Tag: lr circuit

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Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

A $0.21\space H$ inductor and a $12\Omega$ resistance are connected in series to a $220\space V, 50\space Hz$ ac source. The current in the circuit is :

  1. $\displaystyle\frac{220}{\sqrt{4400}}A$
  2. $\displaystyle\frac{22}{3\sqrt5}A$
  3. $\displaystyle\frac{220}{\sqrt{4550}}A$
  4. $\displaystyle\frac{22}{5\sqrt3}A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ 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  $

Multiple choice ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

An A.C voltage $V=5\cos { \left( 1000t \right) V } $ is applied to a L-R series circuit of inductance $3mH$ and resistance $4\Omega$. The value of maximum current in the circuit is  _______ A

  1. $0.8$
  2. $1.0$
  3. $\cfrac{5}{7}$
  4. $\cfrac { 5 }{ \sqrt { 7 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

The standard equation for the voltage is:
$V={ V } _{ 0 }\cos { \omega t } $

So, from the equation, ${ V } _{ 0 }=5volt;\omega =1000rad/s$
$L=3\times { 10 }^{ -3 }H,R=4\Omega $

Maximum current $10=\cfrac { { V } _{ 0 } }{ Z } \quad $
$10=\cfrac { 5 }{ \sqrt { { \omega  }^{ 2 }{ L }^{ 2 }+{ R }^{ 2 } }  } =\cfrac { 5 }{ 5 } =1A\quad \quad $