {{E} _{0}}=125\,J $

Now, the magnetic energy

  $ E=\dfrac{{{E} _{0}}}{4} $

 $ E=\dfrac{125}{4}\,J $

Now, 

  $ E=\dfrac{1}{2}L{{i}^{2}} $

 $ \dfrac{125}{4}=\dfrac{1}{2}L{{i}^{2}} $

 $ \dfrac{125}{2\times 10}={{i}^{2}} $

 $ {{i}^{2}}=\dfrac{25}{2} $

 $ i=\dfrac{5}{2} $

 $ i=2.5\,A $

Now, the time taken to rise current from 0 - 2.5A.

We know that, the instantaneous current during its growth in an L-R circuit.

 $ i={{i} _{0}}\left( 1-{{e}^{-\frac{Rt}{L}}} \right) $

 $ 2.5=5\left( 1-{{e}^{-\frac{Rt}{L}}} \right) $

 $ {{e}^{\frac{-Rt}{L}}}=0.5 $

 $ \dfrac{-Rt}{L}=\ln (0.5) $

 $ \dfrac{Rt}{L}=0.693 $

 $ t=\dfrac{6.93}{2} $

 $ t=3.46 $

 $ t=3.5\sec  $

Hence, the magnetic energy will be increases to $\dfrac{1}{4}$ of the maximum value at $3.5$ sec.

The time constant tau = L/R = 8.4mH / 6 ohms = 1.4ms. The current in an RL circuit is I(t) = (V/R)(1 - e^(-t/tau)). Here V/R = 12/6 = 2A. We want I(t) = 1A, so 1 = 2(1 - e^(-t/tau)), 0.5 = 1 - e^(-t/tau), e^(-t/tau) = 0.5. t = tau * ln(2) = 1.4ms * 0.693 approx 1ms.

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 } $

Accuracy of potentiometer can be easily increased by decreasing the potential gradient across the potentiometer wire and that can be done by increasing its length.

The phase difference theta corresponds to a time difference delta_t. Since 2*pi radians corresponds to one time period T, the ratio is delta_t/T = theta/(2*pi). Thus, delta_t = (T * theta) / (2*pi).