JEE Physics 2026 Question with Solution

Given: $U=16\,J$, $P=32\,W$, $I=2\,A$, and $f=50\,Hz$.

Find: The ratio $\frac{X_L}{R}$.

Use the relations for energy stored in an inductor, power dissipated in resistance, and inductive reactance:

$$U=\frac{1}{2}LI^2$$ $$P=I^2R$$ $$X_L=\omega L=2\pi fL$$

From energy stored:

$$16=\frac{1}{2}L(2)^2$$ $$16=2L$$ $$L=8\,\text{H}$$

From power dissipated:

$$32=(2)^2R$$ $$32=4R$$ $$R=8\,\Omega$$

Now calculate inductive reactance:

$$X_L=2\times 3.14\times 50\times 8=2512\,\Omega$$

Therefore,

$$\frac{X_L}{R}=\frac{2512}{8}=314$$

So, the required ratio is $314$.

Given: $U=16\,J$, $P=32\,W$, $I=2\,A$, and $f=50\,Hz$.

Find: $\frac{X_L}{R}$.

First find $L$ and $R$ separately because the ratio depends on both:

• From $U=\frac{1}{2}LI^2$, we get $L=8\,\text{H}$.
• From $P=I^2R$, we get $R=8\,\Omega$.

Then

$$\frac{X_L}{R}=\frac{2\pi fL}{R}$$

Substitute the values:

$$\frac{X_L}{R}=\frac{2\times 3.14\times 50\times 8}{8}=2\times 3.14\times 50=314$$

The cancellation of $8$ makes the calculation faster. Hence, the ratio is $314$.