Question

A virtual current of $4A$ and $50Hz$ flows in an AC circuit containing a coil. The power consumed in the coil is $250W$. If the virtual voltage across the coil is $100V$, the inductance in the coil will be-
(A). $\dfrac{1}{5\pi }H$
(B). $\dfrac{1}{3\pi }H$
(C). $15H$
(D). $\dfrac{1}{\pi }H$

Answer 1

In AC circuits different components like capacitor, inductor and resistor can be connected together. The current flowing in an AC circuit is generally taken as the root mean squared current which is the average square of values of currents. The power is dissipated across the resistor and the total impedance is the root of squares of reactance and resistor.

Formulae used:
$P=I_{rms}^{2}R$
$V={{I}_{rms}}Z$
$Z=\sqrt{X_{L}^{2}+{{R}^{2}}}$
$\omega L={{X}_{L}}$

Complete step-by-step solution:
Inductance is the tendency to oppose the change of current in a circuit. The current can change by changing its direction or magnitude. It is directly proportional to the rate of change of current I the circuit. Its SI unit is Henry ($H$). The potential developed due to change in current is given by-
$e=-L\dfrac{dI}{dt}$
Here,
$e$ is the potential developed
$L$ is the inductance
$\dfrac{dI}{dt}$ is the rate of change of current

In an AC circuit, the value of current oscillates between mean and extreme positions. The power consumption in an AC circuit takes place over a resistor as in a resistor power is dissipated as heat.

The formula for power is-
$P=I_{rms}^{2}R$
Here,
$P$ is power
${{I}_{rms}}$ is the root mean square current
$R$ is the resistor

Given, ${{I}_{rms}}=4A$, $P=250W$. We substitute given values in the above equation to get,
$\begin{aligned} & 240={{(4)}^{2}}\times R \\\ & \Rightarrow R=\dfrac{240}{16} \\\ & \therefore R=15\Omega \\\ \end{aligned}$
Therefore, the resistance in the circuit is $15\Omega$.

The voltage in the circuit is given by-
$V={{I}_{rms}}Z$
Here,
$V$ is the voltage
$Z$ is the impedance
$\begin{aligned} & \Rightarrow Z=\dfrac{V}{{{I}_{rms}}} \\\ & \Rightarrow Z=\dfrac{100}{4} \\\ & \therefore Z=25\Omega \\\ \end{aligned}$

We know that,
$\begin{aligned} & Z=\sqrt{X_{L}^{2}+{{R}^{2}}} \\\ & \Rightarrow {{Z}^{2}}=X_{L}^{2}+{{R}^{2}} \\\ & \Rightarrow {{(25)}^{2}}=X_{L}^{2}+{{(15)}^{2}} \\\ & \Rightarrow X_{L}^{2}=625-225 \\\ & \Rightarrow X_{L}^{2}=400 \\\ & \therefore {{X}_{L}}=20\Omega \\\ \end{aligned}$
Therefore, the inductive reactance is $20\Omega$

Inductive reactance is also given by-
$\begin{aligned} & \omega L={{X}_{L}} \\\ & \Rightarrow 2\pi fL={{X}_{L}} \\\ & \Rightarrow 2\pi \times 50L=20 \\\ & \therefore L=\dfrac{1}{5\pi }H \\\ \end{aligned}$

Therefore, the inductance in the coil is $\dfrac{1}{5\pi }H$. Hence, the correct option is (A).

Note:
In an inductor potential difference is developed because the magnetic flux due to current associated with the inductor changes when current changes. The negative sign indicates that the direction of potential is opposite to the direction of change of current. If a capacitor was introduced in the circuit, the total reactance would be the difference between inductive and capacitive reactance.