Question

When $10\,V$, $DC$ is applied across a coil current through it is $2.5\,A$ , if $10\,V,\,50\,Hz$ A.C is applied current reduces to $2\,A$. Calculate reluctance of the coil.

Answer 1

Hint: In order to answer this question, to calculate the reluctance of the coil, we will first calculate the resistance that will help to find the reluctance. And then we will find the reluctance by applying the relationship between current, voltage and the inductance.

Complete step by step answer:
The given current passes through the coil, $I = 2.5A$
Given voltage $ = 10V$
So, as we know the relation between Voltage, resistance and the current is:
$\dfrac{V}{R} = I$
Here, $V$ is the voltage, $R$ is the resistance, and $I$ is the current.
$R = \dfrac{{10}}{{2.5}} \\
\Rightarrow R= 4\Omega $
Now, we have resistance, i.e.. $4\Omega $.

Again, now, we will find the reluctance, and to find the reluctance, first we need to find the impedance which is denoted by $Z$ : So, as we know that electric current is directly proportional to the voltage, i.e.. $I\alpha V$.And, also Current is inversely proportional to the Impedance, i.e.. $I\alpha \dfrac{1}{Z}$.From both the above statement, we have:
$\therefore I = \dfrac{V}{Z}$
As we know, the formula of Impedance:
$Z = (R + jwL) \\
\Rightarrow Z= \sqrt {{R^2} + {w^2}{L^2}} \\ $
Here, $R$ is the resistance and $L$ is the reluctance.
$I = \dfrac{V}{Z} \\
\Rightarrow I = \dfrac{V}{{\sqrt {{R^2} + {w^2}{L^2}} }} \\
\Rightarrow 2 = \dfrac{{10}}{{\sqrt {{4^2} + {w^2}{L^2}} }} \\
\Rightarrow 16 + {w^2}{L^2} = {5^2} \\
\Rightarrow {w^2}{L^2} = 9 \\
\Rightarrow {L^2} = \dfrac{9}{{{w^2}}} \\
\Rightarrow L = \dfrac{3}{w} \\ $
[ $\because w = 2\pi f = 2\pi 50 = 100\pi $ ]
where, $f$ is the frequency that is given already.
$\Rightarrow L = \dfrac{3}{{100\pi }} \\
\Rightarrow L = 9.5mH \\ $
Hence, the required reluctance of the coil is $9.5\,mH$.

Note: In inductive reactance, the current across an inductor changes when potential difference develops across it. The potential difference and rate of change of current are proportional to each other. When a capacitor is connected to a circuit with AC supply, there is no simultaneous change in the capacitor voltage and capacitor current. The potential difference across the capacitor is dependent on the AC power supply.