Question:medium

In an AC circuit a resistor $100\Omega$, inductor $0.1\,\text{mH}$ and a capacitor are connected in series across an AC source of $220\,\text{V},\,50\,\text{Hz}$. If the power factor of the circuit is $ \frac{1}{2} $ and $ |X_L - X_C| = \alpha\sqrt{3}\,\Omega $, find $\alpha$.

Updated On: Apr 4, 2026

$100\sqrt{3}$
$100$
$ \frac{100}{\sqrt{3}} $
$1000$

Show Solution

The Correct Option is B

Solution and Explanation

Was this answer helpful?

Questions Asked in JEE Main exam

If $Z$ be a complex number such that $|Z + 2| = |Z - 2|$ and $\text{arg} \left( \frac{Z - 3}{Z + 1} \right) = \frac{\pi}{4}$, then the value of $|Z|^2$ is

JEE Main - 2026
Complex numbers

View Solution

Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}

JEE Main - 2026
Three Dimensional Geometry

View Solution

"X" is an oxoanion of the lightest element of group 7 (in the periodic table). The metal is in +6 oxidation state in "X". The color of the potassium salt of X is

JEE Main - 2026
Oxidation Number And Oxidation States

View Solution

As shown in the figure, the ratio of $T_1$ and $T_2$ is

JEE Main - 2026
Quantum Mechanics

View Solution

In an AC circuit a resistor \(100\Omega\), inductor \(0.1\,\text{mH}\) and a capacitor are connected in series across an AC source of \(220\,\text{V},\,50\,\text{Hz}\). If the power factor of the circuit is \( \frac{1}{2} \) and \( |X_L - X_C| = \alpha\sqrt{3}\,\Omega \), find \(\alpha\).

The Correct Option is B

Solution and Explanation

Top Questions on Current electricity

Questions Asked in JEE Main exam