Question

In a LR circuit of 3Mh inductance and $4\Omega $ resistance ,emf E = $4{\text{ }}\cos (1000{\text{ t)}}$ volt is applied the amplitude of current is
${\text{(A) 0}}{\text{.8 A}}$
${\text{(B) }}\dfrac{4}{7}{\text{ A}}$
${\text{(C) 1}}{\text{.0 A}}$
${\text{(D) }}\dfrac{4}{{\sqrt 7 }}{\text{ A}}$

Answer 1

Hint:In order to solve the question we will use standard equation of emf then we will compare it with the equation given in the question after then we will use the equation of peak of current and peak of emf after solving that we will finally arrives at the answer that is amplitude of current.

Formula Used:
$E = {E_0}\cos (\omega t)$
 E refers to electromotive force (emf) or we can say energy per unit electric charge
${E_0}$ stands peak value of electromotive force
$\omega $ refers to angular frequency
t refers to time
${i_0} = \dfrac{{{E_0}}}{Z}$
Z refers to impedance
${E_0}$ stands peak value of electromotive force
${i_0}$ stands peak value of electromotive force

$Z = \sqrt {{R^2} + {X_L}} $
R refers to resistance
${X_L} = \omega L$
$\omega $ refers to angular frequency
L refers to inductance

Complete step-by-step solution:
Standard emf equation is
$E = {E_0}\cos (\omega t)$
But in the question we are given this emf equation
$E = 4{\text{ }}\cos (1000{\text{ t)}}$
By comparing standard and emf equation we get to know the values of ${E_0}$ And $\omega $
So the values are
Peak value of emf ${E_0} = 4V$
V is volt standard unit of emf
Angular frequency $\omega = 1000Hz$
Hz is hertz unit of angular frequency
${i_0} = \dfrac{{{E_0}}}{Z}$
Now using this equation, we will find the amplitude value of current
As we know $Z = \sqrt {{R^2} + {X_L}} $ so we will replace Z by this and we will get
${i_0} = \dfrac{{{E_0}}}{{\sqrt {{R^2} + {X_L}} }}$
Now we will replace ${X_L}$ by $\omega L$ using the equation ${X_L} = \omega L$
Therefore, the equation becomes

${i_0} = \dfrac{{{E_0}}}{{\sqrt {{R^2} + {{(\omega L)}^2}} }}$
Now we will insert the values of the variable in above equation and we will find the amplitude of current
${E_0} = 4V$
$\omega = 1000Hz$
$R = 4\Omega $
$L = 3mH$
$mH$ is not the standard unit of inductance so we have to change it to H
$1mH = {10^{ - 3}}H$
Therefore $3mH = 3 \times {10^{ - 3}}H$ now $L = 3 \times {10^{ - 3}}H$
${i_0} = \dfrac{{{E_0}}}{{\sqrt {{R^2} + {{(\omega L)}^2}} }}$
$ \Rightarrow \dfrac{4}{{\sqrt {{4^2} + {{(1000 \times 3 \times {{10}^{ - 3}})}^2}} }}$
1000 will be cut off by the ${10^{ - 3}}$
$ \Rightarrow \dfrac{4}{{\sqrt {{4^2} + {3^2}} }}$
After squaring 3 and 4 we will get
$ \Rightarrow \dfrac{4}{{\sqrt {16 + 9} }}$
$ \Rightarrow \dfrac{4}{{\sqrt {25} }}$
$ \Rightarrow \dfrac{4}{5}$
We need the answer in points as given in the question
$ \Rightarrow 0.8{\text{ A}}$
Hence, the correct option is ${\text{(A) 0}}{\text{.8 A}}$

Note:Many of the people will make the mistake by not converting the inductance from Mh to H which create the mistakes of power hence the answer will be wrong so it should be taken care of unit along with comparing and inserting the other equation should also be done carefully.