Tag: ac voltage applied to a series lr circuit

Questions Related to ac voltage applied to a series lr circuit

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 :

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

  1. $\displaystyle\frac{220}{\sqrt{4400}}A$

  2. $\displaystyle\frac{22}{3\sqrt5}A$

  3. $\displaystyle\frac{220}{\sqrt{4550}}A$

  4. $\displaystyle\frac{22}{5\sqrt3}A$

Show answer
Correct Option: B
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  $

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

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

  1. $0.8$

  2. $1.0$

  3. $\cfrac{5}{7}$

  4. $\cfrac { 5 }{ \sqrt { 7 } } $

Show answer
Correct Option: B
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 $